## Tag Archives: simple group

### No simple group of order 720 exists

Suppose $G$ is a simple group of order 720. Find as many properties of $G$ as you can (Sylow numbers, isomorphism type of Sylow subgroups, conjugacy classes, etc.) Is there such a group?

[Disclaimer: I had lots of help from a discussion about an old proof by Burnside, as well as a proof by Derek Holt.]

Note that $720 = 6! = 2^4 \cdot 3^2 \cdot 5$. Sylow’s Theorem forces the following.

1. $n_2(G) \in \{1,3,5,9,15,45\}$
2. $n_3(G) \in \{1,4,10,16,40\}$
3. $n_5(G) \in \{1,6,16,36\}$

Note that $|G|$ does not divide $5!$, so that no proper subgroup has index at most 5.

Suppose $G$ has a subgroup $H$ of index 6. Then via the action of $G$ on $G/H$ we have $G \leq A_6$; however, $|A_6| = 360$, a contradiction. Thus no subgroup has index 6.

Suppose $n_5(G) = 16$. Now if $P_5 \leq G$ is a Sylow 5-subgroup, then $|N_G(P_5)| = 3^2 \cdot 5$. Since 3 does not divide 4 and 5 does not divide 8, $N_G(P_5)$ is abelian. Moreover, if $P_3 \leq N_G(P_5)$ is a Sylow 3-subgroup, then $P_3$ is also Sylow in $G$. Then $P_5 \leq N_G(P_3)$. Thus $n_3(G) = 16$. Now every Sylow 3-subgroup of $G$ normalizes some Sylow 5-subgroup, and no Sylow 5-subgroup is normalized by two Sylow 3-subgroups. Likewise for Sylow 5s and 3s. Thus if $P_5,Q_5 \in \mathsf{Syl}_5(G)$, we have $|N_G(P_5) \cap N_G(Q_5)| \in \{1,3\}$; as otherwise either $P_5 = Q_5$ or some Sylow 3-subgroup normalizes distinct Sylow 5s.

Now $N = N_G(P_5)$ acts on $S = \mathsf{Syl}_5(G)$ by conjugation, and $\mathsf{stab}_{N_G(P_5)}(Q_5) = N_G(P_5) \cap N_G(Q_5)$. Using the Orbit-Stabilizer Theorem, each orbit of this action has order 1, 15, or 45. Only one orbit has order 1; namely $\{P_5\}$, since any other orbit of order 1 consists of a normal Sylow 5-subgroup in $N_G(P_5)$. There are not enough elements left for an orbit of order 45. Thus there is an orbit of order 15, which is precisely $\mathsf{Syl}_5(G) \setminus \{P_5\}$. Thus $N_G(P_5)$ acts transitively on $S \setminus \{P_5\}$, and we have $|N_G(P_5) \cap N_G(Q_5)| = 3$ for all $Q_5 \in S$, $Q_5 \neq P_5$.

Now let $K = N_G(P_5) \cap N_G(Q_5)$, where $Q_5 \in S$ with $Q_5 \neq P_5$. Moreover choose $R_5 \in S$ distinct from both $P_5$ and $Q_5$. Now $R_5 = yQ_5y^{-1}$ for some $y \in N_G(P_5)$. Let $x \in K$. Then $xQ_5x^{-1} = Q_5$. Thus we have $y^{-1}R^5y = xy^{-1}R_5yx^{-1}$, hence $(yxy^{-1})R_5(yx^{-1}y^{-1}) = R_5$. So $yxy^{-1} \in N_G(R_5)$. Since $y \in N_G(P_5)$ and $K \leq N_G(P_5)$ is normal, we have $yKy^{-1} \leq N_G(R_5)$, so that $K \leq N_G(R_5)$.

Now $K \leq \bigcap_{x \in G} N_G(xP_5x^{-1}) = M$, and clearly $M \leq G$ is normal. Since $K$ is nontrivial, $M$ is nontrivial, and $G$ is not simple, a contradiction. Thus $n_5(G) \neq 16$. Now $n_5(G) = 36$, and $G$ has $4 \cdot 36 = 144$ elements of order 5.

Now suppose $n_3(G) = 16$. Then $|N_G(P_3)| = 3^2 \cdot 5$ for each $P_3 \in \mathsf{Syl}_3(G)$; every group of this order is abelian, so that $P_3 \leq N_G(P_5)$ for some Sylow 5-subgroup $P_5 \leq G$. But this is a contradiction since $|N_G(P_5)| = 2 \cdot 3 \cdot 5$. Thus $n_3(G) \neq 16$.

Now suppose $n_3(G) = 40$. Then if $P_3 \in \mathsf{Syl}_3(G)$, then $|N_G(P_3)| = 2 \cdot 3^2$. Now $P_3$ acts on $S = \mathsf{Syl}_3(G)$ by conjugation, an each orbit of this action has order 1, 3, or 9. Clearly $\{P_3\}$ is one orbit of order 1; if there is another, say $P_3$ normalizes $Q_3$, then $P_3 \leq N_G(Q_3)$ is Sylow and we have $P_3 = Q_3$. Thus there is a unique orbit of order 1 under this action. Now if the remaining orbits all have order 9, we have a contradiction, since $1 + 9k = 40$ has no integer solution. Thus there exists an orbit $\mathcal{O} = \{A_3, B_3, C_3\}$ of order 3. Let $Q = \mathsf{stab}_{P_3}(A_3)$, and consider $N_G(Q)$. Now $P_3$ normalizes $Q$ since $Q \leq P_3$ and $P_3$ is (necessarily) abelian. We claim also that $A_3$ normalizes $Q$. To see this, note that $Q \leq N_G(A_3)$, and that $N_G(A_3)$ has a unique Sylow 3-subgroup, namely $A_3$, which then contains $Q$, so that $A_3 \leq N_G(Q)$. Thus $N_G(Q)$ contains more than one Sylow 3-subgroup. Now $|N_G(Q)| \in$ $\{3^2 \cdot 2,$ $3^2 \cdot 2^2,$ $3^2 \cdot 2^3,$ $3^2 \cdot 2^4,$ $3^2 \cdot 5,$ $3^2 \cdot 2 \cdot 5,$ $3^2 \cdot 2^2 \cdot 5,$ $3^2 \cdot 2^3 \cdot 5,$ $3^2 \cdot 2^4 \cdot 5 \}$. In all but the two cases $3^2 \cdot 2^2$ and $3^2 \cdot 2^3$, either $G$ has a normal subgroup, $G$ has a subgroup of sufficiently small index, or $N_G(Q)$ does not have enough Sylow 3-subgroups. We handle these cases in turn. Note that in this discussion, it matters only that $Q$ is a subgroup of order 3 and that $N_G(Q)$ contains more than one Sylow 3-subgroup.

Suppose $|N_G(Q)| = 2^2 \cdot 3^2$. Then $N_G(Q)/Q$ is a group of order 12, and as we saw in this previous exercise, acts on $Q$ by conjugation (by coset representatives). Moreover, $N_G(Q)/Q$ has 4 Sylow 3-subgroups by the Lattice Isomorphism Theorem. By the lemma, $N_G(Q)/Q \cong A_4$. Since $A_4$ has no subgroups of index 2, by the Orbit Stabilizer Theorem, the orbits of this action have order 1 or 3. However, the identity element of $Q$ is in an orbit of order 1; thus all orbits have order 1. Thus this action is trivial; that is, for all $g \in N_G(Q)$ and $h \in Q$, $ghg^{-1} = h$; thus $Q \leq Z(N_G(Q))$. Note that the four Sylow 3-subgroups of $N_G(Q)$ intersect in $Q$, so that $N_G(Q)$ has $8 + 3 \cdot 6 = 26$ elements of 3-power order, and 10 elements remain. Now $n_2(N_G(Q)) \in \{1,3,9\}$. Let $Q = \langle x \rangle$. Suppose $n_2(N_G(Q)) = 9$. If some element of $N_G(Q)$ has order 4, then each Sylow 2-subgroup has two elements of order 4, all distinct. But then $|G| \geq 26 + 18 = 44$, a contradiction. Thus all elements in Sylow 2-subgroups of $N_G(Q)$ have order 2. Now $G$ must contain at least 4 elements of order 2; let these comprise the set $S$. Now $S$, $xS$, and $x^2S$ are 12 mutually distinct elements of order 2, 6, and 6, respectively, so that $|G| \geq 26 + 12 = 38$, a contradiction. Suppose now that $n_2(N_G(Q)) = 3$. Again if some element has order 4, then $N_G(Q)$ has $3 \cdot 2 = 6$ elements of order 4 and 2 \cdot 6 = 12$elements of order 12, so that $|G| \geq 26 + 6 + 12 = 44$, a contradiction. Thus the elements in Sylow 2-subgroups of $N_G(Q)$ have order 2. There are at least 4 of these, say in the set $S$. Now $S$, $xS$, and $x^2S$ contain (at least) 12 distinct elements of order 2, 6, and 6, respectively, so that $|G| \geq 26 + 12 = 38$, a contradiction. Thus $n_2(N_G(Q)) = 1$. Let $Q_2 \leq N_G(Q)$ be the unique Sylow 2-subgroup. Now $Q_2 \leq P_2$ for some Sylow 2-subgroup of $G$, and $Q_2$ is properly contained in $N_{P_2}(Q_2)$. Thus 8 divides $|N_G(Q_2)|$, and since $Q_2$ is normal in $N_G(Q)$, $3^2$ also divides $|N_G(Q_2)|$. Thus $|N_G(Q_2)| \in \{$ $3^2 \cdot 2^3,$ $3^2 \cdot 2^4,$ $3^2 \cdot 2^3 \cdot 5,$ $3^2 \cdot 2^4 \cdot 5 \}$. In all but one case, either $G$ has a normal subgroup or a subgroup of sufficiently small index. Thus we have $|N_G(Q_2)| = 2^3 \cdot 3^2$. Now $N_G(Q) \leq N_G(Q_2)$ has index 2, and thus is normal. Moreover, $Q = O_3(N_G(Q))$ (see this previous exercise) is characteristic in $N_G(Q)$, and thus normal in $N_G(Q_2)$. But then $N_G(Q_2) \leq N_G(Q)$, a contradiction. Suppose $|N_G(Q)| = 2^3 \cdot 3^2$. Note that, by the N/C Theorem, $N_G(Q)/C_G(Q) \leq \mathsf{Aut}(Q) \cong Z_2$. Then we have $|C_G(Q)| \in \{ 2^2 \cdot 3^2, 2^3 \cdot 3^2 \}$; in either case, there exists an element of order 2 which centralizes $Q$. Now write $Q = \langle t \rangle$, and note that $Q$ has 10 conjugates since $[G : N_G(Q)] = 10$. Consider the action of $G$ on the ten cosets in $Q/N_G(Q)$ by left multiplication. Via this action, we have $G \leq S_{10}$. Now $t \in S_{10}$ is a product of 3-cycles. Note that since $N_G(xQx^{-1}) = xN_G(Q)x^{-1}$, every conjugate of $Q$ is contained in 4 Sylow 3-subgroups, and that every Sylow 3-subgroup contains a conjugate of $Q$ (by Sylow’s Theorem, since $Q \leq P_3$). Moreover, in this case there are 40 Sylow 3-subgroups. By the pigeonhole principle, every Sylow 3-subgroup of $G$ contains exactly one conjugate of $Q$. Suppose now that $t \in S_{10}$ fixes $xN_G(Q)$. Then $t \cdot x N_G(Q) = x N_G(Q)$, so that $x^{-1}tx \in N_G(Q)$; thus in fact $x^{-1}Qx = Q$ (since $x^{-1}Qx$ is contained in some Sylow 3-subgroup of $G$ which also contains $Q$), and we have $x \in N_G(Q)$. Thus $t \in S_{10}$ fixes only $N_G(Q)$, and has cycle shape $(3,3,3)$. As shown in the beginning of this paragraph, let $u \in C_G(Q)$ have order 2. Now $u^{-1}tu = t$, so that $t$ is obtained from $t$ by applying $u$ entrywise. In particular, $u$ permutes the orbits of $t$; it must interchange two and fix the third. Thus $u \in S_{10}$ has cycle shape $(2,2,2)$ or $(2,2,2,3)$; in either case, $G$ contains an odd permutation, so that $G$ is not simple. Thus $n_3(G) \neq 40$. We are left with $n_3(G) = 10$. Now $|N_G(P_3)| = 2^3 \cdot 3^2$, and $G$ acts by conjugation on $\mathsf{Syl}_3(G)$, and via this action we have $G \leq S_{10}$. Let $P_3 \in \mathsf{Syl}_3(G)$ and suppose $P_3$ is cyclic; say $P_3 = \langle t \rangle$. Then $t \in S_{10}$ is necessarily a 9-cycle. Moreover, by the N/C Theorem we have $N_G(P_3)/C_G(P_3) \leq \mathsf{Aut}(P_3) \cong Z_6$, so that $[N_G(P_3) : C_G(P_3)] \in \{1,2\}$. In either case, by Cauchy there exists an element $u$ of order 2 such that $u^{-1}tu = t$. However, we saw in the discussion on page 127 of the text that $|C_{S_{10}}(t)| = 9$; a contradiction. Thus $P_3$ (indeed each of the Sylow 3-subgroups) is elementary abelian. Now let $t \in P_3$ be an element of order 3; $t \in S_{10}$ is a product of 3-cycles. Suppose $t$ fixes more than one element- that is, $t$ is a product of one or two 3-cycles. Note first that if $t$ fixes a Sylow 3-subgroup $R_3$, then $t \in N_G(R_3)$ and in fact, since $R_3 \leq N_G(R_3)$ is the unique Sylow 3-subgroup in $N_G(R_3)$, we have $t \in R_3$. Thus, letting $Q = \langle t \rangle$, we have $Q \leq R_3$. Since $R_3$ is abelian, $R_3 \leq N_G(Q)$. Now since $t$ fixes exactly 7 Sylow 3-subgroups of $G$, $N_G(Q)$ contains 7 Sylow 3-subgroups. But $7 \not\equiv 1$ mod 3, a contradiction. Suppose now that $t$ is a product of two 3-cycles. Now $|N_G(Q)|$ is divisible by $3^2$ and, since exactly four Sylow 3-subgroups contain $Q$, is divisible by $2^2$ as well. Thus $|N_G(Q)| \in \{$ $3^2 \cdot 2^2,$ $3^2 \cdot 2^3,$ $3^2 \cdot 2^4,$ $3^2 \cdot 2^2 \cdot 5,$ $3^2 \cdot 2^3 \cdot 5,$ $3^2 \cdot 2^4 \cdot 5 \}$. We can see that in all but the cases $3^2 \cdot 2^2$ and $3^2 \cdot 2^3$, either $Q$ is normal in $G$ or a subgroup of $G$ has sufficiently small index. In the two remaining cases, $N_G(Q)$ contains four Sylow 3-subgroups and has order 36 or 72; we eliminated these possibilities in our discussion about the case $n_3(G) = 40$. Thus $t$ must be a product of three 3-cycles. Suppose now that there exists an element $u \in G$ of order 2 such that $u^{-1}tu = t$. Now $u$ must permute the orbits of $t$, so that it must interchange two orbits and fix the third. Thus $u$ has cycle shape $(2,2,2)$ or $(2,2,2,3)$, and in either case $G$ contains an odd permutation. Thus if $t \in G$ is an element of order 3, there does not exist an element of order 2 which centralizes it. Finally, note that if $R_3 \in \mathsf{Syl}_3(G)$ with $R_3 \neq P_3$, then $t^{-1} R_3 t \neq R_3$. That is, $t \notin N_G(R_3)$. In particular, $t \notin R_3$. Thus, $P_3 \cap R_3 = 1$, and hence $P_3$ intersects every other Sylow 3-subgroup trivially. By this previous exercise, all Sylow 3-subgroups intersect trivially. Thus $G$ contains $10 \cdot 8 = 80$ elements of order 3. Now since $G \leq S_{10}$ and using Lagrange, every element of $G$ has order 1, 2, 3, 4, 5, 6, 8, 9, or 10. We showed previously that no element has order 9. If $x$ has order 6, then $x^2$ is an element of order 3 and $x^3$ is an element of order 2 which centralizes it, a contradiction. Thus no element has order 6. Now let $x \in G$ have order 5; then $x$ is a product of one or two 5-cycles. If $x$ is a single 5-cycle, then $x \in N_G(P_3)$ for some Sylow 3-subgroup $P_3$; this is a contradiction. Thus $x$ is a product of two 5-cycles. Suppose now that there exists an element $y$ which centralizes $x$; that is, $y^{-1}xy = x$. Now $y$ must permute the orbits of $x$. Suppose $y$ fixes each orbit of $x$; the restriction of $y$ to an orbit of $x$ is then either the identity or a 5-cycle, contradicting the fact that $y$ has order 2. Thus $y$ transposes the orbits of $x$ and is thus a product of five 2-cycles; then $y \in G$ is an odd permutation and $G$ is not simple, a contradiction. Thus no element of order 2 centralizes an element of order 5. In particular, no element of $G$ has order 10 (for the same reason that no element has order 6). Thus every element of $G$ is contained in a Sylow subgroup. Now suppose $x$ is an element of order 8 which normalizes a Sylow 3-subgroup; then $x$ must have the cycle shape $(8)$, so that $G$ contains an odd permutation and is not simple. Thus no element of order 8 normalizes a Sylow 3-subgroup. Now $|G| = 720$, every element has order 1, 2, 3, 4, 5, or 8, and there are 80 and 144 elements of order 3 and 5, respectively. Thus the remaining 496 elements are contained in Sylow 2-subgroups. Recall also from our proof classifying the groups of order 20 that only one group of this order does not contain an element of order 10; namely $Z_5 \rtimes_\varphi Z_4$, where $Z_5 = \langle a \rangle$, $Z_4 = \langle b \rangle$, and $\varphi(b)(a) = a^2$. This group is thus isomorphic to the normalizers of Sylow 5-subgroups in $G$. Let $a \in G$ have order 3. Now $a$ is contained in, and thus normalizes, a unique Sylow 3-subgroup $P_3$. Without loss of generality, we can say that $a \in S_{10}$ (via the action of $G$ on $\mathsf{Syl}_3(G)$) is a product of three 3-cycles; say $a = (2\ 3\ 4)(5\ 6\ 7)(8\ 9\ 10)$. Now let $P_3 = \langle a,b \rangle$. We have $b^{-1}ab = a$ and $b \notin \langle a \rangle$, so that $b$ must permute the orbits of $a$. There are 6 choices we can make for $b$, and any one of them generates $P_3$ (together with $a$). So without loss of generality, let $b = (2\ 5\ 8)(3\ 6\ 9)(4\ 7\ 10)$. Since $|N_G(P_3)| = 2^3 \cdot 3^2$, by Cauchy there exists an element of order 2 which normalizes $P_3$, say $u$. There are 7 possibilities for $u^{-1}au$; note that since $u \in N_G(P_3)$, $u(1) = 1$. 1. Suppose $u^{-1}au = b$. If $u(2) = 2$, then $u = (3\ 5)(4\ 8)(7\ 9)$ is an odd permutation. If $u(2) = 3$, then latex u(3) = 6$. If $u(2) = 4$, then $u(4) = 10$. If $u(2) = 5$, then $u(4) = 2$. If $u(2) = 6$, then $u(3) = 9$ and $u(4) = 3$. If $u(2) = 7$, then $u = (2\ 7)(3\ 10)(6\ 8)$ is an odd permutation. If $u(2) = 8$, then $u(3) = 2$. if $u(2) = 9$, then $u = (2\ 9)(4\ 6)(5\ 10)$ is an odd permutation. If $u(2) = 10$, then $u(3) = 4$ and $u(4) = 7$.
2. Suppose $u^{-1}au = b^2 = (2\ 8\ 5)(3\ 9\ 6)(4\ 10\ 7)$. If $u(2) = 2$, then $u = (3\ 8)(4\ 5)(6\ 10)$ is an odd permutation. If $u(2) = 8$, then $u(4) = 2$. If $u(2) = 5$, then $u(3) = 2$. If $u(2) = 3$, then $u(3) = 9$. If $u(2) = 9$, then $u(3) = 6$ and $u(4) = 3$. If $u(2) = 6$, then $u = (2\ 6)(4\ 9)(7\ 8)$ is an odd permutation. If $u(2) = 4$, then $u(4) = 7$. If $u(2) = 10$, then $u = (2\ 10)(3\ 7)(5\ 9)$ is an odd permutation. If $u(2) = 7$, then $u(3) = 4$ and $u(4) = 10$.
3. Suppose $u^{-1}au = ab = (2\ 6\ 10)(3\ 7\ 8)(4\ 5\ 9)$. If $u(2) = 2$, then $u = (3\ 6)(4\ 10)(5\ 8)$ is an odd permutation. If $u(2) = 6$, then we must have $u(6) = 2$; but we also have $u(4) = 2$, a contradiction. Similarly, if $u(2) = 10$, then $u(10) = u(3) = 2$. If $u(2) = 3$, then $u(3) = 2$ and $u(3) = 7$, a contradiction. If $u(2) = 7$, then $u(3) = 8$ and $u(4) = 3$, a contradiction. If $u(2) = 8$, then $u = (2\ 8)(6\ 9)(4\ 7)$ is an odd permutation. If $u(2) = 4$, then $u(4) = 2$ and $u(4) = 9$, a contradiction. If $u(2) = 5$, then $u = (2\ 5)(3\ 9)(7\ 10)$ is an odd permutation. If $u(2) = 9$, then $u(3) = 4$ and $u(4) = 5$, a contradiction.
4. Suppose $u^{-1}au = a^2b$. If $u(2) = 2$, then $u = (3\ 7)(6\ 10)(9\ 4)$ is an odd permutation. If $u(2) = 7$, then $u(5) = 7$, a contradiction. If $u(2) = 9$, then $u(3) = 2$, a contradiction. If $u(2) = 3$, then $u(3) = 5$. If $u(2) = 5$, then $u(3) = 10$ and $u(4) = 3$, a contradiction. If $u(2) = 10$, then $u = (2\ 10)(7\ 8)(4\ 5)$ is an odd permutation. If $u(2) = 4$, then $u(4) = 8$, a contradiction. If $u(2) = 6$, then $u = (2\ 6)(3\ 8)(5\ 9)$ is an odd permutation. If $u(2) = 8$, then $u(3) = 4$ and $u(4) = 6$, a contradiction.
5. Suppose $u^{-1}au = ab^2 = (2\ 9\ 7)(3\ 10\ 5)(4\ 8\ 6)$. If $u(2) = 2$, then $u = (3\ 9)(4\ 7)(5\ 8)$ is an odd permutation. If $u(2) = 9$, then $u(4) = 2$, a contradiction. If $u(2) = 7$, then $u(3) = 2$. If $u(2) = 3$, then $u(3) = 10$. If $u(2) = 10$, then $u(3) = 5$ and $u(4) = 3$. If $u(2) = 5$, then $u = (2\ 5)(4\ 10)(6\ 9)$ is an odd permutation. If $u(2) = 4$, then $u(4) = 6$. If $u(2) = 8$, then $u = (2\ 8)(3\ 6)(7\ 10)$ is an odd permutation. If $u(2) = 6$, then $u(3) = 4$ and $u(4) = 8$.
6. Suppose $u^{-1}au = a^2b^2 = (2\ 10\ 6)(3\ 8\ 7)(4\ 9\ 5)$. If $u(2) = 2$, then $u = (3\ 10)(4\ 6)(7\ 9)$ is an odd permutation. If $u(2) = 10$, then $u(8) = 10$, a contradiction. If $u(2) = 6$, then $u(3) = 2$, a contradiction. If $u(2) = 3$, then $u(3) = 8$, a contradiction. If $u(2) = 8$, then $u(3) = 7$ and $u(4) = 3$, a contradiction. If $u(2) = 7$, then $u = (2\ 7)(5\ 10)(4\ 8)$ is an odd permutation. If $u(2) = 4$, then $u(4) = 5$, a contradiction. If $u(2) = 9$, then $u = (2\ 9)(3\ 5)(6\ 8)$ is an odd permutation. If $u(2) = 5$, then $u(3) = 4$ and $u(4) = 9$, a contradiction.

Thus we must have $u^{-1}au = a^2$; by a similar argument, $u^{-1}bu = b^2$. Hence $u$ corresponds to the matrix $2I \in GL_2(\mathbb{F}_3) \cong \mathsf{Aut}(P_3)$. In particular, if $Q \leq N_G(P_3)$ is a Sylow 2-subgroup, then $|Q| = 8$ and $Q$ contains a unique element of order 2. Moreover, we know that $Q$ contains no element of order 8 since such an element would have cycle shape $(8)$, and thus be odd. We know from our classification of order 8 groups that $Q \cong Q_8$; thus $N_G(P_3)$ contains exactly 9 elements of order 2 and $6 \cdot 9 = 54$ elements of order 4. Moreover, every element of order 2 in $G$ must fix some Sylow 3-subgroup, since otherwise it has cycle shape $(2,2,2,2,2)$. We have shown that every element of order 2 which fixes a Sylow 3-subgroup has cycle shape $(2,2,2,2)$, and thus every element of $G$ of order 2 has this cycle shape.

Consider now $u^{-1}au = a^2$; since $a$ and $a^2$ have the same orbits, $u$ must permute the orbits of $a$. If $u$ fixes an orbit $\mathcal{O}$ (of order 3) then $u|_\mathcal{O}$ is a 2-cycle. If $u$ fixes all of the orbits of $a$, then $u$ has cycle shape $(2,2,2)$ and is thus an odd permutation. So $u$ must transpose two orbits of $a$ and fix the third. Let $(x\ y\ z)$ and $(w\ v\ t)$ be arbitrary disjoint 3-cycles with orbits $A = \{x,y,z\}$ and $B = \{w,v,t\}$; if $u$ fixes $(x\ y\ z)$ then $u|_A$ is one of $(x\ z)$, $(x\ y)$, and $(y\ z)$. If $u$ transposes $A$ and $B$, then $u|_{A \cup B}$ is one of $(x\ w)(y\ v)(z\ t)$, $(x\ v)(y\ t)(z\ w)$, and $(x\ t)(y\ w)(z\ v)$. Choosing which orbit of $a$ to fix and which permutation of each orbit of orbits divide $u$, there are $3 \cdot 3 \cdot 3 = 27$ possibilities for $u$.

• Suppose $u = (2\ 3)(5\ 8)(6\ 10)(7\ 9)$. Then $(u^{-1}bu)(2) = 10$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (2\ 3)(5\ 10)(6\ 9)(7\ 8)$. Then $(u^{-1}bu)(2) = 9$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (2\ 4)(5\ 8)(6\ 10)(7\ 9)$. Then $(u^{-1}bu)(2) = 9$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (2\ 4)(5\ 9)(6\ 8)(7\ 10)$. Then $(u^{-1}bu)(2) = 10$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (3\ 4)(5\ 10)(6\ 9)(7\ 8)$. Then $(u^{-1}bu)(3) = 8$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (3\ 4)(5\ 9)(6\ 8)(7\ 10)$. Then $(u^{-1}bu)(3) = 10$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (5\ 6)(2\ 8)(3\ 10)(4\ 9)$. Then $(u^{-1}bu)(5) = 4$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (5\ 6)(2\ 10)(3\ 9)(4\ 8)$. Then $(u^{-1}bu)(5) = 3$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (5\ 7)(2\ 8)(3\ 10)(4\ 9)$. Then $(u^{-1}bu)(5) = 3$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (5\ 7)(2\ 9)(3\ 8)(4\ 10)$. Then $(u^{-1}bu)(5) = 4$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (6\ 7)(2\ 10)(3\ 9)(4\ 8)$. Then $(u^{-1}bu)(6) = 2$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (6\ 7)(2\ 9)(3\ 8)(4\ 10)$. Then $(u^{-1}bu)(6) = 4$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (8\ 9)(2\ 5)(3\ 7)(4\ 6)$. Then $(u^{-1}bu)(8) = 7$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (8\ 9)(2\ 7)(3\ 6)(4\ 5)$. Then $(u^{-1}bu)(8) = 6$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (8\ 10)(2\ 5)(3\ 7)(4\ 6)$. Then $(u^{-1}bu)(8) = 6$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (8\ 10)(2\ 6)(3\ 5)(4\ 7)$. Then $(u^{-1}bu)(2) = 9$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (9\ 10)(2\ 7)(3\ 6)(4\ 5)$. Then $(u^{-1}bu)(9) = 5$, so that $u^{-1}bu \neq b^2$.
• Suppose $u = (9\ 10)(2\ 6)(3\ 5)(4\ 7)$. Then $(u^{-1}bu)(9) = 7$, so that $u^{-1}bu \neq b^2$.

The remaining nine possibilities,

• $u_1 = (2\ 3)(5\ 9)(6\ 8)(7\ 10)$
• $u_2 = (2\ 4)(5\ 10)(6\ 9)(7\ 8)$
• $u_3 = (3\ 4)(5\ 8)(6\ 10)(7\ 9)$
• $u_4 = (5\ 6)(2\ 9)(3\ 8)(4\ 10)$
• $u_5 = (5\ 7)(2\ 10)(3\ 9)(4\ 8)$
• $u_6 = (6\ 7)(2\ 8)(3\ 10)(4\ 9)$
• $u_7 = (8\ 9)(2\ 6)(3\ 5)(4\ 7)$
• $u_8 = (8\ 10)(2\ 7)(3\ 6)(4\ 5)$
• $u_9 = (9\ 10)(2\ 5)(3\ 7)(4\ 6)$

satisfy $u^{-1}bu = b^2$, and in fact must correspond to the nine elements of order 2 in $N_G(P_3)$. Now for each $u_i$, there are six solutions of the equation $x^2 = u_i$ in $Q_8$. In $S_{10}$, if $x^2 = u_i$, then the cycle shape of $x$ is either $(4,4)$ or $(4,4,2)$, and the latter is not possible. If $u_i = (a\ b)(c\ d)(e\ f)(g\ h)$ and $x^2 = u_i$, then an orbit of $x$ is a union of two orbits of $u_i$. There are 3 different ways to choose pairs of orbits of $u_i$, and for each choice, we get two solutions: $x$ and $x^{-1}$. Thus the only solutions of the equation $x^2 = u_i$ in $S_{10}$ are $x_1 = (a\ c\ b\ d)(e\ g\ f\ h)$, $x_2 = (a\ e\ b\ f)(c\ g\ d\ h)$, $x_3 = (a\ g\ b\ h)(c\ e\ d\ f)$, and their inverses. Fixing $i = 1$, let $x_1 = (2\ 5\ 3\ 9)(6\ 7\ 8\ 10)$, $x_2 = (2\ 6\ 3\ 8)(5\ 7\ 9\ 10)$, and $x_3 = (2\ 7\ 3\ 10)(5\ 6\ 9\ 8)$.

At this point, note the following for all nonidentity elements $x \in G$.

1. If $|x| = 2$, then $x$ has cycle shape $(2,2,2,2)$.
2. If $|x| = 3$, then $x$ has cycle shape $(3,3,3)$.
3. If $|x| = 4$, then $x^2$ has order 2 and thus has cycle shape $(2,2,2,2)$. So $x$ has cycle shape $(4,4)$ or $(4,4,2)$, the latter being odd and thus impossible.
4. If $|x| = 5$, then $x$ does not normalize a Sylow 3-subgroup, and hence has cycle shape $(5,5)$.
5. If $|x| = 8$, then $x$ has cycle shape $(8,2)$ or $(8)$, the latter being impossible.

In no case does $x$ fix more than 3 elements. Recall the elements $x_1$, $x_2$, and $x_3$ for later.

If $n_2(G) = 9$, then there are at most $9 \cdot 15 = 135$ elements of 2-power order, and $G$ does not contain enough elements. Similarly, if $n_2(G) = 15$, then there are at most $15 \cdot 15 = 225$ elements of 2-power order. Thus $n_2(G) = 45$. Let $P_2,Q_2 \in \mathsf{Syl}_2(G)$ be chosen such that $|P_2 \cap Q_2| = k$ is maximal.

• If $k = 1$, then the Sylow 2-subgroups of $G$ are pairwise disjoint and $G$ contains $45 \cdot 15 = 675$ elements of 2-power order, a contradiction.
• If $k = 2$, then consider $N_G(P_2 \cap Q_2)$. Since $P_2 \cap Q_2 \leq N_G(P_2 \cap Q_2)$ is a normal subgroup of order 2, we know also that $P_2 \cap Q_2$ is central in $N_G(P_2 \cap Q_2)$. In particular, there cannot be an element of order 3 in $N_G(P_2 \cap Q_2)$, as otherwise $G$ would have an element of order 6. Moreover, since $P_2 \cap Q_2 \leq P_2$ is contained in a 2-group, it is properly contained in its normalizer in $P_2$, which is then contained in $N_G(P_2 \cap Q_2)$. So $|N_G(P_2 \cap Q_2)|$ is divisible by $2^2$ and not divisible by 3; thus $|N_G(P_2 \cap Q_2)| \in \{$ $2^2,$ $2^3,$ $2^4,$ $2^2 \cdot 5$, $2^3 \cdot 5,$ $2^4 \cdot 5 \}$.
• If $|N_G(P_2 \cap Q_2)| = 2^4$, then $P_2 \cap Q_2 \leq R_2$ is normal for some Sylow 2-subgroup $R_2$. But then $|P_2 \cap R_2| \geq |N_{P_2}(P_2 \cap Q_2) \cap R_2| = 8$, violating the maximalness of $P_2 \cap Q_2$.
• If $|N_G(P_2 \cap Q_2)| = 2^2 \cdot 5$, then by Sylow’s Theorem, $n_5(N_G(P_2 \cdot Q_2)) = 1$. However, in our classification of groups of order 20, we saw that if a group of order 20 has a unique Sylow 2-subgroup, then it is abelian. Thus $G$ has an element of order 10, a contradiction.
• If $|N_G(P_2 \cap Q_2)| = 2^3 \cdot 5$, then by Sylow’s Theorem, we have $n_5(N_G(P_2 \cap Q_2)) = 1$. But then $2^3$ divides $|N_G(P_5)|$ for some Sylow 5-subgroup $P_5 \leq N_G(P_2 \cap Q_2)$, which is also Sylow in $G$; this is a contradiction.
• If $|N_G(P_2 \cap Q_2)| = 2^4 \cdot 5$, then by Sylow’s Theorem, $n_5(N_G(P_2 \cap Q_2)) \in \{1,16\}$. If $n_5(N_G(P_2 \cap Q_2)) = 1$, then as in the previous case, $2^4$ divides $|N_G(P_5)|$ for some Sylow 5-subgroup $P_5 \leq G$. If $n_5(N_G(P_2 \cap Q_2)) = 16$, then since Sylow 5-subgroups intersect trivially, $N_G(P_2 \cap Q_2)$ contains $16 \cdot 4$ elements of order 5. Now suppose $n_2(N_G(P_2 \cap Q_2) = 1$. Then $P_5 \leq N_G(R_2)$ for some Sylow 2- and 5-subgroups $R_2$ and $P_5$, respectively, of $G$, a contradiction since $n_2(G) = 45$.
• Suppose $|N_G(P_2 \cap Q_2)| = 2^3$. Now $N_{P_2}(P_2 \cap Q_2)$ properly contains $P_2 \cap Q_2$, and of course $N_{P_2}(P_2 \cap Q_2) \leq N_G(P_2 \cap Q_2)$; similarly for the normalizer of $P_2 \cap Q_2$ in $Q_2$. Moreover, $N_G(P_2 \cap Q_2) \leq R_2$ for some Sylow 2, and $R_2$ must be distinct from at least one of $P_2$ and $Q_2$. But then either $N_{P_2}(P_2 \cap Q_2) \leq P_2 \cap R_2$ or $N_{Q_2}(P_2 \cap Q_2) \leq Q_2 \cap R_2$ has order 4, violating the maximalness of $P_2 \cap Q_2$.
• If $|N_G(P_2 \cap Q_2)| = 2^2$, then by order considerations we have $N_{P_2}(P_2 \cap Q_2) = N_G(P_2 \cap Q_2) = N_{Q_2}(P_2 \cap Q_2)$, contradicting the maximalness of $P_2 \cap Q_2$.
• If $k = 4$, then since $P_2 \cap Q_2$ is properly contained in $N_{P_2}(P_2 \cap Q_2)$, 8 divides $|N_G(P_2 \cap Q_2)|$. Thus $|N_G(P_2 \cap Q_2)| \in \{$ $2^3,$ $2^4,$ $2^3 \cdot 3$, $2^4 \cdot 3,$ $2^3 \cdot 5,$ $2^4 \cdot 5,$ $2^3 \cdot 3^2,$ $2^4 \cdot 3^2,$ $2^3 \cdot 3 \cdot 5,$ $2^4 \cdot 3 \cdot 5,$ $2^3 \cdot 3^2 \cdot 5,$ $2^4 \cdot 3^2 \cdot 5 \}$.
• If $|N_G(P_2 \cap Q_2)| = 2^3$, then $N_{P_2}(P_2 \cap Q_2) = N_G(P_2 \cap Q_2) = N_{Q_2}(P_2 \cap Q_2)$, contradicting the maximalness of $P_2 \cap Q_2$.
• If $|N_G(P_2 \cap Q_2)| = 2^4$, then we have $P_2 \cap Q_2 \leq R_2$ normal for some Sylow 2-subgroup $R_2$. But then $|P_2 \cap R_2| \geq |N_{P_2}(P_2 \cap Q_2) \cap R_2| = 8$, violating the maximalness of $P_2 \cap Q_2$.
• If $|N_G(P_2 \cap Q_2)| \in \{ 2^4 \cdot 3^2,$ $2^3 \cdot 3 \cdot 5,$ $2^4 \cdot 3 \cdot 5,$ $2^3 \cdot 3^2 \cdot 5 \}$, then $G$ has a subgroup of sufficiently small index.
• If $|N_G(P_2 \cap Q_2)| = 2^4 \cdot 3^2 \cdot 5$, then $P_2 \cap Q_2 \leq G$ is normal.
• If $|N_G(P_2 \cap Q_2)| = 2^3 \cdot 5$, then by Sylow’s Theorem, $n_5(N_G(P_2 \cap Q_2)) = 1$, so that 8 divides $|N_G(P_5)|$ for some Sylow 5-subgroup $P_5$.
• Suppose $|N_G(P_2 \cap Q_2)| = 2^4 \cdot 5$. If $n_5(N_G(P_2 \cap Q_2)) = 1$, then we have $2^4$ dividing $|N_G(P_5)|$ for some Sylow 5-subgroup $P_5$, a contradiction. Thus $n_5(N_G(P_2 \cap Q_2)) = 16$. Since the Sylow 5-subgroups intersect trivially, $N_G(P_5 \cap Q_5)$ contains $4 \cdot 16$ elements of order 5. If $n_2(N_G(P_2 \cap Q_2)) = 5$, then there are at least 16 elements of 2-power order, so that $G$ contains at least $2^4 \cdot 5 + 1$ elements, a contradiction. Thus $n_2(N_G(P_2 \cap Q_2)) = 1$. Now $P_2 \cap Q_2$ has 9 conjugates, and each conjugate of $P_2 \cap Q_2$ is normal in a (unique) Sylow 2-subgroup of $G$. Similarly, every Sylow 2-subgroup of $G$ normalizes a conjugate of $P_2 \cap Q_2$. However, there are 45 Sylow 2-subgroups in $G$, so that some conjugate of $P_2 \cap Q_2$ must be normalized by more than one Sylow 2-subgroup, a contradiction.
• Suppose $|N_G(P_2 \cap Q_2)| = 2^4 \cdot 3$. If $n_2(N_G(P_2 \cap Q_2)) = 1$, then we have 3 dividing $|N_G(R_2)|$ for some Sylow 2-subgroup $R_2$; this is a contradiction since $n_2(G) = 45$. Thus $n_2(N_G(P_2 \cap Q_2)) = 3$. If $A_2$ and $B_2$ are Sylow 2-subgroups in $N_G(P_2 \cap Q_2)$ (which are also Sylow in $G$), we clearly have $P_2 \cap Q_2 \leq A_2 \cap B_2$; if this inclusion is proper, we contradict the maximalness of $P_2 \cap Q_2$. Thus $P_2 \cap Q_2 = A_2 \cap B_2$ for all Sylow 2-subgroups $A_2,B_2 \leq N_G(P_2 \cap Q_2)$. Thus there are $3 + 3 \cdot 11 = 37$ elements of 2-power order in $N_G(P_2 \cap Q_2)$. Now $n_3(N_G(P_2 \cap Q_2)) \in \{1,4,16\}$. Suppose $n_3(N_G(P_2 \cap Q_2)) = 1$, and let $Q_3$ be the (unique) Sylow 3-subgroup in $N_G(P_2 \cap Q_2)$. Now $Q_3 \leq R_3$ is normal for some Sylow 3-subgroup $R_3 \leq G$, so that $|N_G(Q_3)| \in \{2^4 \cdot 3^2 \cdot 5, 2^4 \cdot 3^2 \}$; either case yields a contradiction. If $n_3(N_G(P_2 \cap Q_2)) = 4$, then $N_G(P_2 \cap Q_2)$ contains $4 \cdot 2 = 8$ elements of order 3. But then $G$ contains $37 + 8 + 1 = 46$ elements of order 1, 2, 3, 4, or 8; no other orders are possible, so that $G$ does not contain enough elements. If $n_3(N_G(P_2 \cap Q_2)) = 16$, then $G$ contains $16 \cdot 3 = 48$ elements of order 3, so that $G$ contains too many elements.
• Suppose $|N_G(P_2 \cap Q_2)| = 2^3 \cdot 3$. Now $N_G(P_2 \cap Q_2)$ is a group of order 24 which does not contain an element of order 6; thus, by this previous exercise, $N_G(P_2 \cap Q_2) \cong S_4$. In this previous exercise, we found that $n_2(S_4) = 3$, and since Sylow subgroups of $A_4$ are Sylow in $S_4$, $n_3(S_4) = 4$. We claim that $S_4$ has a unique normal subgroup of order 4 which is isomorphic to $V_4$. To see this, recall that every normal subgroup (of an arbitrary group) is a union of conjugacy classes. If $H \leq S_4$ is a normal subgroup of order 4, then its elements have order 2 or 4. The elements of these orders in $S_4$ have cycle shape $(2)$, $(4)$, or $(2,2)$, which have orders 6, 6, and 3, respectively. Thus there is only one possible conjugacy class of elements of order 2 in $H$, which (with the identity) comprise all of $H$. Thus we have $P_2 \cap Q_2 \cong V_4$, and moreover $P_2 \cap Q_2$ is characteristic in $N_G(P_2 \cap Q_2)$. Now each normalizer of a conjugate of $P_2 \cap Q_2$ contains three subgroups of order 8 (which, incidentally, are isomorphic to $D_8$). Each of these is contained in a Sylow 2-subgroup of $G$, and in fact is contained in a unique Sylow 2-subgroup of $G$, as otherwise we violate the maximalness of $P_2 \cap Q_2$.

Now every Sylow 2-subgroup contains as normal subgroups isomorphic copies of $D_8$, which has 5 elements of order 2, and of $Q_8$, which has 1 element of order 2. Thus there is an element $w$ of order 2 which normalizes $Q_8 = \langle x_1, x_2 \rangle$ but is not in $Q_8$. (Recall the definitions of $x_1$ and $x_2$ above.) If $w$ fixes 1, then $w \in N_G(P_3)$, so that $\langle x_1, x_2, w \rangle = P_2 \leq N_G(P_3)$, a violation of Lagrange. Thus $w$ must not fix 1. By our calculation above of the elements of order 2 which normalizes Sylow 3-subgroups, we may assume that $(1\ 2)$ appears in the cycle decomposition of $w$.

Now $w$ permutes the three order 4 subgroups of $Q_8$, and thus must fix at least one. Note that $(wx_iw)(2) = (wx_i)(1) = w(1)$; however, this implies that some element of $Q_8$ does not fix 1, a contradiction.

• Suppose $|N_G(P_2 \cap Q_2)| = 2^3 \cdot 3^2$. If $n_2(N_G(P_2 \cap Q_2)) = 1$, then we have $N_{P_2}(P_2 \cap Q_2), N_{Q_2}(P_2 \cap Q_2) \leq N_G(P_2 \cap Q_2)$ Sylow 2-subgroups. Thus $N_{P_2}(P_2 \cap Q_2) = N_{Q_2}(P_2 \cap Q_2)$, and this subgroup has order 8, violating the maximalness of $P_2 \cap Q_2$. Thus $n_2(N_G(P_2 \cap Q_2)) \in \{3,9\}$. Since the Sylow 2-subgroups here intersect pairwise in $P_2 \cap Q_2$ (otherwise would violate the maximalness of $P_2 \cap Q_2$), $N_G(P_2 \cap Q_2)$ contains either $3 + 4 \cdot 3 = 15$ or $3 + 4 \cdot 9 = 39$ elements of 2-power order. Similarly, by Sylow’s Theorem and since the Sylow 3-subgroups of $N_G(P_2 \cap Q_2)$ intersect trivially, $N_G(P_2 \cap Q_2)$ contains either $8$ or $4 \cdot 8 = 32$ elements of order 3. Since every element of $N_G(P_2 \cap Q_2)$ is contained in a Sylow subgroup, we see that the only possibility is $n_2(N_G(P_2 \cap Q_2)) = 9$ and $n_3(N_G(P_2 \cap Q_2)) = 4$. Let $Q_3 \leq N_G(P_2 \cap Q_2)$ be a Sylow 3-subgroup. Then $(P_2 \cap Q_2)Q_3 \cong (P_2 \cap Q_2) \rtimes Q_3$ is a group of order 36. In a lemma to this previous exercise, we showed that a group of order 36 which does not have a unique Sylow 3-subgroup contains an element of order 6, a contradiction. Thus $Q_3 \leq (P_2 \cap Q_2) \rtimes Q_3$ is normal, so that in fact $(P_2 \cap Q_2)Q_3 \cong (P_2 \cap Q_2) \times Q_3$, and using Cauchy, we have an element of order 6, for another contradiction.
• If $k = 8$, then since $P_2 \cap Q_2$ is normal in both $P_2$ and $Q_2$, $|N_G(P_2 \cap Q_2)|$ is divisible by $2^4$ and another prime. Now $|N_G(P_2 \cap Q_2)| \in \{$ $2^4 \cdot 3,$ $2^4 \cdot 3^2,$ $2^4 \cdot 5,$ $2^4 \cdot 3 \cdot 5,$ $2^4 \cdot 3^2 \cdot 5\}$. We can see that in all but the cases $2^4 \cdot 3$ and $2^4 \cdot 5$, either $P_2 \cap Q_2$ is normal in $G$ or some subgroup has sufficiently small index. Suppose $|N_G(P_2 \cap Q_2)| = 2^4 \cdot 5$. By Sylow’s theorem, $n_5(N_G(P_2 \cap Q_2)) \in \{1,16\}$. If $n_5(N_G(P_2 \cap Q_2)) = 1$, then since the Sylow 5-subgroups in $N_G(P_2 \cap Q_2)$ are Sylow in $G$, we have $P_2 \leq N_G(P_5)$ for some Sylow 5-subgroup $P_5$. This is a contradiction of Lagrange. If $n_5(N_G(P_2 \cap Q_2)) = 16$, then since the Sylow 5-subgroups of $N_G(P_2 \cap Q_2)$ intersect trivially, $N_G(P_2 \cap Q_2)$ contains $4 \cdot 16$ elements of order 5. However it also contains at least 17 elements of 2-power order, since $P_2, Q_2 \leq N_G(P_2 \cap Q_2)$; this is a contradiction, and thus $|N_G(P_2 \cap Q_2)| \neq 2^4 \cdot 5$. Suppose $|N_G(P_2 \cap Q_2)| = 2^4 \cdot 3$. Note that $n_2(N_G(P_2 \cap Q_2)) = 3$; say that $P_2$, $Q_2$, and $R_2$ are the Sylow 2-subgroups of $N_G(P_2 \cap Q_2)$. Now $P_2 \cap Q_2$ is the pairwise intersection of $P_2$, $Q_2$, and $R_2$, so that $N_G(P_2 \cap Q_2)$ contains precisely $15 + 8 + 8 = 31$ nonidentity elements of 2-power order. By Sylow’s Theorem, $n_3(N_G(P_2 \cap Q_2)) \in \{1,4,16\}$. If $n_3(N_G(P_2 \cap Q_2)) = 1$, then $P_2 \leq N_G(T)$ for some subgroup $T$ of order 3. Since $T$ is also normal in some Sylow 3-subgroup, we have $|N_G(T)| \in \{ 2^4 \cdot 3^2, 2^4 \cdot 3^2 \cdot 5 \}$. Neither of these can occur, either because $T$ would then be normal of $N_G(T)$ would have sufficiently small index. If $n_3(N_G(P_2 \cap Q_2)) = 4$, then since Sylow 3-subgroups intersect trivially, there are precisely $4 \cdot 2 = 8$ elements of order 3 in $N_G(P_2 \cap Q_2)$. Now the elements of order 3 or 2-power number $31 + 8 = 39$, but because $N_G(P_2 \cap Q_2) \leq G$, no other orders are possible. Thus $G$ does not contain enough elements. If $n_3(N_G(P_2 \cap Q_2)) = 16$, then since Sylow 3-subgroups intersect trivially, $N_G(P_2 \cap Q_2)$ contains precisely $16 \cdot 2 = 32$ elements of order 3. But then $G$ contains $31 + 32 = 63$ elements, a contradiction.

Thus $n_3(G) \neq 10$, and no simple group of order 720 exists.

### No simple groups of order 3³·7·13·409 exist

Given the information on the Sylow numbers for a simple group of order $3^3 \cdot 7 \cdot 13 \cdot 409$ found in the previous exercise, prove that there is no such group. [Hint: Work in the permutation representation of degree 819.]

Recall from the previous exercise that $n_{409}(G) = 819$, so that $|N_G(P_{409}| = 3 \cdot 409$. Now $N_G(P_{409}) = N$ acts on $\mathsf{Syl}_{409}(G) = S$ by conjugation. Moreover, since $N$ has a unique Sylow 409-subgroup, this action has precisely one orbit of order 1, namely $\{P_{409} \}$. By the Orbit Stabilizer Theorem and Lagrange, the remaining orbits have order 3, 409, or $3 \cdot 409$. Since $3 \cdot 409 > 819$, no orbit has this order. If some orbit has order 409, then there are $819 - 409 - 1 = 409$ remaining elements of $S$, which are distributed among orbits of order three. However, $409 \equiv 1$ mod 3, so there must be another orbit of order 1 – a contradiction. If no orbit has order 409, then there are $819 - 1 = 818$ elements of $S$ in orbits of order 3; however, $818 \equiv 2$ mod 3, again a contradiction. Thus no simple group of order $3^3 \cdot 7 \cdot 13 \cdot 409$ exists.

### Compute the permissible Sylow numbers for a simple group of order 3³·7·13·409

Let $G$ be a simple group of order $3^3 \cdot 7 \cdot 13 \cdot 409$. Compute all permissible values of $n_p$ for each $p \in \{3,7,13,409\}$ and reduce to the case where there is a unique possible value for each $n_p$.

By Sylow’s Theorem, we have the following.

1. $n_3(G) \in \{1,7,13,7 \cdot 13 = 91,$ $409, 7 \cdot 409,$ $13 \cdot 409,$ $7 \cdot 13 \cdot 409 \}$
2. $n_7(G) \in \{ 1, 3^3 \cdot 13 = 351, 3^2 \cdot 13 \cdot 409 \}$
3. $n_{13}(G) \in \{1, 3^3 = 27, 3^2 \cdot 7 \cdot 409\}$
4. $n_{409}(G) \in \{1,3^2 \cdot 7 \cdot 13\}$

Note that $|G|$ does not divide $408!$, so that no proper subgroup has index at most 408. This forces $n_7(G) = 3^2 \cdot 13 \cdot 409 = 47853$ and $n_{13}(G) = 3^2 \cdot 7 \cdot 409 = 25767$. We also have $n_{409}(G) = 3^2 \cdot 7 \cdot 13 = 819$. Since the Sylow 7-, 13-, and 409-subgroups of $G$ intersect trivially, $G$ contains

1. $6 \cdot 47853 = 287118$ elements of order 7,
2. $12 \cdot 25767 = 309204$ elements of order 13, and
3. $408 \cdot 819 = 334152$ elements of order 409.

This consumes 930473 elements in $G$, whose total order is 1004913; only 74440 elements remain.

Suppose $n_3(G) = 5317$ and let $P_3 \leq G$ be a Sylow 3-subgroup. Then $|N_G(P_3)| = 3^3 \cdot 7$. Sylow’s Theorem then forces $n_7(N_G(P_3)) = 1$, so that $P_7 \leq N_G(P_3)$ is normal for some Sylow 7-subgroup. Note that $P_7$ is also Sylow in $G$. Now $P_3 \leq N_G(P_7)$, a contradiction because $3^3$ does not divide $|N_G(P_7)|$. Thus $n_3(G) \neq 5317$.

Suppose there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $|P_3 \cap Q_3| = 3^2$; then $P_3 \cap Q_3$ is normal in $P_3$ and $Q_3$, so that $|N_G(P_3 \cap Q_3)|$ is divisible by $3^3$ and another prime, and $n_3(N_G(P_3 \cap Q_3)) > 1$. Thus we have $|N_G(P_3 \cap Q_3)| \in \{$ $3^3 \cdot 7,$ $3^3 \cdot 13,$ $3^3 \cdot 409,$ $3^3 \cdot 7 \cdot 13,$ $3^3 \cdot 7 \cdot 409,$ $3^3 \cdot 13 \cdot 409,$ $3^3 \cdot 7 \cdot 13 \cdot 409 \}$. In cases 3, 5, and 6, $N_G(P_3 \cap Q_3)$ has index smaller than 408. In the last case, $P_3 \cap Q_3$ is normal in $G$. In the first case, we have $n_7(N_G(P_3 \cap Q_3)) = 1$, and every Sylow 7-subgroup of $N_G(P_3 \cap Q_3)$ is Sylow in $G$, so that $P_3 \leq N_G(P_7)$ for some Sylow 7- in $G$; this is a contradiction. In the second case, Sylow’s Theorem forces $n_{13}(N_G(P_3 \cap Q_3)) \in \{1, 3^3 \cdot 13 \}$. If 1, then since the Sylow 13-subgroups of $N_G(P_3 \cap Q_3)$ are Sylow in $G$, we have $P_3 \leq N_G(P_{13})$ for some Sylow 13-subgroup of $G$; this is a contradiction. If $3^3 \cdot 13$, then since the Sylow 13-subgroups of $G$ intersect trivially, $N_G(P_3 \cap Q_3)$ has $3^3 \cdot 12$ elements of order 13. Then $27$ elements remain, which must compose a unique Sylow 3-subgroup; this is a contradiction since $P_3, Q_3 \leq N_G(P_3 \cap Q_3)$ are Sylow. In the fourth case, we have $|N_G(P_3 \cap Q_3)| = 3^3 \cdot 7 \cdot 13$. Now Sylow’s Theorem forces $n_7 \in \{1, 3^3 \cdot 13 \}$ and $n_{13} \in \{1,3^3 \}$. If a Sylow 7- or 13-subgroup is normal in $N_G(P_3 \cap Q_3)$, then as before a Sylow 3-subgroup of $G$ normalizes a Sylow 7- or 13-subgroup, which is a contradiction. Thus $n_7(N_G(P_3 \cap Q_3)) = 3^3 \cdot 13$ and $n_{13}(N_G(P_3 \cap Q_3)) = 3^3$, so that $N_G(P_3 \cap Q_3)$ has $3^3 \cdot 13 \cdot 6$ elements of order 7 and $3^3 \cdot 12$ elements of order 13. This consumes all but 27 elements, which must constitute a unique Sylow 3-subgroup; this is a contradiction since $P_3, Q_3 \leq N_G(P_3 \cap Q_3)$ are Sylow. Thus no such subgroups $P_3$ and $Q_3$ may exist.

In particular, if $n_3(G) \neq 1$ mod 9, then by Lemma 13 subgroups $P_3$ and $Q_3$ as described in the previous paragraph must exist; thus $n_3(G) \neq 409, 13 \cdot 409$. Thus $n_3(G) = 7 \cdot 409$.

### There is a unique finite simple group whose order is the product of four primes, three of which are distinct

Let $G$ be a simple group of order $p^2qr$, where $p,q,r$ are primes. Prove that $|G| = 60$.

[Special thanks to Chris Curry for bouncing some ideas around. Thanks Chris! Thanks also to K.-S. Liu for pointing out a fatal flaw in my original solution.]

Sylow’s Theorem forces the following.

1. $n_p(G) \in \{1,q,r,qr\}$
2. $n_q(G) \in \{1,p,p^2,r,pr,p^2r\}$
3. $n_r(G) \in \{1,p,p^2,q,pq,p^2q\}$

Suppose $p > q,r$. Then $|G|$ does not divide $(2p-1)!$, so that no proper subgroup of $G$ has index at most $2p-1$; in particular, we have $n_p(G) = qr$. and $n_q(G), n_r(G) \geq p$. Choose $P_1,P_2 \in \mathsf{Syl}_p(G)$ such that $|P_1 \cap P_2|$ is maximal; if $|P_1 \cap P_2| = 1$, then all Sylow $p$-subgroups intersect trivially, so that $G$ has $qr(p^2-1)$ elements of $p$-power order. Now $G$ has at least $p(q-1)$ and $p(r-1)$ elements of order $q$ and $r$, respectively, so that $|G| \geq p^2qr - qr + pq - p + pr - p$ $> p^2qr$. (Note that one of $q$ and $r$ is not 2, so that $pq > 2p$ or $pr > 2p$, depending, and that $pq,pr > qr$.) This is a contradiction. Now suppose $|P_1 \cap P_2| = p$; then by this previous exercise, $|N_G(P_1 \cap P_2)|$ is divisible by $p^2$ and another prime; but then the index of $N_G(P_1 \cap P_2)$ is too small. Thus $p$ is not greater than both $q$ and $r$.

Suppose without loss of generality that $p,q < r$. Now $|G|$ does not divide $(r-1)!$, and no proper subgroup has index at most $r-1$. In particular, $n_p(G) \neq q$, $n_q(G) \neq p$, and $n_r(G) \not\in \{ p,q \}$. For reference we will write down the new possibilities for the Sylow numbers of $G$.

1. $n_p(G) \in \{1,r,qr\}$
2. $n_q(G) \in \{1,p^2,r,pr,p^2r\}$
3. $n_r(G) \in \{1,p^2,pq,p^2q\}$

Suppose $n_r(G) = p^2q$. Since the Sylow $r$-subgroups of $G$ intersect trivially, $G$ contains $p^2q(r-1)$ elements of order $r$. If $p^2 < r$, then $n_q(G) \geq p^2$. Since the Sylow $q$-subgroups of $G$ intersect trivially, $G$ has at least $p^2(q-1)$ elements of order $q$. Now at least $p^2+1$ elements are in Sylow $p$ subgroups, so that $|G| \geq p^2qr - p^2q + p^2q -p^2 + p^2 + 1 > p^2qr$, a contradiction. Suppose then that $p^2 > r$. If $n_p(G) = r$ and $P \in \mathsf{Syl}_p(G)$, then $|N_G(P)| = p^2q$. Moreover, since $q < r$, there exists an element in a Sylow $p$-subgroup which does not normalize $P$. So $|G| \geq p^2qr - p^2q + p^2q + 1 > p^2qr$, a contradiction. If instead $n_p(G) = qr$, choose $P_1,P_2 \in \mathsf{Syl}_p(G)$ such that $|P_1 \cap P_2|$ is maximal. If $|P_1 \cap P_2| = 1$, then the Sylow $p$-subgroups of $G$ intersect trivially, so that $|G| \geq p^2qr + p^2qr - p^2q - qr$ $= p^2qr + q(p^2r - p^2 - r) > p^2qr$, a contradiction. If $|P_1 \cap P_2| = p$, then by this previous exercise, $|N_G(P_1 \cap P_2)|$ is divisible by $p^2$ and another prime. Then $|N_G(P_1 \cap P_2)| \in$ $\{p^2q, p^2r, p^2qr \}$. We can see that in the second case, the index of $N_G(P_1 \cap P_2)$ is too small, and in the third case, $P_1 \cap P_2$ is normal in $G$. Thus $|N_G(P_1 \cap P_2)| = p^2q$. Now since $q < qr$, there exists an element in a Sylow $p$-subgroup which does not normalize $P_1 \cap P_2$, so that $|G| \geq p^2qr - p^2q + p^2q + 1 > p^2qr$, a contradiction. Thus $n_r(G) \neq p^2q$.

Suppose $n_r(G) = p^2$. Then $rt + 1 = p^2$ by Sylow’s Theorem, and thus $rt = p^2-1 = (p+1)(p-1)$. Since $r$ is prime, $r$ divides $p+1$ or $p-1$. Since $p < r$, we must have $r|p+1$. But then $r=p+1$, which forces $p=2$ and $r=3$. But $q$ must be a prime less than 3 and different from 2, and no such primes exist. Thus $n_r(G) \neq p^2$. We may assume henceforth that $n_r(G) = pq$; say $pq = rm+1$; note that $m$ is positive. Now $|N_G(R)| = pr$ for every Sylow $r$-subgroup $R$; note then that no element of $G$ has order $qr$, $p^2r$, or $pqr$, because such an element would commute with an element of order $r$, a contradiction. Moreover, suppose $P \leq N_G(R)$ is a Sylow $p$-subgroup. If $P$ is normal, then $R \leq N_G(P)$. We also have $P$ normal in some Sylow $p$-subgroup of $G$, so that $|N_G(P)|$ is divisible by $p^2r$. This gives a contradiction, however, as either $P$ is normal in $G$ or $N_G(P)$ has index $q$. Thus $n_p(N_G(R)) = r$. Thus we have $N_G(R) \cong Z_r \rtimes Z_p$. Moreover, $G$ contains no elements of order $pr$, since such an element would commute with an element of order $r$, but the normalizers of Sylow $r$-subgroups are not cyclic.

Suppose $n_q(G) = p^2$. Then $|N_G(Q)| = qr$, where $Q \leq G$ is a Sylow $q$-subgroup. Since $r > q$, $n_r(N_G(Q)) = 1$. But then $Q \leq N_G(R)$, where $R \leq N_G(Q)$ is a Sylow $r$-subgroup and thus Sylow in $G$. This is a contradiction since $|N_G(R)| = pr$. Thus $n_q(G) \neq p^2$. Note now that $G$ contains no elements of order $p^2q$, because such an element would commute with an element of order $q$, while the order of Sylow $q$-subgroup normalizers is not divisible by $p^2$.

We now show that $G$ has a subgroup of index $r$. If $n_p(G) = r$, then $[G : N_G(P)] = r$ where $P$ is a Sylow $p$-subgroup. Now suppose $n_p(G) = qr$. Choose $P_1,P_2 \in \mathsf{Syl}_p(G)$ such that $|P_1 \cap P_2|$ is maximal. if $P_1 \cap P_2 = 1$, then the Sylow $p$-subgroups intersect trivially. Then, counting elements in Sylow $p$– and $r$-subgroups, $|G| \geq qr(p^2-1) + pq(r-1)$ $= p^2qr - qr + pqr - pq$ $= p^2qr + q(pr - p - r) > p^2qr$, a contradiction. Thus $|P_1 \cap P_2| = p$, and $|N_G(P_1 \cap P_2)|$ is divisible by $p^2$ and some other prime. Thus $|N_G(P_1 \cap P_2)| \in \{p^2q, p^2r, p^2qr \}$. In the second case, $G$ has a subgroup of index $q$, and in the third case, $P_1 \cap P_2$ is normal in $G$. Thus $|N_G(P_1 \cap P_2)| = p^2q$, and $[G : N_G(P_1 \cap P_2)] = r$. Thus, in either case, $G$ has a subgroup $H$ of index $r$. Via the permutation representation afforded by the action of $G$ on $G/H$, we have $G \leq A_r$. Note moreover that if $R \leq G$ is a Sylow $r$-subgroup, then $R \leq A_r$ is Sylow, and that $|N_G(R)| = pr$ while $|N_{A_r}(R)| = r(r-1)/2$. Thus, by Lagrange, $p$ divides $(r-1)/2$, and thus $2p$ divides $r-1$. Say $r-1 = 2pk$; note that $k$ is positive. Note moreover that we have $2p < r < pq$, since $pq = rm+1$. In particular, $q$ is odd.

Suppose $n_q(G) = r$. Then we have $r \equiv 1$ mod $q$. Recall that $r = 2pk+1$ for some positive integer $k$. Thus we have $2pk+1 \equiv 1$ mod $q$, so that $2pk \equiv 0$ mod $q$. Since $p \neq q$ and $q$ is odd, we have $k \equiv 0$ mod $q$; say $k = qt$ for some positive integer $t$. Now from the equation $r = 2pk + 1$, we have $r = 2pqt + 1$, so that $r = 2(rm+1)t + 1$, and thus $r = 2rmt + 2t + 1$. However, this is a contradiction since $m$ and $t$ are positive. Thus $n_q(G) \neq r$.

Suppose now that $n_p(G) = qr$. Now choose Sylow $p$-subgroups $P_1$ and $P_2$ so that $|P_1 \cap P_2|$ is maximal. If $|P_1 \cap P_2| = 1$, then the Sylow $p$-subgroups of $G$ intersect trivially; thus $G$ contains $qr(p^2-1)$ elements of $p$-power order. Counting the order $r$ elements as well, $|G| \geq p^2qr - qr + pqr - pq$ $= p^2qr + q(pr - p - r) > p^2qr$, a contradiction. If $|P_1 \cap P_2| = p$, then we have $|N_G(P_1 \cap P_2)| \in \{p^2q, p^2r, p^2qr \}$. We see that only the first case is possible; that is, $|N_G(P_1 \cap P_2)| = p^2q$. Now $n_p(N_G(P_1 \cap P_2)) = q$. If $n_q(N_G(P_1 \cap P_2)) = 1$, then $p^2$ divides $|N_G(Q)|$ for some Sylow $q$-subgroup $Q \leq G$, a contradiction. If $n_q(N_G(P_1 \cap P_2)) = p^2$, then $N_G(P_1 \cap P_2)$ contains $(p^2-p)q$ elements of $p$-power order not in $P_1 \cap P_2$ by this previous exercise and $p^2(q-1)$ elements of order $q$, so that $|G| \geq p^2q -pq + p^2q - p^2$ $= p^2q + p(pq - p - q) > p^2q$, a contradiction. Thus $n_q(N_G(P_1 \cap P_2)) = p$. But now we have $p \equiv 1$ mod $q$ and $q \equiv 1$ mod $p$, so that we have $p > q$ and $q > p$, a contradiction. Thus $n_p(G) \neq qr$.

Thus we have $n_p(G) = r$.

Suppose $n_p = r \not\equiv 1$ mod $p^2$. By Lemma 13 in the text, there exist $P_1$ and $P_2$ Sylow $p$-subgroups of $G$ such that $P_1 \cap P_2$ is nontrivial; thus $|N_G(P_1 \cap P_2)| = p^2q$. We have $n_p(N_G(P_1 \cap P_2)) = q$. As before, the number of Sylow $q$-subgroups in this normalizer is not 1 since $p^2$ does not divide the order of the normalizer of the Sylow $q$-subgroups of $G$, and is not $p^2$ since this yields too many elements. Thus $n_q(N_G(P_1 \cap P_2)) = p$, so that $p > q$ and $q > p$, a contradiction. Thus $n_p = r \equiv 1$ mod $p^2$. In particular, we have $p^2 < r$, and since $r < pq$, we have $p < q$.

Let $P \leq G$ be a Sylow $p$ subgroup. Note that $n_q(N_G(P)) \neq 1$ since then $P$ would normalize a Sylow $q$-subgroup, a contradiction. If $n_q(N_G(P)) = p$, then we have $p \equiv 1$ mod $q$, and hence $p > q$, also a contradiction. Thus $n_q(N_G(P)) = p^2$, so that $p^2 \equiv 1$ mod $q$. Thus we have $(p+1)(p-1) \equiv 0$ mod $q$. If $q$ divides $p-1$, then $q < p$, a contradiction. Thus $q$ divides $p+1$. Since $p < q$, we thus have $p+1 = q$. So $p=2$ and $q=3$. From $pq = rm+1$ we have $r = 5$.

Thus $|G| = 60$.

### There are no finite simple groups of even order less than 500 except for the orders 2, 60, 168, and 360

1. Prove that there are no simple groups of order 420.
2. Prove that there are no simple groups of even order less than 500 except for the orders 2, 60, 168, and 360.

We begin with some lemmas.

Lemma 1: [Adapted from Steve Flink’s notes (PostScript)] If $G$ is a group of order $36 = 2^2 \cdot 3^2$ and $G$ does not have a normal Sylow 3-subgroup, then $G$ contains an element of order 6. Proof: Sylow’s Theorem forces $n_3 \in \{1,4\}$; and we have $n_3 = 4 \not\equiv 1$ mod 9. By Lemma 13, there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $P_3 \cap Q_3$ is nontrivial. Now consider $C_G(P_3 \cap Q_3)$. Since $P_3$ and $Q_3$ are abelian, both centralize $P_3 \cap Q_3$. So $\langle P_3, Q_3 \rangle \leq C_G(P_3 \cap Q_3)$. Now $|\langle P_3, Q_3 \rangle| = 2^t3^2$, and by Sylow and the fact that $P_3$ and $Q_3$ are Sylow subgroups, in fact $\langle P_3, Q_3 \rangle = G$. Thus $P_3 \cap Q_3 \leq Z(G)$. If $x \in G$ has order 2 by Cauchy, and $y \in P_3 \cap Q_3$, then $xy$ has order 6. $\square$

Now to the main results.

First we show that no group of order $420 = 2^2 \cdot 3 \cdot 5 \cdot 7$ is simple. [Adapted (slightly) from Steve Flink’s notes.]

Let $G$ be a simple group of order 420. Note that $|G|$ does not divide $6!$, so that no proper subgroup of $G$ has index at most 6. This and Sylow’s Theorem force $n_7(G) = 15$, so that $G$ contains $6 \cdot 15 = 90$ elements of order 7, and similarly $n_5(G) = 21$, so that $G$ has $4 \cdot 21 = 84$ elements of order 5. Suppose now that $G$ has a subgroup $H$ of index 7; the permutation representation induced by the action of $G$ on $G/H$ yields $G \leq A_7$. However, if $P_7 \leq G$ is a Sylow 7-subgroup, $|N_G(P_7)| = 7 \cdot 2$ while $|N_{A_7}(P_7)| = 7 \cdot 3$, which is absurd; thus no subgroup of $G$ has index 7. In particular, $n_3(G) \neq 7$; thus $n_3(G) \in \{ 10, 28, 70 \}$. Suppose $n_3(G) = 10$. If $P_3 \leq G$ is a Sylow 3-subgroup, then $|N_G(P_3)| = 2 \cdot 3 \cdot 7$. But then the Sylow 7-subgroup $P_7 \leq N_G(P_3)$ is normal and Sylow in $G$, so that $P_3 \leq N_G(P_7)$, a contradiction. Similarly, if $n_3(G) = 28$, then $|N_G(P_3)| = 3 \cdot 5$, so that $P_3 \leq N_G(P_5)$ for some Sylow 5-subgroup $P_5 \leq G$, also a contradiction. Thus we have $n_3(G) = 70$, and $G$ contains $2 \cdot 70 = 140$ elements of order 3. Now let $P_7 \leq G$ be a Sylow 7-subgroup. We have $|N_G(P_7)| = 2^2 \cdot 7$, so that by Sylow’s Theorem, $n_2(N_G(P_7)) \in \{1,7\}$. If a Sylow 2-subgroup $P_2 \leq N_G(P_7)$ (which is Sylow in $G$) is normal, then the normalizer of $P_7$ contains a unique Sylow 2-subgroup of $G$, and each Sylow 2-subgroup of $G$ is contained in a Sylow 7 normalizer. Thus there are at most 15 Sylow 2-subgroups in $G$. By Sylow and the above arguments, there are at least 15 Sylow 2-subgroups in $G$, so that $n_2(G) = 15$. Each Sylow 7-normalizer has the form $P_2P_7$, and so contains elements of order 14 or 21. Note that if some element of order 14 or 21 is contained in two distinct Sylow 7 normalizers, then those two normalizers share an element of order 7, and thus must be equal. There are $3 \cdot 6 = 18$ elements of order 14 or 21 in each Sylow 7 normalizer, and 15 such normalizers, so that $G$ contains $18 \cdot 15 = 270$ elements of order 14 or 21. Thus $|G| \geq 90 + 84 + 140 + 270 = 584$, a contradiction. Thus $n_2(N_G(P_7)) = 7$. By Lemma 13 and a previous exercise, there exist $P_2,Q_2 \in \mathsf{Syl}_2(G)$ (which are also Sylow in $G$) such that $|N_{N_G(P_7)}(P_2 \cap Q_2)|$ is divisible by $2^2$ and another prime; in fact, $P_2 \cap Q_2$ is normal in $N_G(P_7)$. Note that $\mathsf{Aut}(P_2 \cap Q_2)$ is trivial, so that by the N/C theorem, $P_2 \cap Q_2 \leq Z(G)$. Note that $N_G(P_7)$ contains $1 + 7 \cdot 2 = 15$ elements of 2-power order, and products of the nonidentity element in $P_2 \cap Q_2$ with $P_7$ yield 6 elements of order 14. There are 15 Sylow 7-normalizers, and if any two share an element of order 14 then they share an element of order 7, a contradiction. Thus $G$ contains at least $6 \cdot 15 = 90$ elements of order 14, so that the elements of order 3, 5, 7, and 14 consume $140 + 84 + 90 + 90 = 404$ elements. Only 15 nonidentity elements remain, which must necessarily be the 15 elements of 2-power order we counted in $N_G(P_7)$. Thus every Sylow 7-normalizer contains all of the Sylow 2-subgroups of $G$. Note, however, that for every Sylow 7-subgroup $Q_7$, $\langle \bigcup \mathsf{Syl}_2(G) \rangle = N_G(Q_7)$. Thus $G$ has but one Sylow 7-subgroup, a contradiction.

So no simple group of order 420 exists.

As in the previous exercise, we handle each remaining case in turn in the following table.

 $n$ Reasoning $12 = 2^2 \cdot 3$ Let $G$ be a simple group of order 12. Note that $|G|$ does not divide $3!$, since the highest power of 2 dividing $3!$ is $2^1$. Thus no proper subgroup of $G$ has index at most 3. This and Sylow’s Theorem then force $n_2(G) = 1$, a contradiction. $24 = 2^3 \cdot 3$ Let $G$ be a simple group of order 24. Note that $|G|$ does not divide $3!$ since the highest power of $2$ dividing $3!$ is $2^1$; thus no proper subgroup of $G$ has index at most 3. This and Sylow’s Theorem force $n_2(G) = 1$, a contradiction. $30 = 2 \cdot 3 \cdot 5$ Example on page 143 $36 = 2^2 \cdot 3^2$ Let $G$ be a simple group of order 36. Note that $|G|$ does not divide $5!$, so that no proper subgroup of $G$ has index at most 5. This and Sylow’s Theorem force $n_3(G) = 1$, a contradiction. $48 = 2^4 \cdot 3$ Let $G$ be a simple group of order 48. Note that $|G|$ does not divide $5!$ since the largest power of 2 which divides $5!$ is $2^3$; thus no proper subgroup of $G$ has index at most 5. This and Sylow’s Theorem force $n_2(G) = 1$, a contradiction. $56 = 2^3 \cdot 7$ Let $G$ be a simple group of order 56. Sylow’s Theorem forces $n_2(G) = 7$ and $n_7(G) = 8$; since the Sylow 7-subgroups of $G$ intersect trivially, $G$ contains $8 \cdot 6 = 48$ elements of order 7 and at least $8+1=9$ elements of 2-power order, for a total of at least 57 elements. $72 = 2^3 \cdot 3^2$ Let $G$ be a simple group of order 72. Note that $|G|$ does not divide $5!$, since the highest power of 3 dividing $5!$ is $3^1$. Thus no proper subgroup of $G$ has index at most 5. This and Sylow’s Theorem force $n_3(G) = 1$, a contradiction. $80 = 2^4 \cdot 5$ §6.2 #4 $90 = 2 \cdot 3^2 \cdot 5$ Let $G$ be a simple group of order 90. Sylow’s Theorem forces $n_5(G) = 6$, so that via the permutation representation afforded by the action of $G$ on $G/N_G(P_5)$, where $P_5 \leq G$ is a Sylow 5-subgroup, we have $G \leq A_6$. But $|N_G(P_5)| = 5 \cdot 3$, while $|N_{A_6}(P_5)| = 5 \cdot 2$, which is absurd by Lagrange. $96 = 2^5 \cdot 3$ Let $G$ be a simple group of order 96. Note that $|G|$ does not divide $7!$, since the highest power of 2 dividing $7!$ is $2^4$. Thus $G$ does not contain a proper subgroup of index at most 7. However, Sylow’s Theorem then forces $n_2(G) = 1$, a contradiction. $108 = 2^2 \cdot 3^3$ Let $G$ be a simple group of order 108. Note that $|G|$ does not divide $8!$, so that no proper subgroup of $G$ has index at most 8. This and Sylow’s Theorem force $n_3(G) = 1$, a contradiction. $112 = 2^4 \cdot 7$ Let $G$ be a simple group of order 112. Sylow’s Theorem forces $n_2(G) = 7 \not\equiv 1$ mod 4, so that by Lemma 13 and a previous example, there exist $P_2,Q_2 \in \mathsf{Syl}_2(G)$ such that $P_2 \cap Q_2 \neq 1$ and $|N_G(P_2 \cap Q_2)|$ is divisible by $2^4$ and another prime; but then $P_2 \cap Q_2$ is normal in $G$, a contradiction. $120 = 2^3 \cdot 3 \cdot 5$ Let $G$ be a simple group of order 120. Sylow’s Theorem forces $n_5(G) = 6$, so that via the permutation representation induced by the action of $G$ on $G/N_G(P_5)$ for some Sylow 5-subgroup $P_5 \leq G$ we have $G \leq A_6$. Now $|N_G(P_5)| = 5 \cdot 2^2$ while $|N_{A_6}(P_5)| = 5 \cdot 2$, implying that 2 divides 1, which is absurd. This $n_5(G) = 1$, a contradiction. $132 = 2^2 \cdot 3 \cdot 11$ Let $G$ be a simple group of order 132. Note that $|G|$ does not divide $10!$, so that no proper subgroup of $G$ has index at most 10. This and Sylow’s Theorem force $n_{11}(G) = 12$ and $n_3(G) = 22$. Since the Sylow 3- and 11-subgroups of $G$ intersect trivially, $G$ contains at least $12 \cdot 10$ $+ 2 \cdot 22 = 164$ elements, a contradiction. $144 = 2^4 \cdot 3^2$ §6.2 #14 $150 = 2 \cdot 3 \cdot 5^2$ Let $G$ be a simple group of order 150. Note that $|G|$ does not divide $9!$, so that no proper subgroup of $G$ has index at most 9. This and Sylow’s Theorem force $n_5(G) = 1$, a contradiction. $160 = 2^5 \cdot 5$ Let $G$ be a simple group of order 160. Note that $|G|$ does not divide $7!$ since the highest power of 2 dividing $7!$ is $2^4$. Thus no proper subgroup of $G$ has index at most 7. However, Sylow’s Theorem then forces $n_2(G) = 1$, a contradiction. $180 = 2^2 \cdot 3^2 \cdot 5$ Let $G$ be a simple group of order 180. Note that $|G|$ does not divide $5!$, so that no proper subgroup has index at most 5. This and Sylow’s Theorem imply that $n_3(G) = 10$. Suppose now that $n_5(G) = 6$. Then via the permutation representation afforded by the action of $G$ on $G/N_G(P_5)$, where $P_5$ is a Sylow 5-subgroup, we have $G \leq A_6$. Note that Sylow 5-subgroups of $G$ are Sylow in $A_6$. But we have $|N_G(P_5)| = 5 \cdot 2 \cdot 3$ while $|N_{A_6}(P_5)| = 5 \cdot 2$; this is a contradiction by Lagrange. Thus $n_5(G) = 36$, and since the Sylow 5-subgroups of $G$ intersect trivially, $G$ has $4 \cdot 36 = 144$ elements of order 5. Now choose $P_3,Q_3 \in \mathsf{Syl}_3(G)$ such that $|P_3 \cap Q_3|$ is maximal. If $|P_3 \cap Q_3| = 1$, then all Sylow 3-subgroups intersect trivially, so that $G$ contains $8 \cdot 10 = 80$ nonidentity elements of 3-power order; but then $|G| \geq 80 + 144 = 224$, a contradiction. Suppose now that $|P_3 \cap Q_3| = 3$. By a previous exercise, $|N_G(P_3 \cap Q_3)|$ is divisible by $3^2$ and another prime, and $P_3 \cap Q_3$ is normal in $P_3$ and $Q_3$. We have $|N_G(P_3 \cap Q_3)| \in$ $3^2 \cdot 2,$ $3^2 \cdot 2^2,$ $3^2 \cdot 5$, $3^2 \cdot 2 \cdot 5,$ $3^2 \cdot 2^2 \cdot 5 \}$. In all but the first case, either $P_3 \cap Q_3$ is normal in $G$ or its normalizer has sufficiently small index. Thus $|N_G(P_3 \cap Q_3)| = 3^2 \cdot 2$. However, we now have $n_3(N_G(P_3 \cap Q_3)) = 1$, a contradiction since we know that $P_3$ and $Q_3$ are Sylow subgroups of $N_G(P_3 \cap Q_3)$. $192 = 2^6 \cdot 3$ Let $G$ be a simple group of order 192. Note that $|G|$ does not divide $7!$ since the highest power of 2 which divides $7!$ is $2^4$; thus no proper subgroup of $G$ has index at most 7. This and Sylow’s Theorem force $n_2(G) = 1$, a contradiction. $210 = 2 \cdot 3 \cdot 5 \cdot 7$ Let $G$ be a simple group of order 210. Note that $|G|$ does not divide $6!$, so that no proper subgroup of $G$ has index at most 6. This and Sylow’s Theorem force $n_7(G) = 15$, $n_5(G) = 21$, and $n_3(G) \in \{10,70\}$. If $n_3(G) = 70$, then because the Sylow subgroups of $G$ intersect trivially, $G$ contains at least $6 \cdot 15$ $+ 4 \cdot 21$ $+ 2 \cdot 70$ $= 314$ elements, a contradiction. Thus we have $n_3(G) = 10$. If $P_3 \leq G$ is a Sylow 3-subgroup, we have $|N_G(P_3)| = 3 \cdot 7$. Let $P_7 \leq N_G(P_3)$ be a Sylow 7-subgroup; now $P_7$ is also Sylow in $G$. Moreover, $n_7(N_G(P_3)) = 1$, so that $P_7$ is normal in $N_G(P_3)$. Thus $P_3 \leq N_G(P_7)$. But we know that $|N_G(P_7) = 2 \cdot 7$, a contradiction by Lagrange. $216 = 2^3 \cdot 3^3$ Let $G$ be a simple group of order 216. Note that $|G|$ does not divide $8!$, since the highest power of 3 dividing $8!$ is $3^2$. Thus no proper subgroup of $G$ has index at most 8. This and Sylow’s Theorem force $n_3(G) = 1$, a contradiction. $224 = 2^5 \cdot 7$ Let $G$ be a simple group of order 224. Note that $|G|$ does not divide $7!$ since the largest power of 2 which divides $7!$ is $2^4$. Thus no proper subgroup of $G$ has index at most 7; this and Sylow’s Theorem imply that $n_2(G) = 1$, a contradiction. $240 = 2^4 \cdot 3 \cdot 5$ Let $G$ be a simple group of order 240. Note that $|G|$ does not divide $5!$, so that no proper subgroup of $G$ has index at most 5. This and Sylow’s Theorem force $n_2(G) = 15 \not\equiv 1$ mod 4, so that by Lemma 13 and §6.2 #13 there exist $P_2,Q_2 \in \mathsf{Syl}_2(G)$ such that $|N_G(P_2 \cap Q_2)|$ is divisible by $2^4$ and another prime. Thus $|N_G(P_2 \cap Q_2)| \in$ $\{2^4 \cdot 3,$ $2^4 \cdot 5,$ $2^4 \cdot 3 \cdot 5 \}$. In each case we either have $P_2 \cap Q_2$ normal in $G$, or its normalizer having sufficiently small index. $252 = 2^2 \cdot 3^2 \cdot 7$ Let $G$ be a simple group of order 252. Note that $|G|$ does not divide $6!$, so that no proper subgroup of $G$ has index at most 6. This and Sylow’s Theorem imply that $n_3(G) \in \{7,28\}$ and $n_7(G) = 36$. Since Sylow 7-subgroups intersect trivially, $G$ contains $6 \cdot 36 = 216$ elements of order 7. Let $P_3,Q_3 \in \mathsf{Syl}_3(G)$ be chosen such that $|P_3 \cap Q_3|$ is maximal. If $|P_3 \cap Q_3| = 1$, then all Sylow 3-subgroups intersect trivially and $G$ contains at least $8 \cdot 6 = 48$ elements of 3-power order, and $|G| \geq 216 + 48 = 264$, a contradiction. If $|P_3 \cap Q_3| = 3$, then since $P_3 \cap Q_3$ is normal in both $P_3$ and $Q_3$, $|N_G(P_3 \cap Q_3)|$ is divisible by $3^2$ and another prime. Thus $|N_G(P_3 \cap Q_3)| \in$ $\{3^2 \cdot 2,$ $3^2 \cdot 2^2,$ $3^2 \cdot 7,$ $3^2 \cdot 2 \cdot 7,$ $3^2 \cdot 2^2 \cdot 7 \}$. In the last three cases, either $P_3 \cap Q_3$ is normal in $G$ or its normalizer has sufficiently small index. If $|N_G(P_3 \cap Q_3)| = 3^2 \cdot 2$, then $n_3(N_G(P_3 \cap Q_3)) = 1$, a contradiction since $P_3$ and $Q_3$ are distinct Sylow 3-subgroups of $N_G(P_3 \cap Q_3)$. Thus $|N_G(P_3 \cap Q_3)| = 3^2 \cdot 2^2$, and $n_3(N_G(P_3 \cap Q_3)) = 4$. Thus $N_G(P_3 \cap Q_3)$ contains four Sylow 3-subgroups; since $P_3 \cap Q_3$ (as a 3-subgroup) is conjugate (in $N_G(P_3 \cap Q_3)$) to a subgroup of each Sylow 3-subgroup, it is in fact contained in each Sylow 3-subgroup, and is moreover the intersection of all the Sylow 3-subgroups of $N_G(P_3 \cap Q_3)$. Thus the Sylow 3-subgroups of $N_G(P_3 \cap Q_3)$ contribute $8 + 6+6+6 = 26$ nonidentity elements of 3-power order to $G$. Let $R_3 \leq G$ be any Sylow 3-subgroup not contained in $N_G(P_3 \cap Q_3)$; now $|R_3 \cap N_G(P_3 \cap Q_3)|$ is at most 3, so that $R_3$ contributes at least 6 new nonidentity elements of 3-power order. Finally, note that $G$ contains at least 4 elements of 2-power order. Including the identity, $|G| \geq 216 +$ $26 + 6 +$ $4 + 1 = 253$, a contradiction. Thus $n_3(G) = 1$, a contradiction. $264 = 2^3 \cdot 3 \cdot 11$ Example on page 205 $270 = 2 \cdot 3^3 \cdot 5$ Let $G$ be a simple group of order 270. Note that $|G|$ does not divide $8!$, since the highest power of 3 dividing $8!$ is $3^2$. Thus no proper subgroup of $G$ has index at most 8. This and Sylow’s Theorem force $n_5(G) = 1$, a contradiction. $280 = 2^3 \cdot 5 \cdot 7$ Let $G$ be a simple group of order 280. Sylow’s Theorem forces $n_7(G) \in \{1,8\}$. If $n_7(G) = 8$, then via the permutation representation induced by the action of $G$ on $G/N_G(P_7)$ by left multiplication, where $P_7 \leq G$ is some Sylow 7-subgroup, we have $G \leq S_8$. In fact, $G \leq A_8$. Nowever, $|N_G(P_7)| = 5 \cdot 7$ while $|N_{A_8}(P_{7})| = 7 \cdot 3$, which is absurt. Thus $n_7(G) = 1$, a contradiction. $288 = 2^5 \cdot 3^2$ Let $G$ be a simple group of order 288. Note that $|G|$ does not divide $7!$, so that no proper subgroup of $G$ has index at most 7. This and Sylow’s Theorem force $n_2(G) = 9$ and $n_3(G) = 16$. Since $n_3(G) \not\equiv 1$ mod 9, there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $|N_G(P_3 \cap Q_3)|$ is divisible by $3^2$ and another prime. Thus $|N_G(P_3 \cap Q_3)| \in$ $\{ 2 \cdot 3,$ $2^2 \cdot 3^2,$ $2^3 \cdot 3^2,$ $2^4 \cdot 3^2,$ $2^5 \cdot 3^2 \}$. In The last three cases, either $P_3 \cap Q_3$ is normal in $G$ or its normalizer has sufficiently small index. In the first case, the normalizer of $P_3 \cap Q_3$ has only one Sylow 3-subgroup, a contradiction since $P_3$ and $Q_3$ normalize $P_3 \cap Q_3$. Thus $|N_G(P_3 \cap Q_3)| = 2^2 \cdot 3^2$. This subgroup has index 8, so that via the permutation representation induced by the action of $G$ on $G/N_G(P_3 \cap Q_3)$, we have $G \leq S_8$. In fact, since $G$ is simple, $G \leq A_8$. In particular, $G$ contains no elements of order 6, since the elements of order 6 in $S_8$ are all odd permutations. Moreover, by Sylow we have $n_3(N_G(P_3 \cap Q_3)) = 4$. By Lemma 1, this group of order 36 has a unique (hence normal) Sylow 2-subgroup of order 4; call this subgroup $K_2$. Thus $P_3 \leq N_G(K_2)$. Now $K_2 \leq P_2$ for some Sylow 2-subgroup $P_2 \leq G$, and moreover $K_2 < N_{P_2}(K_2) \leq N_G(P_2)$ is proper. Thus $2^3 \cdot 3^2$ divides $|N_G(K_2)|$; this subgroup then has index 4, a contradiction. $300 = 2^2 \cdot 3 \cdot 5^2$ Let $G$ be a simple group of order 300. Note that $|G|$ does not divide $9!$, so that no proper subgroup of $G$ has index at most 9. This and Sylow’s Theorem force $n_5(G) = 1$, a contradiction. $306 = 2 \cdot 3^2 \cdot 17$ Let $G$ be a simple group of order 306. Note that $|G|$ does not divide $16!$, so that no proper subgroup of $G$ has index at most 16. This and Sylow’s Theorem force $n_2(G) \geq 17$, where the Sylow 2-subgroups of $G$ intersect trivially. Thus $G$ has at least 17 elements of order 2. Similarly, Sylow forces $n_{17} = 18$, so that $G$ has $16 \cdot 18 = 288$ elements of order 17. Including a Sylow 3-subgroup, $G$ has at least $288 + 17$ $+ 9 = 314$ elements, a contradiction. $320 = 2^6 \cdot 5$ Let $G$ be a simple group of order 320. Note that $|G|$ does not divide $7!$ since the highest power of 2 which divides $7!$ is $2^4$. Thus $G$ has no proper subgroup of index at most 7. This and Sylow’s Theorem imply that $n_2(G) = 1$, a contradiction. $324 = 2^2 \cdot 3^4$ Let $G$ be a simple group of order 324. Note that $|G|$ does not divide $8!$ since the highest power of 3 which divides $8!$ is $3^2$; thus no proper subgroup of $G$ has index at most 8. This and Sylow’s Theorem force $n_3(G) = 1$, a contradiction. $336 = 2^4 \cdot 3 \cdot 7$ §6.2 #9 $380 = 2^2 \cdot 5 \cdot 19$ §6.2 #3 $384 = 2^7 \cdot 3$ Let $G$ be a simple group of order 384. Note that $|G|$ does not divide $7!$, since the highest power of 2 dividing $7!$ is $2^4$. Thus no proper subgroup of $G$ has index at most 7. However, Sylow’s Theorem then forces $n_2(G) = 1$, a contradiction. $392 = 2^3 \cdot 7^2$ Let $G$ be a simple group of order 392. Note that $|G|$ does not divide $13!$; thus no proper subgroup of $G$ has index at most 13. This and Sylow’s Theorem force $n_7(G) = 1$, a contradiction. $396 = 2^2 \cdot 3^3 \cdot 11$ Example on page 204 $400 = 2^4 \cdot 5^2$ Let $G$ be a simple group of order 400. Note that $|G|$ does not divide $9!$, so that no proper subgroup of $G$ has index at most 9. This and Sylow’s Theorem force $n_5(G) = 16 \not\equiv 1$ mod 25, so that by Lemma 13 and a previous exercise, there exist $P_5,Q_5 \in \mathsf{Syl}_5(G)$ such that $|N_G(P_5 \cap Q_5)|$ is divisible by $5^2$ and another prime. Thus $|N_G(P_5 \cap Q_5)| \in$ $\{ 5^2 \cdot 2,$ $5^2 \cdot 2^2,$ $5^2 \cdot 2^3,$ $5^2 \cdot 2^4 \}$. In each case, either $P_5 \cap Q_5$ is normal in $G$ or its normalizer has sufficiently small index. $420 = 2^2 \cdot 3 \cdot 5 \cdot 7$ Part (a) $432 = 2^4 \cdot 3^3$ Let $G$ be a simple group of order 432. Note that $|G|$ does not dividie $8!$; thus no proper subgroup of $G$ has index at most 8. Sylow’s Theorem forces $n_3(G) = 16 \not\equiv 1$ mod 9, so that Lemma 13 and a previous exercise imply that there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $|N_G(P_3 \cap Q_3)|$ is divisible by $3^3$ and another prime. Thus $|N_G(P_3 \cap Q_3)| \in$ $\{ 3^3 \cdot 2,$ $3^3 \cdot 2^2,$ $3^3 \cdot 2^2,$ $3^3 \cdot 2^3,$ $3^3 \cdot 2^4 \}$. In each case either $P_3 \cap Q_3$ is normal in $G$ or its normalizer has sufficiently small index. $448 = 2^6 \cdot 7$ Let $G$ be a simple group of order 448. Note that $|G|$ does not divide $7!$ since the highest power of 2 which divides $7!$ is $2^4$. Thus no proper subgroup of $G$ has index at most 7. This and Sylow’s Theorem then force $n_2(G) = 1$, a contradiction. $450 = 2 \cdot 3^2 \cdot 5^2$ Let $G$ be a simple group of order 450. Note that $|G|$ does not divide $9!$. This and Sylow’s Theorem force $n_5(G) = 1$, a contradiction. $480 = 2^5 \cdot 3 \cdot 5$ Let $G$ be a simple group of order 480. Note that $|G|$ does not divide $7!$, so that no proper subgroup of $G$ has index at most 7. This and Sylow’s Theorem force $n_2(G) = 15 \not\equiv 1$ mod 4, so that by Lemma 13 and a previous exercise there exist $P_2,Q_2 \in \mathsf{Syl}_2(G)$ such that $|N_G(P_2 \cap Q_2)|$ is divisible by $2^5$ and another prime. Thus $|N_G(P_2 \cap Q_2)| \in$ $\{ 2^5 \cdot 3,$ $2^5 \cdot 5,$ $2^5 \cdot 3 \cdot 5 \}$. Thus either $P_2 \cap Q_2$ is normal in $G$ or its normalizer has sufficiently small index.

### There are no nonabelian simple groups of odd order less than 10000

Prove that there are no nonabelian simple groups of odd order less than 10000.

In this previous exercise, we found a list of 60 odd positive integers $n$ less than 10000 such that a group of order $n$ is not forced to have a normal Sylow subgroup merely by the congruence and divisibility criteria of Sylow’s Theorem. We prove that no group of any of these orders is simple in the following table.

 $n$ Reasoning $105 = 3 \cdot 5 \cdot 7$ Done here $315 = 3^2 \cdot 5 \cdot 7$ Let $G$ be a simple group of order 315. Sylow’s Theorem forces $n_3(G) = 7$ and $n_5(G) = 21$. Now the left action of $G$ on $G/N_G(P_3)$ for some Sylow 3-subgroup $P_3$ induces an injective permutation representation $G \leq S_7$. Let $P_5 \leq G$ be a Sylow 5-subgroup; note that $P_5 \leq S_7$ is Sylow. We have $|N_G(P_5)| = 5 \cdot 3$, and $|N_{S_7}(P_5)| = 5 \cdot 4 \cdot 2$; but this implies that 3 divides 8, which is absurd. $351 = 3^3 \cdot 13$ Done here $495 = 3^2 \cdot 5 \cdot 11$ Let $G$ be a simple group of order 495. Sylow’s Theorem forces $n_3(G) = 55$, $n_5(G) = 11$, and $n_{11}(G) = 45$. Since the Sylow 5- and 11-subgroups of $G$ are cyclic, $G$ contains at least $10 \cdot 45 + 4 \cdot 11 + 9 = 503$ elements, a contradiction. $525 = 3 \cdot 5^2 \cdot 7$ Done here $735 = 3 \cdot 5 \cdot 7^2$ Let $G$ be a simple group of order 735. Note that $|G|$ does not divide $13!$, so that no proper subgroup of $G$ has index at most 13. Sylow’s Theorem forces $n_7(G) = 15 \not\equiv 1$ mod 49, so that by Lemma 13 and this previous exercise there exist $P_7,Q_7 \in \mathsf{Syl}_7(G)$ such that $|N_G(P_7 \cap Q_7)|$ is divisible by $7^2$ and some other prime. Thus $[G : N_G(P_7 \cap Q_7)] \in \{1,3,5\}$; in any case we have a contradiction, either because $P_7 \cap Q_7$ is normal in $G$ or because some subgroup has sufficiently small index. $945 = 3^3 \cdot 5 \cdot 7$ Let $G$ be a simple group of order 945. Sylow’s Theorem forces $n_3(G) = 7$ and $n_7(G) = 15$. The left action of $G$ on $G/N_G(P_3)$, where $P_3 \leq G$ is some Sylow 3-subgroup, yields an injective permutation representation $G \leq S_7$. Now let $P_7 \leq G$ be a Sylow 7-subgroup; then $P_7$ is also Sylow in $S_7$. Now $|N_G(P_7)| = 7 \cdot 3^2$, while $|N_{S_7}(P_7)| = 7 \cdot 6$. This implies that 3 divides 2, which is absurd. $1053 = 3^4 \cdot 13$ Example on page 207 $1365 = 3 \cdot 5 \cdot 7 \cdot 13$ Suppose $G$ is simple of order 1365. Then Sylow’s Theorem forces $n_{13} = 105$, $n_{7} = 15$, and $n_5 \geq 21$. Since all Sylow subgroups are cyclic, $G$ has at least $12 \cdot 105 + 6 \cdot 15$ $+ 4 \cdot 21 = 1434$ elements, a contradiction. $1485 = 3^3 \cdot 5 \cdot 11$ Let $G$ be a simple group of order 1485. Sylow’s Theorem forces $n_5(G) = 11$ and $n_{11}(G) = 45$. The left action of $G$ on $G/N_G(P_5)$, where $P_5$ is some Sylow 5-subgroup, induces an injective permutation representation $G \leq S_{11}$. Let $P_{11} \leq G$ be a Sylow 11-subgroup; note that $P_11$ is also Sylow in $S_{11}$. Now $|N_G(P_{11})| = 11 \cdot 3$, while $|N_{S_{11}}(P_{11})| = 11 \cdot 10$; this implies that 3 divides 10, which is absurd. $1575 = 3^2 \cdot 5^2 \cdot 7$ Let $G$ be a simple group of order 1575. Note that $|G|$ does not divide $9!$ since the highest power of 5 dividing $9!$ is $5^1$; thus no proper subgroup of $G$ has index at most 9. Now Sylow’s Theorem forces $n_5(G) = 21 \not\equiv 1$ mod 25, so that by Lemma 13 and this previous exercise, there exist $P_5, Q_5 \in \mathsf{Syl}_5(G)$ such that $P_5$ and $Q_5$ normalize $P_5 \cap Q_5$ and $|N_G(P_5 \cap Q_5)|$ is divisible by $5^2$ and another prime. Thus $|N_G(P_5 \cap Q_5)| \in$ $\{ 5^2 \cdot 3, 5^2 \cdot 7,$ $5^2 \cdot 3^2, 5^2 \cdot 3 \cdot 7,$ $5^2 \cdot 3^2 \cdot 7 \}$. In all but the first case either $P_5 \cap Q_5$ is normal in $G$ or its normalizer has sufficiently small index. Thus $|N_G(P_5 \cap Q_5)| = 5^2 \cdot 3$, and Sylow’s Theorem forces $n_5(N_G(P_5 \cap Q_5)) = 1$. However we know of at least two distinct Sylow 5-subgroups in $N_G(P_5 \cap Q_5)$, namely $P_5$ and $Q_5$. $1755 = 3^3 \cdot 5 \cdot 13$ Done here $1785 = 3 \cdot 5 \cdot 7 \cdot 17$ Example on page 205 $2025 = 3^4 \cdot 5^2$ Done here $2205 = 3^2 \cdot 5 \cdot 7^2$ Done here $2457 = 3^3 \cdot 7 \cdot 13$ Let $G$ be simple of order 2457. Sylow’s Theorem forces $n_7(G) = 351$ and $n_{13}(G) = 27$. Now let $P_{13}$ be a Sylow 13-subgroup of $G$; $|N_G(P_{13})| = 7 \cdot 13$, so that $P_7 \leq N_G(P_{13})$ for some Sylow 7-subgroup $P_7$. Now $P_7P_{13}$ is a subgroup of $G$, and since 7 does not divide 12, $P_7P_{13}$ is abelian. Thus $P_{13} \leq N_G(P_7)$. But we know that $|N_G(P_7)| = 7$, a contradiction. $2475 = 3^2 \cdot 5^2 \cdot 11$ Let $G$ be a simple group of order 2475. Note that $|G|$ does not divide $10!$, so that no proper subgroup of $G$ has index at most 10. Now Sylow’s Theorem forces $n_5(G) = 11 \not\equiv 1$ mod 25, so that by Lemma 13 and this previous exercise there exist $P_5,Q_5 \in \mathsf{Syl}_5(G)$ such that $P_5$ and $Q_5$ normalize $P_5 \cap Q_5$ and $|N_G(P_5 \cap Q_5)|$ is divisible by $5^2$ and some other prime. Thus $|N_G(P_5 \cap Q_5)| \in$ $5^2 \cdot 3,$ $5^2 \cdot 3^2,$ $5^2 \cdot 3 \cdot 11,$ $5^2 \cdot 3^2 \cdot 11 \}$. In the last two cases, either $P_5 \cap Q_5$ is normal in $G$ or its normalizer has sufficiently small index. Thus $|N_G(P_5 \cap Q_5)| \in$ $\{ 5^2 \cdot 3,$ $5^2 \cdot 3^2 \}$. In either case, Sylow’s Theorem forces $n_5(N_G(P_5 \cap Q_5)) = 1$; this is a contradiction since we know of at least two Sylow 5-subgroups in $N_G(P_5 \cap Q_5)$, namely $P_5$ and $Q_5$. $2625 = 3 \cdot 5^3 \cdot 7$ Let $G$ be a simple group of order 2625. Note that $|G|$ does not divide $14!$, so that no proper subgroup of $G$ has index at most 14. Sylow’s Theorem forces $n_5(G) = 21 \not\equiv 1$ mod 25, so that by Lemma 13 and this previous exercise there exist $P_5, Q_5 \in \mathsf{Syl}_5(G)$ such that $|N_G(P_5 \cap Q_5)|$ is divisible by $5^3$ and some other prime. This implies that $[G : N_G(P_5 \cap Q_5)] \in$ $\{ 1,3,7 \}$. In any case we have a contradiction, either because $P_5 \cap Q_5 \leq$ $G$ is normal or some subgroup has sufficiently small index. $2775 = 3 \cdot 5^2 \cdot 37$ Let $G$ be simple of order 2775. Note that $|G|$ does not divide $36!$, so that no proper subgroup of $G$ has index at most 36. Now Sylow’s Theorem forces $n_{37}(G) = 75$, $n_5(G) = 111$, and $n_3(G) \geq 37$. Since the Sylow 13- and 3-subgroups of $G$ are cyclic, $G$ has at least $2 \cdot 37 + 36 \cdot 75$ $+ 25 = 2799$ elements, a contradiction. $2835 = 3^4 \cdot 5 \cdot 7$ Let $G$ be simple of order 2835. Note that $|G|$ does not divide $8!$ since the highest power of 3 which divides $8!$ is $3^2$. On the other hand, Sylow’s Theorem forces $n_3(G) = 7$, so that $G$ has a subgroup of index 7, a contradiction. $2907 = 3^2 \cdot 17 \cdot 19$ Let $G$ be simple of order 2907. Sylow’s Theorem forces $n_{17}(G) = 171$ and $n_{19}(G) = 153$. Since Sylow 17- and 19-subgroups are cyclic, $G$ has at least $18 \cdot 153 + 16 \cdot 171 = 5490$ elements, a contradiction. $3159 = 3^5 \cdot 13$ Done here $3393 = 3^2 \cdot 13 \cdot 29$ Example on page 203 $3465 = 3^2 \cdot 5 \cdot 7 \cdot 11$ Let $G$ be a simple group of order 3465. Sylow’s Theorem forces $n_7(G) \in \{15,99\}$ and $n_{11}(G) = 45$. Let $P_{11} \leq G$ be a Sylow 11-subgroup. Now $|N_G(P_{11})| = 7 \cdot 11$, so that $P_7 \leq N_G(P_{11})$ for some Sylow 7-subgroup P_7. Thus $P_7P_{11} \leq G$ is a subgroup, and moreover because 7 does not divide 10, $P_7P_{11}$ is abelian. Thus $P_{11} \leq N_G(P_7)$, and we have $n_7(G) = 15$. The left action of $G$ on $G/N_G(P_7)$ induces an injective permutation representation $G \leq S_{15}$. Note that $P_{11}$ is Sylow in $S_{11}$. Now $|N_G(P_{11})| = 7 \cdot 11$ while $N_{S_{11}}(P_{11})| = 11 \cdot 10 \cdot 4 \cdot 3 \cdot 2$; this is a contradiction since $N_G(P_{11}) \leq N_{S_{11}}(P_{11})$. $3675 = 3 \cdot 5^2 \cdot 7^2$ Example on page 206 $3875 = 5^3 \cdot 31$ Done here $4095 = 3^2 \cdot 5 \cdot 7 \cdot 13$ Done here $4125 = 3 \cdot 5^3 \cdot 11$ Done here $4389 = 3 \cdot 7 \cdot 11 \cdot 19$ Done here $4455 = 3^4 \cdot 5 \cdot 11$ Let $G$ be a simple group of order 4455. Note that $|G|$ does not divide $10!$; thus no proper subgroup of $G$ has index at most 10. Now Sylow’s Theorem forces $n_3(G) = 55$ and $n_{11}(G) = 45$, and $n_5(G) \in \{11,81,891\}$. Suppose now that $G$ has a subgroup of index 11. Then via the premutation representation induced by left multiplication of the cosets of this subgroup, we have $G \leq S_{11}$, and notably the Sylow 11-subgroups of $G$ are Sylow in $S_{11}$. Now let $P_{11} \leq G$ be a Sylow 11-subgroup. We have $|N_G(P_{11})| = 11 \cdot 3^2$ and $|N_{S_{11}}(P_{11})| = 11 \cdot 10$, but this implies that 9 divides 10 by Lagrange, which is absurd. In particular, $n_5(G) \neq 11$. Suppose now that $n_5(G) = 81$ and let $P_5 \leq G$ be a Sylow 5-subgroup. Then $|N_G(P_5)| = 11 \cdot 5$. By Sylow, we have $n_{11}(N_G(P_5)) = 1$. If $P_{11} \leq N_G(P_5)$ is a Sylow 11-subgroup, it is Sylow in $G$ as well, and we have $P_5 \leq N_G(P_{11})$. However, we know that $|N_G(P_{11})| = 11 \cdot 3^2$, a contradiction by Lagrange. Thus $n_5(G) \neq 81$, and moreover, $n_5(G) = 891$. In particular, note that $G$ contains $10 \cdot 45 + 4 \cdot 891 = 4014$ elements of order 5 or 11; there are 441 elements remaining. Now choose distinct $P_3,Q_3 \in \mathsf{Syl}_3(G)$ such that $|P_3 \cap Q_3|$ is maximal. Now $|P_3 \cap Q_3| \in \{1,3,$ $3^2, 3^3 \}$. We consider each case in turn. If $|P_3 \cap Q_3| = 1$, then all Sylow 3-subgroups of $G$ intersect trivially, and $G$ contains $80 \cdot 55 = 4400$ elements in Sylow 3-subgroups, a contradiction. If $|P_3 \cap Q_3| = 3$, we can carefully fill up some of the 55 Sylow 3-subgroups of $G$ with the remaining elements. Let $A_3 \leq G$ be a Sylow 3-subgroup; note that $|A_3| = 81$. Now let $B_3 \leq G$ be a second Sylow 3-subgroup. Since the largest possible intersection of Sylow 3-subgroups contains 3 elements, there are (at least) $81 - 3 = 78$ elements in $B_3$ which are not also in $A_3$. Now let $C_3 \leq G$ be a third Sylow 3-subgroup; in the worst case, the intersections $C_3 \cap A_3$ and $C_3 \cap B_3$ do not overlap. Thus there are at least $81 - 2 \cdot 3 = 75$ elements in $C_3$ which are not in $A_3$ or $B_3$. Carrying on in this manner, after filling up seven Sylow 3-subgroups, there are at least 504 elements in Sylow 3-subgroups of $G$, a contradiction. If $|P_3 \cap Q_3| = 3^2$, then using this previous exercise, $|N_G(P_3 \cap Q_3)|$ is divisible by $3^3$ and another prime. Thus $|N_G(P_3 \cap Q_3)| \in$ $3^3 \cdot 5,$ $3^3 \cdot 11,$ $3^3 \cdot 5 \cdot 11,$ $3^4 \cdot 5,$ $3^4 \cdot 11,$ $3^4 \cdot 5 \cdot 11 \}$. We can see that in the last four cases, either $P_3 \cap Q_3$ is normal in $G$, or its normalizer has sufficiently small index – either $\leq 10$ or 11. If $|N_G(P_3 \cap Q_3)| = 3^3 \cdot 5$, then by Sylow, $n_5(N_G(P_3 \cap Q_3)) = 1$. Since the Sylow 5-subgroups of $N_G(P_3 \cap Q_3)$ are Sylow in $G$, this yields P_3 \cap Q_3 \leq N_G(P_5)\$ for some Sylow 5-subgroup $P_5 \leq G$. But Sylow’s Theorem forces $|N_G(P_5)| = 5$, a contradiction. Thus we have $|N_G(P_3 \cap Q_3)| = 3^3 \cdot 11$. In this case, Sylow forces $n_{11}(N_G(P_3 \cap Q_3))$ $= n_3(N_G(P_3 \cap Q_3))$ $= 1$. Now let $R_3 \leq N_G(P_3 \cap Q_3)$ be the unique Sylow 3-subgroup and $P_{11}$ the unique Sylow 11-subgroup. In particular, $P_{11}$ normalizes $R_3$, and $P_3 \cap Q_3 < R_3$. Note that, since $n_3(G) = 55$, $P_{11}$ does not normalize $P_3$; thus if $x \in P_{11}$, $P_3 \neq xP_3x^{-1}$. But $P_{11}$ does normalize $R_3$, so that we have $P_3 \cap Q_3 < R_3 \leq P_3 \cap xP_3x^{-1}$. This is a contradiction since $|P_3 \cap Q_3|$ is maximal among the intersections of distinct Sylow 3-subgroups of $G$. If $|P_3 \cap Q_3| = 3^3$, we have $|N_G(P_3 \cap Q_3)| \in$ $\{ 3^4 \cdot 5,$ $3^4 \cdot 11,$ $3^4 \cdot 5 \cdot 11\}$. In each case either $P_3 \cap Q_3$ is normal in $G$, the index of $N_G(P_3 \cap Q_3)$ is too small, or $G$ has a subgroup of index 11, each of which is a contradiction. $4563 = 3^3 \cdot 13^2$ Let $G$ be a simple group of order 4563. Note that $|G|$ does not divide $25!$, so that no proper subgroup of $G$ has index at most 25. Sylow’s Theorem then forces $n_3(G) = 169 \not\equiv 1$ mod 9, so that by Lemma 13 and this previous exercise there exist Sylow 3-subgroups $P_3$ and $Q_3$ in $G$ such that $|N_G(P_3 \cap Q_3)|$ is divisible by $3^3$ and another prime. But then $[G:N_G(P_3 \cap Q_3)] \in \{1,13\}$, which yields either $P_3 \cap Q_3$ normal in $G$ or a subgroup of sufficiently small index. $4725 = 3^3 \cdot 5^2 \cdot 7$ Let $G$ be a simple group of order 4725. Note that $|G|$ does not divide $9!$ since the largest power of 5 dividing $9!$ is $5^1$; thus no proper subgroup of $G$ has index at most 9. Now Sylow’s Theorem forces $n_5(G) = 21 \not\equiv 1$ mod 25, so that by Lemma 13 and this previous exercise, there exist $P_5, Q_5 \in \mathsf{Syl}_5(G)$ such that $P_5$ and $Q_5$ both normalize $P_5 \cap Q_5$ and $|N_G(P_5 \cap Q_5)|$ is divisible by $5^2$ and some other prime. Thus $|N_G(P_5 \cap Q_5)| \in \{ 5^2 \cdot 3, 5^2 \cdot 7,$ $5^2 \cdot 3^2, 5^2 \cdot 3 \cdot 7,$ $5^2 \cdot 3^3, 5^2 \cdot 3^2 \cdot 7,$ $5^2 \cdot 3^3 \cdot 7 \}$. We can see that in all but the first three cases, either $P_5 \cap Q_5$ is normal in $G$ or its normalizer has sufficiently small index. Thus $|N_G(P_5 \cap Q_5)| \in \{5^2 \cdot 3,$ $5^2 \cdot 7, 5^2 \cdot 3^2 \}$. In each case, Sylow’s Theorem forces $n_5(N_G(P_5 \cap Q_5)) = 1$, but this is a contradiction since $P_5,Q_5 \leq N_G(P_5 \cap Q_5)$ are distinct Sylow 5-subgroups. $4851 = 3^2 \cdot 7^2 \cdot 11$ Done here $5103 = 3^6 \cdot 7$ Done here $5145 = 3 \cdot 5 \cdot 7^3$ Done here $5265 = 3^4 \cdot 5 \cdot 13$ Done here $5313 = 3 \cdot 7 \cdot 11 \cdot 23$ Done here $5355 = 3^2 \cdot 5 \cdot 7 \cdot 17$ Let $G$ be a simple group of order 5355. Note that $|G|$ does not divide $16!$, so that no proper subgroup of $G$ has index at most 16. Sylow’s Theorem forces $n_{17}(G) = 35$. Let $P_{17} \leq G$ be a Sylow 17-subgroup; we have $|N_G(P_{17})| = 3^2 \cdot 17$. Let $Q_3 \leq N_G(P_{17})$ be a Sylow 3-subgroup; $Q_3$ is also Sylow in $G$, and $Q_3P_{17} \leq G$ is a subgroup. Moreover, because 3 does not divide 16 and 17 does not divide 8, $Q_3P_{17}$ is abelian. Thus $P_{17} \leq N_G(Q_3)$. However, Sylow’s Theorem and the minimum index condition force $|N_G(Q_3)| \in \{3^2, 3^2 \cdot 7\}$, a contradiction. $5445 = 3^2 \cdot 5 \cdot 11^2$ Let $G$ be a simple group of order 5445. Note that $|G|$ does not divide $21!$ since the largest power of 11 which divides $21!$ is $11^1$; thus no proper subgroup of $G$ has index at most 21. Sylow’s Theorem forces $n_{11}(G) = 45$. Since $45 \not\equiv 1$ mod 121, by Lemma 13 and this previous exercise there exist Sylow 11-subgroups $P_{11}$ and $Q_{11}$ in $G$ such that $|N_G(P_{11} \cap Q_{11})|$ is divisible by $11^2$ and some other prime. Thus the index of $N_G(P_{11} \cap Q_{11})$ in $G$ is either 1, 3, 5, or 15; in each case we have a contradiction, either because $P_{11} \cap Q_{11}$ is normal in $G$ or a subgroup has sufficiently small index. $6075 = 3^5 \cdot 5^2$ Let $G$ be a simple group of order 6075. Note that $|G|$ does not divide $11!$, since the highest power of 3 which divides $11!$ is $3^4$. Thus no proper subgroup of $G$ has index at most 11. Sylow’s Theorem forces $n_3(G) = 25$, and since $25 \not\equiv 1$ mod 9, by Lemma 13 and this previous exercise there exist Sylow 3-subgroups $P_3$ and $Q_3$ in $G$ such that $|N_G(P_3 \cap Q_3)|$ is divisible by $3^5$ and 5. But then either $P_3 \cap Q_3$ is normal in $G$ or $G$ has a subgroup of index 5, a contradiction. $6375 = 3 \cdot 5^3 \cdot 17$ Let $G$ be a simple group of order 6375. Sylow’s Theorem forces $n_{17}(G) = 375$ and $n_{5}(G) = 51$, with $n_3(G) \in \{25,85,2125\}$. If $n_3(G) = 2125$, then since the Sylow 17- and 3-subgroups of $G$ intersect trivially, $G$ contains at least $16 \cdot 375 + 2 \cdot 2125$ $= 10250$ elements, a contradiction. If $n_3(G) = 25$, let $P_3 \leq G$ be a Sylow 3-subgroup. Then $|N_G(P_3)| = 3 \cdot 5 \cdot 17$, so that by Cauchy there is a Sylow 17-subgroup $P_{17}$ of $G$ which normalizes $P_3$. Now $P_3P_{17} \leq G$ is a subgroup and 3 does not divide 16, so that $P_3P_{17}$ is abelian. Thus $P_3 \leq N_G(P_{17})$. But we already know that $|N_G(P_{17})| = 17$, a contradiction. Thus $n_3(G) = 85$. Now $G$ contains $16 \cdot 375 = 6000$ elements of order 17 and $2 \cdot 85 = 170$ elements of order 3, so that there are at most $205$ elements of order a power of 5. Let $P_5, Q_5 \leq G$ be Sylow 5-subgroups. The largest possible intersection of $P_5$ and $Q_5$ contains 25 elements, so that $|P_5 \cup Q_5| \geq 225$, a contradiction. $6435 = 3^2 \cdot 5 \cdot 11 \cdot 13$ Done here $6545 = 5 \cdot 7 \cdot 11 \cdot 17$ Done here $6615 = 3^3 \cdot 5 \cdot 7^2$ Let $G$ be a simple group of order 6615. Note that $|G|$ does not divide $8!$ since the highest power of 3 dividing $8!$ is $3^2$. Thus no proper subgroup of $G$ has index at most 8; this and Sylow’s Theorem force $n_3(G) = 49$. Now let $P_3 \leq G$ be a Sylow 3-subgroup. We have $|N_G(P_3)| = 3^3 \cdot 5$. By Sylow again, $n_5(N_G(P_3)) = 1$; let $P_5 \leq N_G(P_3)$ be the unique Sylow 5-subgroup; note that $P_5$ is Sylow in $G$ as well. Now $P_5,P_3 \leq N_G(P_3)$ are both normal and intersect trivially, and $P_3P_5 = N_G(P_3)$ by Lagrange. Thus $N_G(P_3) \cong P_3 \times P_5$ by the recognition theorem for direct products. In particular, $P_3 \leq N_{N_G(P_3)}(P_5) \leq N_G(P_5)$. However, Sylow also forces $|N_G(P_5)| \in \{ 3^2 \cdot 5 \cdot 7, 3 \cdot 5 \}$, a contradiction by Lagrange. $6669 = 3^3 \cdot 13 \cdot 19$ Done here $6825 = 3 \cdot 5^2 \cdot 7 \cdot 13$ Let $G$ be a simple group of order 6825. Sylow’s Theorem forces $n_7(G) = 15$ and $n_{13}(G) = 105$. Let $P_7 \leq G$ be a Sylow 7-subgroup; we have $|N_G(P_7)| = 5 \cdot 7 \cdot 13$, so that by Cauchy, $P_{13} \leq N_G(P_7)$ for some Sylow 13-subgroup $P_{13}$. Now $P_7P_{13} \leq G$ is a subgroup, and since 7 does not divide 12, $P_7P_{13}$ is abelian. Thus $P_7 \leq N_G(P_{13})$; but we already know that $|N_G(P_{13})| = 5 \cdot 13$, a contradiction. $7371 = 3^4 \cdot 7 \cdot 13$ Let $G$ be a simple group of order 7371. Sylow’s Theorem forces $n_7(G) = 351$ and $n_{13}(G) = 27$. Let $P_{13} \leq G$ be a Sylow 13-subgroup. Now $|N_G(P_{13})| = 3 \cdot 7 \cdot 13$. By Cauchy, some Sylow 7-subgroup $P_7 \leq G$ normalizes $P_{13}$. Thus $P_7P_{13} \leq G$ is a subgroup, and moreover because 7 does not divide 12, $P_7P_{13}$ is abelian. Thus $P_{13} \leq N_G(P_7)$. But we already know that $|N_G(P_7)| = 3 \cdot 7$, a contradiction. $7425 = 3^3 \cdot 5^2 \cdot 11$ Let $G$ be a simple group of order 7425. Note that $|G|$ does not divide $10!$, so that no proper subgroup of $G$ has index at most 10. Sylow’s Theorem forces $n_5(G) = 11 \not\equiv 1$ mod 25, so that by Lemma 13 and this previous exercise there exist $P_5,Q_5 \in \mathsf{Syl}_5(G)$ such that $P_5$ and $Q_5$ normalize $P_5 \cap Q_5$ and $|N_G(P_5 \cap Q_5)|$ is divisible by $5^2$ and another prime. Thus $|N_G(P_5 \cap Q_5)| \in \{ 5^2 \cdot 3,$ $5^2 \cdot 3^2, 5^2 \cdot 3^3,$ $5^2 \cdot 11, 5^2 \cdot 3 \cdot 11,$ $5^2 \cdot 3^2 \cdot 11,$ $5^2 \cdot 3^3 \cdot 11 \}$. We can see that in the last three cases, either $P_5 \cap Q_5$ is normal in $G$ or its normalizer has sufficiently small index. Moreover, in the first three cases, $n_5(N_G(P_5 \cap Q_5)) = 1$ by Sylow, a contradiction since $P_5,Q_5 \leq N_G(P_5 \cap Q_5)$ are distinct Sylow 5-subgroups. Thus $|N_G(P_5 \cap Q_5)| = 5^2 \cdot 11$. If $n_5(N_G(P_5 \cap Q_5)) = 1$, then we have a contradiction since $P_5,Q_5 \leq N_G(P_5 \cap Q_5)$ are Sylow, thus Sylow’s Theorem forces $n_5(N_G(P_5 \cap Q_5)) = 11$. Since Sylow 5-subgroups of $N_G(P_5 \cap Q_5)$ are Sylow in $G$, and $G$ has 11 Sylow 5-subgroups, every Sylow 5-subgroup of $G$ normalizes $P_5 \cap Q_5$. We now consider the subgroup $H = \langle \bigcup \mathsf{Syl}_5(G) \rangle$. Note that $|H|$ is divisible by $5^2$ and by another prime since $H$ contains more than one Sylow 5-subgroup. We saw previously that there are seven possible orders for $H$, taking into account the necessary value $n_5(H) = 11$ and the minimal index condition. Thus $|H| \in \{ 5^2 \cdot 11, 5^2 \cdot 3^3 \cdot 11 \}$. Note that if $|H| = 5^2 \cdot 11$, then $n_{11}(H) = 1$ by Sylow. Moreover, the unique Sylow 11-subgroup $P_{11} \leq H$ is also Sylow in $G$. Thus (for instance) the Sylow 5-subgroup $P_5 \leq H$ $\leq G$ normalizes $P_{11}$. However, since $n_{11}(G) = 45$, this is absurd. Thus $|H| = 3^3 \cdot 5^2 \cdot 11$, and in fact $H = G$. In particular, $G$ is generated by its Sylow 5-subgroups. But then $N_G(P_5 \cap Q_5) = G$, since every element of $G$ can be written as a product of elements in Sylow 5-subgroups and every elements of a Sylow 5-subgroup normalizes $P_5 \cap Q_5$. This is a contradiction becasue $P_5 \cap Q_5 \leq G$ is a proper nontrivial subgroup. $7875 = 3^2 \cdot 5^3 \cdot 7$ Let $G$ be a simple group of order 7875. Note that $|G|$ does not divide $14!$, since the largest power of 5 dividing $14!$ is $5^2$. Thus no proper subgroup of $G$ has index at most 14. Now Sylow’s Theorem forces $n_5(G) = 21 \not\equiv 1$ mod 25, so that by Lemma 13 and this previous exercise there exist $P_5,Q_5 \in \mathsf{Syl}_5(G)$ such that $P_5$ and $Q_5$ normalize $P_5 \cap Q_5$ and $|N_G(P_5 \cap Q_5)|$ is divisible by $5^2$ and another prime. Thus we have $|N_G(P_5 \cap Q_5)| \in \{5^3 \cdot 3, 5^3 \cdot 7,$ $5^3 \cdot 3^2, 5^3 \cdot 3 \cdot 7,$ $5^3 \cdot 3^2 \cdot 7 \}$. We can see that in every case except $|N_G(P_5 \cap Q_5)| = 5^3 \cdot 3$, either $P_5 \cap Q_5$ is normal in $G$ or its normalizer has sufficiently small index. Now $n_5(N_G(P_5 \cap Q_5)) = 1$ by the congruence and divisibility criteria of Sylow’s Theorem, but we know that $N_G(P_5 \cap Q_5)$ has at least two Sylow 5-subgroups- namely $P_5$ and $Q_5$. $8325 = 3^2 \cdot 5^2 \cdot 37$ Let $G$ be a simple group of order 8325. Note that $|G|$ does not divide $36!$, so that no proper subgroup of $G$ has index at most 36. Now Sylow’s Theorem forces $n_5(G) = 111 \not\equiv 1$ mod 25. Then by Lemma 13 and this previous exercise, there exist $P_5,Q_5 \in \mathsf{Syl}_5(G)$ such that $P_5$ and $Q_5$ normalize $P_5 \cap Q_5$ and $|N_G(P_5 \cap Q_5)|$ is divisible by $5^2$ and some other prime. Thus $|N_G(P_5 \cap Q_5)| \in \{ 5^2 \cdot 3, 5^2 \cdot 3^2,$ $5^2 \cdot 37, 5^2 \cdot 3 \cdot 37$, $5^2 \cdot 3^2 \cdot 37 \}$. We can see that in all except the first two cases, either $P_5 \cap Q_5$ is normal in $G$ or its normalizer has sufficiently small index. In either of the remaining cases, $|N_G(P_5 \cap Q_5)| \in \{ 5^2 \cdot 3, 5^2 \cdot 3^2\}$, we have $n_5(N_G(P_5 \cap Q_5)) = 1$; this is a contradiction since we know of at least two Sylow 5-subgroups in $N_G(P_5 \cap Q_5)$, namely $P_5$ and $Q_5$. $8505 = 3^5 \cdot 5 \cdot 7$ Let $G$ be a simple group of order 8505. Note that $|G|$ does not divide $11!$ since the largest power of 3 which divides $11!$ is $3^4$; thus $G$ has no proper subgroups of index at most 11. Now Sylow’s Theorem forces $n_3(G) = 7$; since $7 \not\equiv 1$ mod 9, by Lemma 13 and this previous exercise there exist Sylow 3-subgroups $P_3,Q_3 \leq G$ such that $|N_G(P_3 \cap Q_3)|$ is divisible by $3^5$ and some other prime; there are three cases to consider, and we either have $P_3 \cap Q_3 \leq G$ normal, $G$ has a subgroup of index 5, or $G$ has a subgroup of index 7, all of which are contradictions. $8721 = 3^3 \cdot 17 \cdot 19$ Let $G$ be a simple group of order 8721. Sylow’s Theorem forces $n_3(G) = 19$; if $P_3$ is a Sylow 3-subgroup we have an injective permutation representation $G \leq S_{19}$ induced by the left action of $G$ on $G/N_G(P_3)$. Let $P_{17} \leq G$ be a Sylow 17-subgroup; note that $P_{17}$ is also Sylow in $S_{19}$. Now $|N_G(P_{17})| = 17 \cdot 3$, while $N_{S_{19}}(P_{17})| = 19 \cdot 18 \cdot 2$. This yields a contradiction. $8775 = 3^3 \cdot 5^2 \cdot 13$ Let $G$ be a simple group of order 8775. Sylow’s Theorem forces $n_5(G) = 351$ and $n_{13}(G) = 27$. Let $P_{13} \leq G$ be a Sylow 13-subgroup. Now $|N_G(P_{13})| = 5^2 \cdot 13$. By Sylow’s Theorem, there exists a Sylow 5-subgroup $Q_5 \leq N_G(P_{13})|$, and by cardinality considerations $Q_5$ is Sylow in $G$ as well. Thus $P_{13}Q_5 \leq G$ is a subgroup, and moreover because 5 does not divide 12 and 13 does not divide 24, $P_{13}Q_5$ is abelian. Thus $P_{13} \leq N_G(Q_5)$. However, we already know that $|N_G(Q_5)| = 5^2$, a contradiction. $8883 = 3^3 \cdot 7 \cdot 47$ Let $G$ be simple of order 8883. Sylow’s Theorem forces $n_7(G) = 141$ and $n_{47}(G) = 189$. Since Sylow 7- and 47-subgroups of $G$ intersect trivially, $G$ contains at least $6 \cdot 141 + 46 \cdot 189 = 9540$ elements, a contradiction. $8925 = 3 \cdot 5^2 \cdot 7 \cdot 17$ Let $G$ be a simple group of order 8925. Note that Sylow’s Theorem forces $n_{17}(G) = 35$, $n_7(G) \in \{15,85,1275\}$, and $n_5(G) \in \{21,51\}$. Let $P_5 \leq G$ be a Sylow 5-subgroup. If $n_5(G) = 21$, then $|N_G(P_5)| = 5^2 \cdot 17$. By Cauchy, we have $P_{17} \leq N_G(P_5)$ for some Sylow 17-subgroup $P_{17}$. Thus $P_5P_{17} \leq G$ is a subgroup. Now 5 does not divide 16 and 17 does not divide 24, so that $P_5P_{17}$ is abelian. Thus $P_5 \leq N_G(P_{17})$. But we know that $|N_G(P_{17})| = 3 \cdot 5 \cdot 17$, a contradiction. If $n_5(G) = 51$, then $|N_G(P_5)| = 5^2 \cdot 7$. By Cauchy, we have $P_7 \leq N_G(P_5)$ for some Sylow 7-subgroup $P_7$. Thus $P_5P_7 \leq G$ is a subgroup. Again because 5 does not divide 6 and 7 does not divide 24, $P_5P_7$ is abelian, hence $P_5 \leq N_G(P_7)$. However, we know that $|N_G(P_7)| \in \{7, 5 \cdot 7 \cdot 17, 3 \cdot 5 \cdot 7\}$, a contradiction. $9045 = 3^3 \cdot 5 \cdot 67$ Let $G$ be simple of order 9045. Sylow’s Theorem forces $n_5(G) = 201$ and $n_{67}(G) = 135$. Since the Sylow 5- and 67-subgroups of $G$ intersect trivially, $G$ contains at least $4 \cdot 201 + 66 \cdot 135$ $= 9714$ elements, a contradiction. $9405 = 3^2 \cdot 5 \cdot 11 \cdot 19$ Let $G$ be a simple group of order 9405. Sylow’s Theorem forces $n_{11}(G) = 45$ and $n_{19}(G) = 495$. Let $P_{11} \leq G$ be a Sylow 11-subgroup. Now $|N_G(P_{11})| = 11 \cdot 19$. By Cauchy, there exists a Sylow 19-subgroup $P_{19} \leq G$ which normalizes $P_{11}$; now $P_{11}P_{19} \leq G$ is a subgroup, and because 11 does not divide 18, $P_{11}P_{19}$ is abelian. Thus $P_{11} \leq N_G(P_{19})$. But we already know that $|N_G(P_{19})| = 19$, a contradiction. $9477 = 3^6 \cdot 13$ Let $G$ be a simple group of order 9477. Sylow’s Theorem forces $n_3(G) = 13$. Since $13 \not\equiv 1$ mod 9, Lemma 13 and this previous exercise imply that there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $P_3 \cap Q_3$ is maximal in $P_3$ (hence nontrivial) and $|N_G(P_3 \cap Q_3)|$ is divisible by $3^6$ and 13. But then $P_3 \cap Q_3$ is normal in $G$, a contradiction. $9555 = 3 \cdot 5 \cdot 7^2 \cdot 13$ Done here $9765 = 3^2 \cdot 5 \cdot 7 \cdot 31$ Let $G$ be a simple group of order 9765. Note that $|G|$ does not divide $30!$, so that no proper subgroup of $G$ has index at most 30. This and Sylow’s Theorem force $n_3(G) \in \{31,217\}$. In either case, 5 divides $|N_G(P_3)|$, where $P_3 \leq G$ is some Sylow 3-subgroup. By Cauchy, there exists a Sylow 5-subgroup $P_5 \leq N_G(P_3)$. Now $P_3P_5 \leq G$ is a subgroup, and because 3 does not divide 4 and 5 does not divide 8, $P_3P_5$ is abelian. Thus $P_3 \leq N_G(P_5)$. This, Sylow’s Theorem, and our observation on minimal subgroup indices imply that $n_5(G) = 31$. Thus $|N_G(P_5)| = 3^2 \cdot 5 \cdot 7$, and by Cauchy we have $P_7 \leq N_G(P_5)$ for some Sylow 7-subgroup $P_7 \leq G$. Now $P_5P_7 \leq G$ is a subgroup and since 5 does not divide 6, $P_5P_7$ is abelian. Thus $P_5 \leq N_G(P_7)$; but we already know that $|N_G(P_7)| = 7 \cdot 3^2$, so this is a contradiction.

### No simple groups of order 144, 525, 2025, or 3159 exist

Prove that there are no simple groups of order 144, 525, 2025, or 3159.

1. Note that $144 = 2^4 \cdot 3^2$. Let $G$ be a simple group of order 144. By Sylow’s Theorem, we have $n_2 \in \{1,3,9\}$ and $n_3 \in \{1,4,16\}$. Note that $|G|$ does not divide $5!$ since the largest power of 2 which divides $5!$ is $2^3$. Thus no proper subgroup of $G$ has index less than or equal to 5; in particular, $n_2(G) = 9$ and $n_3(G) = 16$. Note that $n_3 \not\equiv 1$ mod 9; by Lemma 13, there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $P_3 \cap Q_3$ is normal in $P_3$ and $Q_3$. In particular, $3^2$ divides $|N_G(P_3 \cap Q_3)|$, and since $N_G(P_3 \cap Q_3)$ contains more than one Sylow 3-subgroup (namely $P_3$ and $Q_3$) its order is divisible by 2. In particular, $[G : N_G(P_3 \cap Q_3)] \in \{1,2,4,8\}$. Since $G$ is simple and has no proper subgroups of index less than 6, in fact $[G : N_G(P_3 \cap Q_3)] = 8$, and we have $|N_G(P_3 \cap Q_3)| = 2 \cdot 3^2$.

However, note that this implies that $n_3(N_G(P_3 \cap Q_3)) = 1$, by Sylow’s Theorem, but that we know $N_G(P_3 \cap Q_3)$ contains at least two Sylow 3-subgroups. Thus we have a contradiction.

2. Note that $525 = 3 \cdot 5^2 \cdot 7$. Let $G$ be a simple group of order 525. By Sylow’s Theorem, we have $n_3 \in \{1,7,25,175\}$, $n_5 \in \{1,21\}$, and $n_7 \in \{1,15\}$. Note that $|G|$ does not divide $9!$ since the highest power of 5 dividing $9!$ is $5^1$. Thus $G$ has no proper subgroups of index at most 9. Moreover, $n_5(G) = 21 \not\equiv 1$ mod 25, so that by a previous exercise and Lemma 13, there exist $P_5, Q_5 \in \mathsf{Syl}_5(G)$ such that $P_5 \cap Q_5 \neq 1$ and $N_G(P_5 \cap Q_5)$ is divisible by $5^2$ and some other prime.

If $|N_G(P_5 \cap Q_5)| \in \{ 3 \cdot 5^2, 7 \cdot 5^2\}$, then $G$ has a proper subgroup of index less than 9, a contradiction. If $|N_G(P_5 \cap Q_5)| = 3 \cdot 5^2 \cdot 7$, then $P_5 \cap Q_5$ is normal in $G$, a contradiction.

3. Note that $2025 = 3^4 \cdot 5^2$. Let $G$ be a simple group of order 2025. By Sylow’s Theorem, we have $n_3(G) = 25$ and $n_5(G) = 81$. Note that $|G|$ does not divide $9!$, since the highest power of 5 dividing $9!$ is $5^1$.

Now since $n_3(G) = 25 \not\equiv 1$ mod 9, by Lemma 13 and a previous exercise there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $|N_G(P_3 \cap Q_3)|$ is divisible by $3^4$ and 5. But then $N_G(P_3 \cap Q_3)$ is either all of $G$ or has index 5; in either case we have a contradiction.

4. Note that $3159 = 3^5 \cdot 13$. Let $G$ be a simple group of order 3159. By Sylow’s Theorem we have $n_3(G) = 13$ and $n_{13}(G) = 27$. Note that $n_3(G) = 13 \not\equiv 1$ mod 9; thus by Lemma 13 and a previous exercise, there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $|N_G(P_3 \cap Q_3)|$ is divisible by $3^5$ and 13- thus $N_G(P_3 \cap Q_3) = G$, a contradiction.

### No simple groups of order 9555 exist

Prove that there are no simple groups of order 9555.

Note that $9555 = 3 \cdot 5 \cdot 7^2 \cdot 13$. Suppose $G$ is a simple group of order 9555. By Sylow’s Theorem, we have $n_7(G) = 15$ and $n_{13}(G) = 105$. Let $P_7 \in \mathsf{Syl}_7(G)$. We compute that $|N_G(P_7)| = 7^2 \cdot 13$. By Cauchy, there exists a Sylow 13-subgroup (of $G$) $P_{13} \leq N_G(P_7)$. Now $P_7P_{13} \leq G$ is a subgroup. Moreover, since 7 does not divide 12 and 13 does not divide 48, $P_7P_{13}$ is abelian. Thus in fact $P_{7} \leq N_G(P_{13})$. However, we also know that $|N_G(P_{13})| = 7 \cdot 13$, a contradiction since $|P_7| = 49$.

### No simple groups of order 4851 or 5145 exist

Prove that there are no simple groups of order 4851 or 5145.

1. Note that $4851 = 3^2 \cdot 7^2 \cdot 11$. Suppose $G$ is a simple group of order 4851. By Sylow’s Theorem, we have $n_3(G) \in \{7,49\}$ and $n_{11}(G) = 441$. Note that $|G|$ does not divide $10!$, so that no subgroup of $G$ has index less than or equal to 10. In particular, $n_3(G) = [G:N_G(P_3)] \neq 7$, so that $n_3(G) = 49$. Let $P_3 \in \mathsf{Syl}_3(G)$. We have $|N_G(P_3)| = 3^2 \cdot 11$. By Cauchy, there exists a Sylow 11-subgroup $P_{11} \leq N_G(P_3)$. Now $P_3P_{11} \leq G$ is a subgroup, and since 3 does not divide 10 and 11 does not divide 8, by this previous exercise, $P_3P_{11}$ is abelian. Thus we have $P_3 \leq N_G(P_{11})$. But we also know that $|N_G(P_{11})| = 11$, a contradiction.
2. Note that $5145 = 3 \cdot 5 \cdot 7^3$. Suppose $G$ is a simple group of order 5145. Note that $|G|$ does not divide $20!$ since the largest power of 7 dividing $20!$ is $7^2$. Thus $G$ has no proper subgroups of index less than or equal to 20. In particular, from Sylow’s Theorem we have $n_7(G) = 1$, a contradiction.

### No simple groups of order 4095, 4389, 5313, or 6669 exist

Prove that there are no simple groups of order 4095, 4389, 5313, or 6669.

1. Note that $4095 = 3^2 \cdot 5 \cdot 7 \cdot 13$. Suppose $G$ is a simple group of order 4095. By Sylow’s Theorem, we have $n_7(G) = 15$ and $n_{13}(G) = 105$. Let $P_7 \in \mathsf{Syl}_7(G)$. Now $[G:N_G(P_7)] = 15$, so that $|N_G(P_7)| = 3 \cdot 7 \cdot 13$. By Cauchy, there exists a Sylow 13-subgroup $P_{13}$ with $P_{13} \leq N_G(P_7)$. Thus $P_7P_{13} \leq G$ is a subgroup. Moreover, since 7 does not divide 12, $P_7P_{13}$ is abelian by this previous exercise. Thus in fact we have $P_7 \leq N_G(P_{13})$. However, from $n_{13}(G) = 105$ we deduce that $|N_G(P_{13})| = 13 \cdot 3$, a contradiction by Lagrange.
2. Note that $4389 = 3 \cdot 7 \cdot 11 \cdot 19$. Let $G$ be a simple group of order 4389. By Sylow’s Theorem we have $n_{11}(G) = 133$ and $n_{7}(G) = 57$. Let $P_7 \in \mathsf{Syl}_7(G)$. We have $|N_G(P_7)| = 7 \cdot 11$, so that by Cauchy there exists a Sylow 11-subgroup $P_{11} \leq N_G(P_7)$. Now $P_7P_{11} \leq G$ is a subgroup, and since 7 does not divide 10, $P_7P_{11}$ is abelian. Thus $P_7 \leq N_G(P_{11})$. But we also have $|N_G(P_{11})| = 3 \cdot 11$, a contradiction.
3. Note that $5313 = 3 \cdot 7 \cdot 11 \cdot 23$. Suppose $G$ is a simple group of order 5313. By Sylow’s Theorem, we have $n_{23}(G) = 231$, $n_{11}(G) = 23$, and $n_7(G) = 253$. Since all Sylow subgroups intersect trivially, $G$ contains $22 \cdot 231 = 5082$ elements of order 23, $10 \cdot 23 = 230$ elements of order 11, and $6 \cdot 253 = 1518$ elements of order 7, for a total of at least $5082 + 230 + 1518 = 6830$ elements, a contradiction.
4. Note that $6669 = 3^3 \cdot 13 \cdot 19$. Suppose $G$ is a simple group of order 6669. By Sylow’s Theorem we have $n_{13}(G) = 27$ and $n_{19}(G) = 39$. Now $|N_G(P_{13})| = 3 \cdot 13 \cdot 19$, so that by Cauchy there exists a Sylow 19-subgroup $P_{19} \leq N_G(P_{13})$. Now $P_{13}P_{19} \leq G$ is a subgroup, and since 13 does not divide 18, $P_{13}P_{19}$ is abelian. Thus $P_{13} \leq N_G(P_{19})$. But we also have $|N_G(P_{19})| = 3 \cdot 19$, a contradiction.