## Tag Archives: finite group

### Finite abelian groups are not naturally QQ-modules

Every finite abelian group $M$ is naturally a $\mathbb{Z}$-module. Can this ring action be extended to make $M$ a $\mathbb{Q}$-module?

Let $M$ be a (multiplicative) finite abelian group of order $k$. Suppose further that $M$ is a $\mathbb{Q}$ module and that the action of $\mathbb{Q}$ on $M$ extends the natural action of $\mathbb{Z}$ given by $k \cdot m = m^k$. Let $x \in M$ be any nonidentity element. Now $\frac{1}{k} \cdot x = y$ for some $y \in M$. then $k \cdot (\frac{1}{k} \cdot x) = k \cdot y$, and so $x = y^k = 1$, a contradiction. Thus no such module structure exists.

### Prove that the augmentation ideal of a given group ring is nilpotent

Let $p$ be a prime and let $G$ be a finite $p$-group. Prove that the augmentation ideal in the group ring $R = (\mathbb{Z}/(p))[G]$ is a nilpotent ideal. (Note that this ring may be noncommutative.)

[With a hint from this forum post and from the book Group Rings and their Augmentation Ideals by Passi.]

We begin with some lemmas.

Lemma 1: Let $\pi : G \rightarrow H$ be a group homomorphism, and let $R$ be a ring. Then the mapping $\varphi : R[G] \rightarrow R[H]$ given by $\varphi(\sum r_i g_i) = \sum r_i \pi(g_i)$ is a ring homomorphism. Moreover, if $\pi$ is surjective, $\varphi$ is surjective. Finally, $\mathsf{ker}\ \varphi = \mathsf{Aug}(R[H]) R[G] = R[G] \mathsf{Aug}(R[H])$. Proof: It is clear that if $\pi$ is surjective then $\varphi$ is surjective. Now If $a = \sum r_i g_i$ and $b = \sum s_i g_i$ are in $R[G]$, we have $\varphi(a+b) = \varphi((\sum r_ig_i)+(\sum s_ig_i))$ $= \varphi(\sum (r_i+s_i)g_i)$ $= \sum (r_i+s_i)\pi(g_i)$ $= (\sum r_i \pi(g_i)) + (\sum s_i \pi(g_i))$ $= \varphi(a) + \varphi(b)$. Similarly, $\varphi(ab) = \varphi((\sum r_i g_i)(\sum s_i g_i))$ $= \varphi(\sum_i \sum_j r_is_j g_ig_j)$ $= \sum_i \sum_j r_is_j \pi(g_ig_j)$ $= \sum_i \sum_j r_is_j \pi(g_i)\pi(g_j)$ $= (\sum r_i \pi(g_i))(\sum s_i\pi(g_i))$ $= \varphi(a)\varphi(b)$. Thus $\varphi$ is a ring homomorphism. Now we show that $\mathsf{ker}\ \varphi = \mathsf{Aug}(R[H]) R[G]$; the other equality is similar since $H$ is normal. $(\subseteq)$ Note that we can write every element of $G$ in the form $g = hk$ where $h \in H$; write $H = \{h_i\}$ and let $K = \{k_j\}$ be a set of coset representatives of $G/H$, and let $g_{i,j} = h_ik_j$. Suppose $\sum r_{i,j}g_{i,j} \in \mathsf{ker}\ \varphi$; then we have $\sum_{j} (\sum_{i} r_{i,j}) k_jH = 0$; comparing coefficients, we see that $\sum_{i} r_{i,j} = 0$ for each $j$. Then $\sum_{i,j} r_{i,j} g_{i,j} = \sum r_{i,j} h_i k_j$ $\sum_j (\sum_i r_{i,j} h_i) k_j$ $\in \mathsf{Aug}(R[H]) R[G]$. $(\supseteq)$ Note that if $\sum r_i h_i \in \mathsf{Aug}(R[H])$, then $\varphi(\sum r_i h_i) = \sum r_i \cdot 1$ $= (\sum r_i) \cdot 1$ $= 0$. Clearly then this ideal is contained in the kernel. $\square$

Lemma 2: Let $\varphi : R \rightarrow S$ be a surjective ring homomorphism and let $A \subseteq R$. Then $\varphi[(A)] \subseteq (\varphi[A])$. Proof: Elements of $\varphi[(A)]$ have the form $\varphi(\sum r_ia_is_i) = \sum \varphi(r_i)\varphi(a_i)\varphi(s_i)$ where $r_i,s_i \in R$ and $a_i \in A$. $\square$

Lemma 3: Let $R$ be a ring and let $I,J \subseteq R$ be ideals. If $I \subseteq Z(R)$, then $IJ = JI$ and $(IJ)^n = I^nJ^n$. Proof: Arbitrary elements of $IJ$ have the form $\sum a_ib_i$ where $a_i \in I$ and $b_i \in J$; since $I \subseteq Z(R)$, we have $\sum a_ib_i = \sum b_ia_i \in JI$. Similarly, $JI \subseteq IJ$. Thus $IJ = JI$. The second conclusion then follows. $\square$

Now we move to the main result.

Since $G$ is a finite $p$-group, $|G| = p^n$ for some $n \geq 1$. We proceed by induction on $n$. We will also provide a bound on the exponent $m$ such that $\mathsf{Aug}(R)^m = 0$; namely, $p^n$.

For the base case, let $|G| = p$. Then $G = \langle \alpha \rangle \cong Z_p$ is cyclic of order $p$. Moreover, $R$ has characteristic $p$ and is commutative. By this previous exercise, $\mathsf{Aug}(R)$ is generated by $\alpha - 1$. By this previous exercise, we have $(\alpha - 1)^p = \alpha^p - 1^p$ $= 1-1 = 0$. Thus $\mathsf{Aug}(R)$ is nilpotent.

For the inductive step, suppose that for some $m \geq 1$, for all groups $H$ of order $p^m$, the augmentation ideal of $(\mathbb{Z}/(p))[H]$ is a nilpotent ideal. Let $G$ be a group of order $p^{m+1}$. Since $G$ is a $p$-group, it has a nontrivial center. By Cauchy’s Theorem, let $x \in Z(G)$ have order $p$. Now $H = \langle x \rangle$ is normal in $G$. As in the lemma above, let $\varphi : (\mathbb{Z}/(p))[G] \rightarrow (\mathbb{Z}/(p))[G/H]$ be the ring homomorphism obtained by letting $\pi$ be the natural projection $G \rightarrow G/H$.

Note that because the augmentation ideal of a group ring is generated by $\{g_i-1\}$ and $\pi$ is surjective, we have $\varphi[\mathsf{Aug}((\mathbb{Z}/(p))[G])] = \mathsf{Aug}((\mathbb{Z}/(p))[G/H])$ By the induction hypothesis, the augmentation ideal of $R[G/H]$ is nilpotent of exponent at most $p^m$. Thus we have $\mathsf{Aug}((\mathbb{Z}/(p))[G])^{p^n} \subseteq \mathsf{ker}\ \varphi$ $= \mathsf{Aug}((\mathbb{Z}/(p))[H]) (\mathbb{Z}/(p))[R]$, so that by the base case, $\mathsf{Aug}((\mathbb{Z}/(p))[G])^{p^{n+1}} \subseteq (\mathsf{Aug}((\mathbb{Z}/(p))[H]) (\mathbb{Z}/(p))[R])^p$ $= \mathsf{Aug}((\mathbb{Z}/(p))[H])^p (\mathbb{Z}/(p))[R]^p$ $= 0$. Thus the augmentation ideal of $R[G]$ is nilpotent with exponent at most $p+1$.

### Find a generating set for the augmentation ideal of a group ring

Let $R$ be a commutative ring with $1 \neq 0$ and $G$ a finite group. Prove that the augmentation ideal in the group ring $R[G]$ is generated by $\{ g-1 \ |\ g \in G \}$. Prove that if $G = \langle \sigma \rangle$ is cyclic then the augmentation ideal is generated by $\sigma - 1$.

Recall that the augmentation ideal of $R[G]$ is the kernel of the ring homomorphism $R[G] \rightarrow R$ given by $\sum r_ig_i \mapsto \sum r_i$; that is, it consists of all elements in $R[G]$ whose coefficients sum to 0 in $R$.

Let $S = \{ g-1 \ |\ g \in G \}$, and let $A$ denote the augmentation ideal of $R[G]$. First, note that $g-1 \in A$ for each $g \in G$, so that $(g-1) \subseteq A$. Thus $(S) \subseteq A$. Now let $\alpha = \sum_{g_i \in G} r_ig_i \in A$; then we have $\sum r_i = 0$. Consider the following.

 $\sum_{g_i \in G} r_i(g_i - 1)$ = $\sum_{g_i \in G} r_ig_i - r_i$ = $= \left( \sum_{g_i \in G} r_ig_i \right) - (\sum_{g_i \in G} r_i)$ = $\alpha - 0$ = $\alpha$

Thus $A \subseteq (S)$, and we have $A = (S)$.

Suppose further that $G = \langle \sigma \rangle$ is cyclic. We still have $(\sigma - 1) \subseteq A$. Now note that $(\sigma - 1)\left( \sum_{t=0}^k \sigma^t \right) = \sigma^{k+1} - 1$, so that $\sigma^k - 1 \in (\sigma - 1)$ for all $k$. Thus $A = (\sigma - 1)$.

### 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.

### 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 $n_3(G) = 7 \cdot 409$.

### Construct a nonsimple group of order 168 with more than one Sylow 7-subgroup

Prove or construct a counterexample to the assertion: if $G$ is a group of order 168 with more than one Sylow 7-subgroup, then $G$ is simple.

Recall that $GL_3(\mathbb{F}_2)$ has order $7 \cdot 6 \cdot 4$, and thus by Cauchy there exists a nontrivial group homomorphism $\varphi : Z_7 \rightarrow \mathsf{Aut}(Z_2^3)$. Then $Z_3 \times (Z_2^3 \rtimes_\varphi Z_7)$ has order 168, is not simple, and does not have a normal Sylow 7-subgroup by Proposition 5.11 and this previous exercise.

### 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 $|G| = 60$.

### A finite group whose order is the product of three distinct primes has a normal Sylow subgroup of largest order

Let $G$ be a group of order $pqr$ where $p < q < r$ and $p$, $q$, and $r$ are primes. Prove that a Sylow $r$-subgroup of $G$ is normal.

Recall that some Sylow subgroup of $G$ is normal. Let $P$, $Q$, and $R$ denote Sylow $p$-, $q$-, and $r$-subgroups of $G$, respectively.

Suppose $n_p(G) = 1$. Note that $n_r(G/P) \in \{1,q\}$, and since $q < r$, $n_r(G/P) = 1$. Let (via the lattice isomorphism theorem) $\overline{R} \leq G$ be the subgroup whose quotient is the unique Sylow $r$-subgroup in $G/P$; we have $|\overline{R}| = pr$, and $P \leq \overline{R}$ is normal. Moreover, because $p < r$, $\overline{R}$ has a unique Sylow $r$-subgroup $R$, and we have $\overline{R} \cong P \times R$. Now if $R^\prime \leq G$ is a Sylow $r$-subgroup, then $R^\prime P/P = \overline{R}/P$, so that $\overline{R} = R^\prime P$. Since $R^\prime \leq \overline{R}$ is Sylow, $R^\prime = R$. Thus $n_r(G) = 1$.

The same argument works if $n_q(G) = 1$.

### A group of order 12 with no subgroup of order 6 is isomorphic to Alt(4)

Show that a group of order 12 with no subgroup of order 6 is isomorphic to $A_4$.

Note that $12 = 2^2 \cdot 3$, so that Sylow’s Theorem forces $n_2(G) \in \{1,3\}$ and $n_3(G) \in \{1,4\}$. Suppose $n_3(G) = 1$. If $n_2(G) = 1$, then by the recognition theorem for direct products, $G \cong P_2 \times P_3$, where $P_2$ and $P_3$ are Sylow 2- and 3-subgroups of $G$, respectively. Let $x \in P_2$ have order 2 and $y \in P_3$ have order 3 by Cauchy; then $xy$ has order 6, a contradiction. Now suppose $n_2(G) = 3$. Since $n_2(G) \not\equiv 1$ mod 4, there exist $P_2,Q_2 \in \mathsf{Syl}_2(G)$ such that $P_2 \cap Q_2$ is nontrivial and $|N_G(P_2 \cap Q_2)|$ is divisible by $2^2$ and another prime; thus we have $P_2 \cap Q_2$ normal in $G$. Recall that every normal group of order 2 in a finite group is central; thus we have an element $x$ of order 2 in the center of $G$. Let $y \in G$ have order 3 by Cauchy; then $xy$ has order 6, a contradiction.

Thus $n_3(G) = 4$. Let $P_3 \leq G$ be a Sylow 3-subgroup and let $N = N_G(P_3)$. Now $N$ has index 4 and is self-normalizing by a previous result; in particular, $N$ is not normal in $G$. Now $G$ acts on $G/N$ by left multiplication, yielding a permutation representation $G \rightarrow S_{G/N}$ whose kernel $K$ is contained in $N$. Note that $|N| = 3$, so that either $K = 1$ or $K = N$. If $K = N$, then for all $g \in G$ and $h \in N$, we have $hgN = gN$, so that $g^{-1}hg \in N$. Thus, for all $g \in G$, $gNg^{-1} \leq N$; since $G$ is finite, this implies that $N \leq G$ is normal, a contradiction. Thus $K = 1$ and in fact $G \leq S_4$. Since $G$ contains $4 \cdot 2 = 8$ elements of order 3, it contains all elements of order 3 in $S_4$. The subgroup generated by the three cycles in $S_4$ is $A_4$, so that $A_4 \leq G$; since these subgroups are finite and have the same cardinality, in fact $G \cong A_4$.

### Every group of order 36 has a normal Sylow subgroup

Prove that if $G$ is a group of order 36, then $G$ has either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup.

Note that $36 = 2^2 \cdot 3^2$, so that Sylow’s Theorem forces $n_2(G) \in \{1,3,9\}$ and $n_3(G) \in \{1,4\}$.

Suppose $n_3(G) \neq 1$. Then $n_3(G) = 4 \not\equiv 1$ mod 9, so that by Lemma 13 and this previous exercise, there exist $P_3,Q_3 \in \mathsf{Syl}_3(G)$ such that $P_3 \cap Q_3$ is nontrivial. Consider now $C_G(P_3 \cap Q_3)$; since $P_3$ and $Q_3$ are abelian, $\langle P_3,Q_3 \rangle$ centralizes $P_3 \cap Q_3$, and moreover we can see by Sylow’s Theorem that $G = \langle P_3, Q_3 \rangle$. Thus $P_3 \cap Q_3 \leq Z(G)$, and in fact $P_3 \cap Q_3$ is the intersection of all Sylow 3-subgroups of $G$. Thus $G$ contains $2 + 6 + 6 + 6 + 6 = 26$ elements of 3-power order. Let $P_2 \leq G$ be a Sylow 2-subgroup; $P_2$ contains three nonidentity elements, whose products with the nonidentity elements in $P_3 \cap Q_3$ yield 6 elements of order 6 or 12. There are 4 elements left, which necessarily comprise a unique Sylow 2-subgroup.