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.