Tag Archives: finite group

Finite abelian groups are not naturally QQ-modules

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


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

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

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


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

We begin with some lemmas.

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

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

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

Now we move to the main result.

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

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

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

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

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

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


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

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

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

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

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

No simple group of order 720 exists

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


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

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

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

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

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

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

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

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

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

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

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

Suppose |N_G(Q)| = 2^2 \cdot 3^2. Then N_G(Q)/Q is a group of order 12, and as we saw in this previous exercise, acts on Q by conjugation (by coset representatives). Moreover, N_G(Q)/Q has 4 Sylow 3-subgroups by the Lattice Isomorphism Theorem. By the lemma, N_G(Q)/Q \cong A_4. Since A_4 has no subgroups of index 2, by the Orbit Stabilizer Theorem, the orbits of this action have order 1 or 3. However, the identity element of Q is in an orbit of order 1; thus all orbits have order 1. Thus this action is trivial; that is, for all g \in N_G(Q) and h \in Q, ghg^{-1} = h; thus Q \leq Z(N_G(Q)). Note that the four Sylow 3-subgroups of N_G(Q) intersect in Q, so that N_G(Q) has 8 + 3 \cdot 6 = 26 elements of 3-power order, and 10 elements remain. Now n_2(N_G(Q)) \in \{1,3,9\}. Let Q = \langle x \rangle. Suppose n_2(N_G(Q)) = 9. If some element of N_G(Q) has order 4, then each Sylow 2-subgroup has two elements of order 4, all distinct. But then |G| \geq 26 + 18 = 44, a contradiction. Thus all elements in Sylow 2-subgroups of N_G(Q) have order 2. Now G must contain at least 4 elements of order 2; let these comprise the set S. Now S, xS, and x^2S are 12 mutually distinct elements of order 2, 6, and 6, respectively, so that |G| \geq 26 + 12 = 38, a contradiction. Suppose now that n_2(N_G(Q)) = 3. Again if some element has order 4, then N_G(Q) has 3 \cdot 2 = 6 elements of order 4 and 2 \cdot 6 = 12$ elements of order 12, so that |G| \geq 26 + 6 + 12 = 44, a contradiction. Thus the elements in Sylow 2-subgroups of N_G(Q) have order 2. There are at least 4 of these, say in the set S. Now S, xS, and x^2S contain (at least) 12 distinct elements of order 2, 6, and 6, respectively, so that |G| \geq 26 + 12 = 38, a contradiction. Thus n_2(N_G(Q)) = 1. Let Q_2 \leq N_G(Q) be the unique Sylow 2-subgroup. Now Q_2 \leq P_2 for some Sylow 2-subgroup of G, and Q_2 is properly contained in N_{P_2}(Q_2). Thus 8 divides |N_G(Q_2)|, and since Q_2 is normal in N_G(Q), 3^2 also divides |N_G(Q_2)|. Thus |N_G(Q_2)| \in \{ 3^2 \cdot 2^3, 3^2 \cdot 2^4, 3^2 \cdot 2^3 \cdot 5, 3^2 \cdot 2^4 \cdot 5 \}. In all but one case, either G has a normal subgroup or a subgroup of sufficiently small index. Thus we have |N_G(Q_2)| = 2^3 \cdot 3^2. Now N_G(Q) \leq N_G(Q_2) has index 2, and thus is normal. Moreover, Q = O_3(N_G(Q)) (see this previous exercise) is characteristic in N_G(Q), and thus normal in N_G(Q_2). But then N_G(Q_2) \leq N_G(Q), a contradiction.

Suppose |N_G(Q)| = 2^3 \cdot 3^2. Note that, by the N/C Theorem, N_G(Q)/C_G(Q) \leq \mathsf{Aut}(Q) \cong Z_2. Then we have |C_G(Q)| \in \{ 2^2 \cdot 3^2, 2^3 \cdot 3^2 \}; in either case, there exists an element of order 2 which centralizes Q. Now write Q = \langle t \rangle, and note that Q has 10 conjugates since [G : N_G(Q)] = 10. Consider the action of G on the ten cosets in Q/N_G(Q) by left multiplication. Via this action, we have G \leq S_{10}. Now t \in S_{10} is a product of 3-cycles. Note that since N_G(xQx^{-1}) = xN_G(Q)x^{-1}, every conjugate of Q is contained in 4 Sylow 3-subgroups, and that every Sylow 3-subgroup contains a conjugate of Q (by Sylow’s Theorem, since Q \leq P_3). Moreover, in this case there are 40 Sylow 3-subgroups. By the pigeonhole principle, every Sylow 3-subgroup of G contains exactly one conjugate of Q. Suppose now that t \in S_{10} fixes xN_G(Q). Then t \cdot x N_G(Q) = x N_G(Q), so that x^{-1}tx \in N_G(Q); thus in fact x^{-1}Qx = Q (since x^{-1}Qx is contained in some Sylow 3-subgroup of G which also contains Q), and we have x \in N_G(Q). Thus t \in S_{10} fixes only N_G(Q), and has cycle shape (3,3,3). As shown in the beginning of this paragraph, let u \in C_G(Q) have order 2. Now u^{-1}tu = t, so that t is obtained from t by applying u entrywise. In particular, u permutes the orbits of t; it must interchange two and fix the third. Thus u \in S_{10} has cycle shape (2,2,2) or (2,2,2,3); in either case, G contains an odd permutation, so that G is not simple. Thus n_3(G) \neq 40.

We are left with n_3(G) = 10. Now |N_G(P_3)| = 2^3 \cdot 3^2, and G acts by conjugation on \mathsf{Syl}_3(G), and via this action we have G \leq S_{10}. Let P_3 \in \mathsf{Syl}_3(G) and suppose P_3 is cyclic; say P_3 = \langle t \rangle. Then t \in S_{10} is necessarily a 9-cycle. Moreover, by the N/C Theorem we have N_G(P_3)/C_G(P_3) \leq \mathsf{Aut}(P_3) \cong Z_6, so that [N_G(P_3) : C_G(P_3)] \in \{1,2\}. In either case, by Cauchy there exists an element u of order 2 such that u^{-1}tu = t. However, we saw in the discussion on page 127 of the text that |C_{S_{10}}(t)| = 9; a contradiction. Thus P_3 (indeed each of the Sylow 3-subgroups) is elementary abelian. Now let t \in P_3 be an element of order 3; t \in S_{10} is a product of 3-cycles. Suppose t fixes more than one element- that is, t is a product of one or two 3-cycles. Note first that if t fixes a Sylow 3-subgroup R_3, then t \in N_G(R_3) and in fact, since R_3 \leq N_G(R_3) is the unique Sylow 3-subgroup in N_G(R_3), we have t \in R_3. Thus, letting Q = \langle t \rangle, we have Q \leq R_3. Since R_3 is abelian, R_3 \leq N_G(Q). Now since t fixes exactly 7 Sylow 3-subgroups of G, N_G(Q) contains 7 Sylow 3-subgroups. But 7 \not\equiv 1 mod 3, a contradiction. Suppose now that t is a product of two 3-cycles. Now |N_G(Q)| is divisible by 3^2 and, since exactly four Sylow 3-subgroups contain Q, is divisible by 2^2 as well. Thus |N_G(Q)| \in \{ 3^2 \cdot 2^2, 3^2 \cdot 2^3, 3^2 \cdot 2^4, 3^2 \cdot 2^2 \cdot 5, 3^2 \cdot 2^3 \cdot 5, 3^2 \cdot 2^4 \cdot 5 \}. We can see that in all but the cases 3^2 \cdot 2^2 and 3^2 \cdot 2^3, either Q is normal in G or a subgroup of G has sufficiently small index. In the two remaining cases, N_G(Q) contains four Sylow 3-subgroups and has order 36 or 72; we eliminated these possibilities in our discussion about the case n_3(G) = 40. Thus t must be a product of three 3-cycles. Suppose now that there exists an element u \in G of order 2 such that u^{-1}tu = t. Now u must permute the orbits of t, so that it must interchange two orbits and fix the third. Thus u has cycle shape (2,2,2) or (2,2,2,3), and in either case G contains an odd permutation. Thus if t \in G is an element of order 3, there does not exist an element of order 2 which centralizes it. Finally, note that if R_3 \in \mathsf{Syl}_3(G) with R_3 \neq P_3, then t^{-1} R_3 t \neq R_3. That is, t \notin N_G(R_3). In particular, t \notin R_3. Thus, P_3 \cap R_3 = 1, and hence P_3 intersects every other Sylow 3-subgroup trivially. By this previous exercise, all Sylow 3-subgroups intersect trivially. Thus G contains 10 \cdot 8 = 80 elements of order 3.

Now since G \leq S_{10} and using Lagrange, every element of G has order 1, 2, 3, 4, 5, 6, 8, 9, or 10. We showed previously that no element has order 9. If x has order 6, then x^2 is an element of order 3 and x^3 is an element of order 2 which centralizes it, a contradiction. Thus no element has order 6. Now let x \in G have order 5; then x is a product of one or two 5-cycles. If x is a single 5-cycle, then x \in N_G(P_3) for some Sylow 3-subgroup P_3; this is a contradiction. Thus x is a product of two 5-cycles. Suppose now that there exists an element y which centralizes x; that is, y^{-1}xy = x. Now y must permute the orbits of x. Suppose y fixes each orbit of x; the restriction of y to an orbit of x is then either the identity or a 5-cycle, contradicting the fact that y has order 2. Thus y transposes the orbits of x and is thus a product of five 2-cycles; then y \in G is an odd permutation and G is not simple, a contradiction. Thus no element of order 2 centralizes an element of order 5. In particular, no element of G has order 10 (for the same reason that no element has order 6). Thus every element of G is contained in a Sylow subgroup. Now suppose x is an element of order 8 which normalizes a Sylow 3-subgroup; then x must have the cycle shape (8), so that G contains an odd permutation and is not simple. Thus no element of order 8 normalizes a Sylow 3-subgroup. Now |G| = 720, every element has order 1, 2, 3, 4, 5, or 8, and there are 80 and 144 elements of order 3 and 5, respectively. Thus the remaining 496 elements are contained in Sylow 2-subgroups. Recall also from our proof classifying the groups of order 20 that only one group of this order does not contain an element of order 10; namely Z_5 \rtimes_\varphi Z_4, where Z_5 = \langle a \rangle, Z_4 = \langle b \rangle, and \varphi(b)(a) = a^2. This group is thus isomorphic to the normalizers of Sylow 5-subgroups in G.

Let a \in G have order 3. Now a is contained in, and thus normalizes, a unique Sylow 3-subgroup P_3. Without loss of generality, we can say that a \in S_{10} (via the action of G on \mathsf{Syl}_3(G)) is a product of three 3-cycles; say a = (2\ 3\ 4)(5\ 6\ 7)(8\ 9\ 10). Now let P_3 = \langle a,b \rangle. We have b^{-1}ab = a and b \notin \langle a \rangle, so that b must permute the orbits of a. There are 6 choices we can make for b, and any one of them generates P_3 (together with a). So without loss of generality, let b = (2\ 5\ 8)(3\ 6\ 9)(4\ 7\ 10). Since |N_G(P_3)| = 2^3 \cdot 3^2, by Cauchy there exists an element of order 2 which normalizes P_3, say u. There are 7 possibilities for u^{-1}au; note that since u \in N_G(P_3), u(1) = 1.

  1. Suppose u^{-1}au = b. If u(2) = 2, then u = (3\ 5)(4\ 8)(7\ 9) is an odd permutation. If u(2) = 3, then latex u(3) = 6$. If u(2) = 4, then u(4) = 10. If u(2) = 5, then u(4) = 2. If u(2) = 6, then u(3) = 9 and u(4) = 3. If u(2) = 7, then u = (2\ 7)(3\ 10)(6\ 8) is an odd permutation. If u(2) = 8, then u(3) = 2. if u(2) = 9, then u = (2\ 9)(4\ 6)(5\ 10) is an odd permutation. If u(2) = 10, then u(3) = 4 and u(4) = 7.
  2. Suppose u^{-1}au = b^2 = (2\ 8\ 5)(3\ 9\ 6)(4\ 10\ 7). If u(2) = 2, then u = (3\ 8)(4\ 5)(6\ 10) is an odd permutation. If u(2) = 8, then u(4) = 2. If u(2) = 5, then u(3) = 2. If u(2) = 3, then u(3) = 9. If u(2) = 9, then u(3) = 6 and u(4) = 3. If u(2) = 6, then u = (2\ 6)(4\ 9)(7\ 8) is an odd permutation. If u(2) = 4, then u(4) = 7. If u(2) = 10, then u = (2\ 10)(3\ 7)(5\ 9) is an odd permutation. If u(2) = 7, then u(3) = 4 and u(4) = 10.
  3. Suppose u^{-1}au = ab = (2\ 6\ 10)(3\ 7\ 8)(4\ 5\ 9). If u(2) = 2, then u = (3\ 6)(4\ 10)(5\ 8) is an odd permutation. If u(2) = 6, then we must have u(6) = 2; but we also have u(4) = 2, a contradiction. Similarly, if u(2) = 10, then u(10) = u(3) = 2. If u(2) = 3, then u(3) = 2 and u(3) = 7, a contradiction. If u(2) = 7, then u(3) = 8 and u(4) = 3, a contradiction. If u(2) = 8, then u = (2\ 8)(6\ 9)(4\ 7) is an odd permutation. If u(2) = 4, then u(4) = 2 and u(4) = 9, a contradiction. If u(2) = 5, then u = (2\ 5)(3\ 9)(7\ 10) is an odd permutation. If u(2) = 9, then u(3) = 4 and u(4) = 5, a contradiction.
  4. Suppose u^{-1}au = a^2b. If u(2) = 2, then u = (3\ 7)(6\ 10)(9\ 4) is an odd permutation. If u(2) = 7, then u(5) = 7, a contradiction. If u(2) = 9, then u(3) = 2, a contradiction. If u(2) = 3, then u(3) = 5. If u(2) = 5, then u(3) = 10 and u(4) = 3, a contradiction. If u(2) = 10, then u = (2\ 10)(7\ 8)(4\ 5) is an odd permutation. If u(2) = 4, then u(4) = 8, a contradiction. If u(2) = 6, then u = (2\ 6)(3\ 8)(5\ 9) is an odd permutation. If u(2) = 8, then u(3) = 4 and u(4) = 6, a contradiction.
  5. Suppose u^{-1}au = ab^2 = (2\ 9\ 7)(3\ 10\ 5)(4\ 8\ 6). If u(2) = 2, then u = (3\ 9)(4\ 7)(5\ 8) is an odd permutation. If u(2) = 9, then u(4) = 2, a contradiction. If u(2) = 7, then u(3) = 2. If u(2) = 3, then u(3) = 10. If u(2) = 10, then u(3) = 5 and u(4) = 3. If u(2) = 5, then u = (2\ 5)(4\ 10)(6\ 9) is an odd permutation. If u(2) = 4, then u(4) = 6. If u(2) = 8, then u = (2\ 8)(3\ 6)(7\ 10) is an odd permutation. If u(2) = 6, then u(3) = 4 and u(4) = 8.
  6. Suppose u^{-1}au = a^2b^2 = (2\ 10\ 6)(3\ 8\ 7)(4\ 9\ 5). If u(2) = 2, then u = (3\ 10)(4\ 6)(7\ 9) is an odd permutation. If u(2) = 10, then u(8) = 10, a contradiction. If u(2) = 6, then u(3) = 2, a contradiction. If u(2) = 3, then u(3) = 8, a contradiction. If u(2) = 8, then u(3) = 7 and u(4) = 3, a contradiction. If u(2) = 7, then u = (2\ 7)(5\ 10)(4\ 8) is an odd permutation. If u(2) = 4, then u(4) = 5, a contradiction. If u(2) = 9, then u = (2\ 9)(3\ 5)(6\ 8) is an odd permutation. If u(2) = 5, then u(3) = 4 and u(4) = 9, a contradiction.

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

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

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

The remaining nine possibilities,

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

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

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

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

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

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

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

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

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

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

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

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

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


By Sylow’s Theorem, we have the following.

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

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

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

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

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

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

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

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

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


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

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

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


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

Sylow’s Theorem forces the following.

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

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

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

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

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

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

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

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

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

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

Thus we have n_p(G) = r.

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

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

Thus |G| = 60.

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

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


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

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

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

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

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


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

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

Every group of order 36 has a normal Sylow subgroup

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


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

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