Tag Archives: sylow’s theorem

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.

A group of order 24 with no elements of order 6 is isomorphic to Sym(4)

Show that a group of order 24 with no element of order 6 is isomorphic to S_4.


Note that 24 = 2^3 \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. By Cauchy, there exist elements x \in P_2 and y \in P_3 of order 2 and 3, so that xy has order 6, a contradiction. Suppose now that 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 its normalizer has order 2^3 \cdot 3, by a previous theorem. Thus P_2 \cap Q_2 \leq G is normal. Note that |P_2 \cap Q_2| is either 2 or 4.

Suppose |P_2 \cap Q_2| = 4. Then at most 7 + 3 + 7 nonidentity elements of G are contained in Sylow 2-subgroups, and 3 elements are contained in Sylow 3-subgroups. This leaves 4 elements not of prime power order, one of which must have order 6- a contradiction.

Suppose now that |P_2 \cap Q_2| = 2. By the N/C Theorem, G/C_G(P_2 \cap Q_2) \leq \mathsf{Aut}(P_2 \cap Q_2) \cong 1, so that P_2 \cap Q_2 is central in G. By Cauchy, there exist elements x \in P_2 \cap Q_2 and y \in G of order 2 and 3, so that xy has order 6, a contradiction.

Thus we may assume n_3(G) = 4. Let P_3 \leq G be a Sylow 3-subgroup and let N = N_G(P_3). The action of G on G/N yields a permutation representation G \rightarrow S_4 whose kernel K is contained in N. Recall that normalizers of Sylow subgroups are self normalizing, so that N is not normal in G. Moreover, we have |N| = 6. We know from the classification of groups of order 6 that N is isomorphic to either Z_6 or D_6; however, in the first case we have an element of order 6, a contradiction. Thus N \cong D_6. We know also that the normal subgroups of D_6 have order 1, 3, or 6. If |K| = 6, then K = N is normal in G, a contradiction. If |K| = 3, then by the N/C theorem we have G/C_G(K) \leq \mathsf{Aut}(Z_3) \cong Z_2. In particular, C_G(K) contains an element of order 2, so that G contains an element of order 6, a contradiction.

Thus K = 1, and in fact G \leq S_4. Since |G| = |S_4| = 24 is finite, G \cong S_4.

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.

No simple groups of order 336 exist

Prove that there are no simple groups of order 336.


Note that 336 = 2^4 \cdot 3 \cdot 7. Suppose G is a simple group of order 336. By Sylow’s Theorem, we have n_7(G) = 8. The permutation representation afforded by the action of G on G/N_G(P_7) by left multiplication, where P_7 is some Sylow 7-subgroup, has a proper, normal kernel in G; thus G \leq S_8. Note that G has no subgroup of index 2 since such a subgroup would be normal. Thus, by Proposition 12, G \leq A_8.

As usual, we compute that |N_G(P_7)| = 7 \cdot 3 \cdot 2. Now by Lemma 1 to this previous exercise, we have |N_{S_8}(P_7)| = 7 \cdot 6, and by Proposition 12, |N_{A_8}(P_7)| = 7 \cdot 3. But, since Sylow 7-subgroups of G are Sylow 7-subgroups of A_8, we have N_G(P_7) \leq N_{A_8}(P_7), which implies that 2 divides 1, a contradiction.

No simple groups of order 792 or 918 exist

Prove that there are no simple groups of order 792 or 918.


  1. Note that 792 = 2^3 \cdot 3^2 \cdot 11. Suppose G is a simple group of order 792. By Sylow’s Theorem, n_{11}(G) = 12. Consider the permutation representation G \rightarrow S_{12} afforded by the action of G on G/N_G(P_{11}) by left multiplication, where P_{11} is some Sylow 11-subgroup. The kernel of this action is a proper normal subgroup, and thus is trivial; hence G \leq S_{12}. Since [G:N_G(P_{11})] = 12, we have |N_G(P_11)| = 6 \cdot 11. Moreover, P_{11} is also a Sylow 11-subgroup of S_{12}. By Lemma 1 of the previous exercise, |N_{S_{12}}(P_11)| = 10 \cdot 11. Now Lemma 2 of the previous exercise implies that 6 divides 10, a contradiction.
  2. Note that 918 = 2 \cdot 3^3 \cdot 17. Suppose G is a simple group of order 918. By Sylow’s Theorem, n_{17}(G) = 18. The permutation representation afforded by the action of G on N_G(P_{17}) gives us that G \leq S_{18}. Now |N_G(P_{17})| = 3 \cdot 17, while N_{S_{18}}(P_{17})| = 16 \cdot 17, which implies that 3 divides 16, a contradiction.

No simple groups of order 1755 or 5265 exist

Prove that there are no simple groups of order 1755 or 5265.


We begin by formalizing some lemmas which are sketched in the text.

Lemma 1: Let p be a prime and let p \leq n < 2p. Then n_p(S_n) = n!/p(p-1)(n-p)!. Proof: p^1 is the largest power of p dividing |S_n|, so that the Sylow p-subgroups of S_n are cyclic and have prime order. Thus the Sylow p-subgroups intersect trivially; every p-cycle in S_n is contained in a unique Sylow p-subgroup, and each Sylow p-subgroup is generated by p-1 p-cycles. We computed the number of p-cycles in S_n in a previous theorem; using that result, the number of Sylow p-subgroups in S_n is as desired. \square

Lemma 2: If p is a prime, G a finite group with p dividing |G|, G \leq S_n for some p \leq n < 2p, and P \leq G a Sylow p-subgroup, then P \leq S_n is a Sylow p-subgroup and N_G(P) \leq N_{S_n}(P). Proof: This follows because Sylow p-subgroups of G and S_n have the same order. \square

  1. Note that 1755 = 3^3 \cdot 5 \cdot 13. Suppose G is a simple group of order 1755. By Sylow’s Theorem, n_3(G) = 13, so that we have a permutation representation G \rightarrow S_{13} via the action of G on G/N_G(P_3) for some Sylow 3-subgroup P_3. The kernel of this representation is proper and normal in G, so that it is trivial, hence we have G \leq S_{13}. Let P_{13} be a Sylow 13-subgroup of G (and thus also of S_{13}). By Sylow’s Theorem, we have n_{13}(G) = 27, so that [G:N_G(P_{13})] = 27, and |N_G(P_{13})| = 13 \cdot 5. By Lemma 1, n_{13}(S_{13}) = 11!, so that [S_{13}:N_{S_{13}}(P_{13})] = 11! and |N_{S_{13}}(P_{13})| = 13 \cdot 12. But then Lemma 2 implies that 5 | 12, a contradiction.
  2. Note that 5265 = 3^4 \cdot 5 \cdot 13. Suppose G is a simple group of order 5265. By Sylow’s Theorem, n_3(G) = 13 so that, via the (necessarily injective) permutation representation afforded by the action of G on G/N_G(P_3) for some P_3 \in \mathsf{Syl}_3(G), we have G \leq S_{13}. As before, since Sylow’s Theorem forces n_{13}(G) = 27, we have |N_G(P_{13}| = 13 \cdot 15 where P_{13} is a Sylow 13-subgroup. As before, we have |N_{S_{13}}(P_{13})| = 13 \cdot 12; but then Lemma 2 implies that 15 divides 12, a contradiction.