## 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.