So I've been trying to prove that there's no simple group of order $300$. This is what I did and I was wondering if it was enough.

$|G|=2^2 \cdot 3 \cdot 5^2$. Suppose $G$ is simple. Then there would be $6$ Sylow $5$-subgroups, one of which will have an index of $6$. But then $|G|=300$ does not divide $6!$ which leads to a contradiction. So, the number of Sylow $5$-subgroups is $1$ and $\exists$ a proper normal Sylow $5$-subgroup in $G$. Hence $G$ is not simple.

  • $\begingroup$ What do u want to ask? $\endgroup$ – anonymous Dec 5 '10 at 16:24
  • $\begingroup$ I'd like to ask if I've done enough to deserve full credit. $\endgroup$ – Nana Dec 5 '10 at 16:35
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    $\begingroup$ I think you might have added a little more explanation of why $|G|$ not dividing 6! leads to a contradiction. Incidentally, even $|G|$ not dividing $6!/2 = 360$ would have led to a contradiction - do you see why? $\endgroup$ – Derek Holt Dec 5 '10 at 17:12
  • $\begingroup$ @Derek Will this explanation be enough. If $G$ contains a subgroup H of index $n$, then it contains a normal subgroup $K$ in $H$ such that [$G:K$] is finite and divides $n!$ and If $G$ were to be simple, then $K$ would be $e$ or $G$. I'd be happy if you could explain why |G| not dividing $6!/2$ would also lead to a contradiction. Thanks. $\endgroup$ – Nana Dec 5 '10 at 17:47
  • $\begingroup$ @Derek, @John: Could you post the hint as an answer in the interest of having fewer unanswered questions? $\endgroup$ – Arturo Magidin Dec 6 '10 at 3:04

OK, the following results lead to a solution to this and similar problems.

Theorem. Let $G$ be a finite nonabelian simple group with a subgroup $H$ of index $n>1$. Then $n \ge 5$, and $|G|$ divides $n!/2$.

Proof. Let $\phi: G \rightarrow S_n$ be the permutation representation of $G$ acting by (left or right depending on whether you use left or right group actions) multiplication on the set of (left or right) cosets of $H$ in $G$. Then $G/{\rm Ker}(\phi) \cong {\rm im}(\phi) \le S_n$. Since $n>1$ and ${\rm im}(\phi)$ is transitive, $|{\rm im}(\phi)| > 1$ and so $G$ simple implies ${\rm Ker}(\phi) = 1$, and hence $G \cong {\rm im}(\phi)$. Now $S_n$ is solvable for $n < 5$, so we must have $n \ge 5$. Furthermore, we must have ${\rm im}(\phi) \le A_n$, since otherwise ${\rm im}(\phi) \cap A_n$ would be a normal subgroup of ${\rm im}(\phi)$ of index 2, and so $G$ would not be simple. Hence $|G|$ divides $|A_n| = n!/2$.

Corollary. Let $G$ be a finite simple group and $n = |{\rm Syl}_p(G)|$ for some prime $p$ dividing $|G|$. Then $n \ge 5$ and $|G|$ divides $n!/2$.

Proof. Let $P \in {\rm Syl}_p(G)$. We cannot have $n=1$ because then $P$ would be normal in $G$. Now apply the theorem to the subgroup $N_G(P)$ of index $n$ in $G$.

  • $\begingroup$ Are you assuming that $G$ is non abelian ? Because otherwise why we must have $n>4$ ? $\endgroup$ – JeSuis Nov 7 '15 at 16:38
  • $\begingroup$ Yes, I was indeed assuming that $G$ is nonabelian *which is OK, because there is no abelian simple group of order $300$). $\endgroup$ – Derek Holt Nov 7 '15 at 17:30

Assume $G$ is simple. Then the existence of 6 Sylow 5-groups implies $G$ embeds in $S_{6}$ (let $G$ act on the Sylow 5-subgroups by conjugation and use the assumption that $G$ is simple). But 300 does not divide 6 factorial. So $G$ is not simple.

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    $\begingroup$ Did you read Derek's answer? $\endgroup$ – j.p. Dec 6 '10 at 16:57

Ok..I got something here..John.this is betty from class..let me know what you guys think of this. Proof: Let $G$ be a simple group of order 300. i.e, $|G|=300$ assume ----><----- that G is simple. Since $|G|=300=2^2\times3\times 5^2$, then there exists a sylow 5-subgroup, say $P$ that has order $25$. Now $s_5$ is congruent to $1$(mod 5) and $s_5$ divides $300/25=12$. Hence $s_5=1$ or $6$. But if $s_5=1$, then $P\trianglelefteq G$, which is impossible. Thus $s_5=6$ and $|G:N_G(P)|=6$. Now recall a thm, proved from class. "Let $G$ be a group. If there exists a subgroup of $G$ s.t $[G:H]=n$, then there exists a $N\trianglelefteq G$, s.t $N$ C $H$, and $[G:N] | n!$." We apply this thm to $|G:N_G(P)|=6$ now. If $|G:N_G(P)|=6$, then there exists a normal subgroup of $G$ that is contained in $N_G(P)$, say this subgroup is $P$ (again), where $P\trianglelefteq G$ and $P$ C $N_G(P)$, then $[G:P] | 6!$. But this is impossible, since $|G|=300$ does not divide $6!$. Also, if $G$ were simple, then $P$ would be $1$ or $G$, and we already showed above that $P$ cannot be equal to $1$, and $P$ is not equal to $G$. Thus we have reached a contradiction. Therefore $G$ cannot be simple, and there is no simple group of order $300$.


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