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Show that if $G$ is a simple group with a subgroup $H$ of index $n>1$, then $|G| \leq n!$.

Hence show that a group of order $2^k \times 3$ can never be simple for $k>1$.

So I have let $X$ be the set of all left cosets of $H$ in $G$, which has order $n$. I know I should look for a homomorphism from $G$ to $S_n$ but this at this point I am stuck.

For the second part Sylow III will tell me that there must be 1 or 3 Sylow 2-groups, so I'm guessing that I need to rule out there being 3?

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  • $\begingroup$ What is t? some characters for a comment $\endgroup$ – kesa Nov 1 '17 at 12:21
  • $\begingroup$ Related: math.stackexchange.com/questions/88719/… $\endgroup$ – lhf Nov 1 '17 at 12:26
  • $\begingroup$ Was a typo, now corrected $\endgroup$ – David Forn Nov 1 '17 at 12:36
  • $\begingroup$ haha oops sorry! fixed $\endgroup$ – David Forn Nov 1 '17 at 15:45
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Consider $\phi: G \to \text{Sym}(X)$ given by $\phi(x)(aH)=(xa)H$.

Then $\phi$ is a homomorphism and so $\ker \phi$ is a normal subgroup of $G$ contained in $H$.

Since $G$ is simple and $H \ne G$, we must have $\ker \phi=1$.

Therefore, $\phi$ is injective and $|G| = |\phi(G)| \le |\text{Sym}(X)|= n!$.

For the second part, take $H$ to be the $2$-Sylow subgroup.

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