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?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.