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I'm not sure how to go about this question:

Let $G$ be a finite nonabelian simple group. Let $H\leq G$. Use the action $(G,G/H)$ to show that $|G:H|\geq 5$.

Now very good on group actions, could anyone help?

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up vote 5 down vote accepted

$G$ acts on the left cosets of $H$ by left multiplication. This action is nontrivial (assuming $H$ is a proper subgroup) so, since $G$ is simple, the action has trivial kernel. It thus defines an embedding of $G$ into $S_{|G:H|}$. Since $S_4$ has no nonabelian simple subgroups, we must have $|G:H|\ge 5$.


Of course, there's nothing special about this particular action. What we've actually proved is that if a nonabelian simple group acts nontrivially on a set, then that set must have size at least $5$. In a similar way, we can show that, if a simple group $G$ (other than $\mathbb{Z}/2\mathbb{Z}$) acts nontrivially on a set of size $i$, then $|G|$ divides $\frac{i!}{2}$. Thus a "large" simple group can never act nontrivially on a "small" set. This has lots of implications: large simple groups can't have (proper) subgroups of small index, can't have (nonidentity) elements with small numbers of conjugates, can't have (nontrivial) subgroups with small numbers of conjugates, and so on.

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+1 for the last paragraph. –  user21436 Apr 20 '12 at 18:05
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