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Let $G$ be a finite non-abelian simple group and $p$ the largest prime divisor of $|G|$. Show that if $H < G $ then $|G : H | \geq p $.

This is from a chapter of a book about group actions, specifically a section on conjugacy classes. Just before the question it also explains a theorem showing that a finite non-abelian simple group, such as $G$, must be divisible by at least $2$ different primes.

I'm guessing the question can be answered with these facts but I'm not sure how to go about it.

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    $\begingroup$ The action on cosets of $H$ yields an isomorphism from $G$ to a subgroup of $S_n$, where $n = |G:H|$, but if $n < p$ then $p$ does not divide $S_n$, contradiction. $\endgroup$ – Derek Holt Nov 10 '18 at 18:09
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Take $X$ to be the set of distinct left cosets of $G$, i.e. $X = \{g_1H, g_2H, \dots, g_nH\}$. Associate to $a \in G$ the permutation $\sigma_a \in S_X$ given by $\sigma_a(g_kH) = ag_kH$. Then define the homomorphism $\varphi : G \to S_X$ by $\varphi(g) = \sigma_g$ (you can check that this is indeed a homomorphism). Since $G$ is simple, the kernel of $\varphi$ is trivial. Thus $G \cong \text{im } \varphi$ by the first isomorphism theorem. So $p$ divides $|\text{im } \varphi|$, but $\text{im } \varphi$ is a subgroup of $S_X$, so $|X| \geq p$.

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  • $\begingroup$ Thanks, got it now. $\endgroup$ – John Doe Nov 10 '18 at 19:11

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