# Minimum size of index of a proper subgroup of a finite, non-abelian simple group $G$

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.

• 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. – Derek Holt Nov 10 '18 at 18:09

## 1 Answer

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 $$|X| \geq p$$.

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