# Finite simple group with prime index subgroup

## Homework Statement

If G is a finite simple group and
H is a subgroup of prime index p
Then
1. p is the largest prime divisor of $$\left|G\right|$$ (the order of G)
2. p2 doesnt divide $$\left|G\right|$$

I think I have this proved, but want to confirm my reasoning is sound.
this problem is from intro to group theory by rose. it is problem 192 on p 75 in the chapter on group actions on sets, including embedding finite groups in symmetric groups.

## Homework Equations

related theorem: if H is of finite index in G, then G/HG, where HG is the largest normal subgroup of G contained in H, (that is the core of H in G) can be embedded in the symmetric group on $$\left|G: H \right|$$ objects.

## The Attempt at a Solution

the core must be = 1 or G since G is simple (yes?).
and it cant be G since G isnt contained in H.
..
the theorem above implies, since HG is going to be 1 , that G/HG = G can be embedded in S$$\left|G: H \right|$$, which has order p!.

from here I think I am supposed to assume that there is a prime q that divides G. but then it would divide p!, and that means it would have to be less that p.

I believe this implies conclusion 1. Yes???
As to part 2, ...
Since the core is normal, the subgroup isomorphism theorem (I think) gives
$$\left|G/ H_{G}\right|= \left[G \right] \left[H _{G}\right] = p \left[H _{G}\right]$$
therefore, since the core is trivial.
$$\left|G\right|= p \left[H\right]$$
now the order of G has to be less that p!
so the order of H has to less that p-1!, which means that p cant divide it, so p2 cant divide G.

Did I get it, or have I fooled myself?

Related Calculus and Beyond Homework Help News on Phys.org
For part 2, you are thinking too hard. If $$p$$ is prime, and you know that $$|G|$$ divides $$p!$$, can $$p^2$$ divide $$|G|$$?

duh! Thinking too hard is what I do. I think.

(I was fooled because I was following copying the steps of a related proof on the same page...)

otherwise proof ok?