- #1

asllearner

- 2

- 0

## 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 [tex]\left|G\right|[/tex] (the order of G)

2. p

^{2}doesnt divide [tex]\left|G\right|[/tex]

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.

thanks in advance for help

## Homework Equations

related theorem: if H is of finite index in G, then G/H

_{G}, where H

_{G}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 [tex]\left|G: H \right|[/tex] 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 H

_{G}is going to be 1 , that G/H

_{G}= G can be embedded in S

_{[tex]\left|G: H \right|[/tex]}, 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

[tex]\left|G/ H_{G}\right|= \left[G\right] \left[H_{G}\right] = p \left[H_{G}\right][/tex]

therefore, since the core is trivial.

[tex]\left|G\right|= p \left[H\right][/tex]

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 p

^{2}cant divide G.

Did I get it, or have I fooled myself?