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Finite simple group with prime index subgroup

  • Thread starter asllearner
  • Start date
  • #1

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. p2 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/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 [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 HG is going to be 1 , that G/HG = 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:H\right] \left[H:H_{G}\right] = p \left[H: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 p2 cant divide G.

Did I get it, or have I fooled myself?
 

Answers and Replies

  • #2
352
0
For part 2, you are thinking too hard. If [tex]p[/tex] is prime, and you know that [tex]|G|[/tex] divides [tex]p![/tex], can [tex]p^2[/tex] divide [tex]|G|[/tex]?
 
  • #3
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?

thanks tads for the reply...much obliged!
 
  • #4
352
0
Assuming the "related theorem" you cite, yes, your proof is correct.
 

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