- #1
asllearner
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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 doesn't 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 can't be G since G isn't 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\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 can't divide it, so p2 can't divide G.
Did I get it, or have I fooled myself?