Re: algebra with Sylow....


"mina_world" <mina_world@xxxxxxxxxxx> wrote in message
news:eq6egd$hh9$1@xxxxxxxxxxxxxxxxxxx
Hello sir~

If H is a subgroup of a finite group G and
|H| is a power of a prime p,

then H is contained in some Sylow p-subgroup of G.

--------------------------------------

I found a simple proof.

lemma)
Let G be a group of order p^n and
Let X be a finite G-set.
Then |X| = |X_G| (mod p)
(X_G = {x in X | gx = x for all g in G})
------------------------------------------------

Let S be a Sylow p-subgroup of G.
Let X = {xS | x in G}
Mapping H x X -> X , (h, xS) -> hxS.
So, X is a H-set.

Since |X| = |G:S|, p does not divide |X|.
Since |X| = |X_H| (mod p) by lemma, |X_H| > 0.
Namely, there is a xS in X such that h(xS) = xS for all h in H.
so, x^{-1}.hx in S.
so, h in x.S.x^{-1} for all h in H.
so, H subset x.S.x^{-1}

Since x.S.x^{-1} is a Sylow p-subgroup of G,
H is contained in some Sylow p-subgroup of G.


.

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