Normal subgroup of prime order in the center

[antiemptyv]

Homework Statement

Let H be a normal subgroup of prime order p in a finite group G. Suppose that p is the smallest prime dividing |G|. Prove that H is in the center Z(G).

Homework Equations

the Class Equation?
Sylow theorems are in the next section, so presumably this is to be done without them.

The Attempt at a Solution

Not completely sure of a solution, but here's (at least some of) what we know:

1. Since H is normal, ghg^{-1} \in H.
2. Since |H| is prime, H is cyclic and abelian.
3. G is finite, with order |G| = p^nq.
4. The normalizer N(H) (stabilizer under conjugation) is all of G...
5. ...so |G| = |N(H)| ??
6. Probably some more relevant properties.

And we want to show that H \subseteq Z(G), i.e. H \subseteq \{g \in G | gx = xg \forall x \in G\}

 

[antiemptyv]

So would these things imply that H = G is cyclic, thus abelian and is the center?

 

[Kreizhn]

Well, from the facts you've given, since H has prime order, it's cyclic, and every cyclic group is abelian. Now since H is normal, like you've also shown we have

ghg^{-1} \in H

It can quite easily be shown that this is true if

\forall g \in G \forall h \in H, gh \in Hg

that is, \exists h' \in H such that

gh = h'g

but H is abelian so...make some conclusion.

Since this holds for every member of H when applied to every member of G, the result follows

 

[antiemptyv]

I misinterpreted the problem. i was thinking of G acting on H, as a subgroup, and not of G acting on H element-wise, by just permuting the elements around in H.

this means if H was not in Z(G), then its orbit would have order 2,...,p-1, none of which divide |H| and |G| since p is prime and it's the least prime that divides |G|.

i believe this does it.