normal subgroup of prime order in the center

antiemptyv

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

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

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

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

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

[tex]ghg^{-1} \in H[/tex]

It can quite easily be shown that this is true if

[tex] \forall g \in G[/tex] [tex] \forall h \in H, gh \in Hg[/tex]

that is, [tex] \exists h' \in H[/tex] such that

[tex] gh = h'g[/tex]

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.