# Normal subgroup of prime order in the center

1. Oct 25, 2007

### 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, $$ghg^{-1} \in H$$.
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. $$H \subseteq \{g \in G | gx = xg \forall x \in G\}$$

Last edited: Oct 25, 2007
2. Oct 25, 2007

### antiemptyv

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

3. Oct 25, 2007

### 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

4. Oct 25, 2007

### 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.