Normal group contained in the center.

[Barre]

I am doing exercises from Hungerford's text 'Algebra', and would appreciate if someone took the time to verify my write-up for me, and possibly provide me with tips how this could be done more efficiently (using less mathematical machinery)

Homework Statement

If a normal subgroup N of order p ( p prime) is contained in a group G of order $p^n$, then N is in the center of G.

Homework Equations

Orbit-Stabilizer Theorem: http://www.proofwiki.org/wiki/Orbit-Stabilizer_Theorem

The Attempt at a Solution

I construct the group action $f: G \times H \rightarrow H$ by this following rule: $f((g,h)) = ghg^{-1}$. This is well-defined by normality of H of course.
By Orbit-Stabilizer Theorem I know that the size of orbit of any $h \in H$ must have cardinality dividing $|G| = p^n$. So cardinality of orbits is 1 or p (since anything bigger
would imply more elements in an orbit than there are in H) and H is the disjoint union of orbits of it's elements, so all orbits cardinalities add up to p. But we see that the orbit of the identity in H must be of size 1 (itself), since for any $g \in G$ , $geg^{-1} = e$. This means we have p-1 other elements in orbits, but orbit cardinalities have to divide p, so they are all of size 1.
This means that for any $h \in H$ and all $g \in G$ we have $ghg^{-1} = h$, so all elements of H are in the center of G.

 

[Dick]

That's very nicely written. And the Orbit-Stabilizer Theorem is about the only piece of machinery you used. I can't think how you would do it more simply.