$H$ prime order, normal subgroup of group $G$. Prove $H$ in center $Z(G)$. I am looking at the following question: "Let $H$ be a normal subgroup of prime order $p$ in a finite group $G$. Suppose that $p$ is the smallest prime that divides the order of $G$. Prove that $H$ is in the center $Z(G)$." 
Thoughts: We know that $H$ is a normal subgroup of $G$ and that $H$ has prime order. So, $H$ is cyclic and generated by some element $x \in G$. So, $H=\langle x\rangle$. Since $H$ is normal $gH = Hg$ for all $g \in G$. Thus, $x$ is in the center $Z(G)$. Since $H = \langle x\rangle$, then $H \in Z(G)$. 
Is this the correct logic? Thanks! 
 A: Suppose $g$ does not commute with $x$. Then you get a non-trivial aciton of $K=\langle g\rangle$ on $H$ by conjugation. This means a non-trivial morphism $K\to Aut(H)$. 
Now try to see that $Aut(H)\simeq \mathbb{Z}/(p-1)\mathbb{Z}$, and try to show that any morphism $K\to \mathbb{Z}/(p-1)\mathbb{Z}$ must be trivial (using the fact that $p$ is the smallest factor of the order of $G$).
A: Another argument: we can let $G$ act on $H$ by conjugation. Then the orbit-stabilizer theorem gives us
$$p = |H| = \text{# of one-point orbits} + \sum (\text{sizes of nontrivial orbits})$$
Now $h \in H$ is in a one-point orbit iff $ghg^{-1} = h$ for every $g \in G$, iff $gh = hg$ for every $g \in G$, if and only if $h \in H \cap Z(G)$. Therefore,
$$p = |H| = |H \cap Z(G)| + \sum (\text{sizes of nontrivial orbits})$$
Now we argue that there cannot in fact be any nontrivial orbits. This is true because the size of any orbit cannot exceed $p$ and must divide $|G|$. But $|G|$ has no prime divisors smaller than $p$, hence no integer divisors strictly between $1$ and $p$. So any nontrivial orbit must be of size $p$, hence all of $H$. But $\{e\}$ is a one-point orbit, so there is no room for an orbit of size $p$.
This means that the orbit-stabilizer equation reduces to
$$p = |H| = |H \cap Z(G)|,$$
hence $H = H \cap Z(G)$, which means that $H \leq Z(G)$.
