Subgroup of a p-group in its center Suppose $p$ is prime, $n\in\mathbb Z^+$, and $G$ is a group of order $p^n$. If $H$ is a subgroup of order $p$ and $ghg^{-1}\in H$ for all $g\in G$ and $h\in H$,I can't seem to show that $H \subseteq Z(G)$. I have shown that $[G:Z(G)]$ can't be prime and that $Z(G)$ is divisible by p.
 A: There is a more general result, often referred to as the "N/C Theorem", which goes as follows. The theorem comes in very handy when analyzing subgroups. If $H$ is a subgroup of $G$, then $N_G(H):=\{g \in G : g^{-1}Hg=H \}$, the normalizer of $H$ in $G$, and $C_G(H):=\{g \in G : g^{-1}hg=h \text { for all } h \in H\}$, the centralizer of $H$ in $G$. Observe that $C_G(H) \unlhd N_G(H)$.
Theorem Let $G$ be a group and $H$ a subgroup, then $N_G(H)/C_G(H)$ is isomorphic to a subgroup of Aut$(H)$.
Proof (sketch) Define a homomorphism $\alpha:  N_G(H) \rightarrow \text{ Aut}(H)$, by $\alpha(x)=\phi_x$, the inner automorphism induced by $x \in N_G(H)$. Check for yourself that $\alpha$ is well-defined, that it is a homomorphism and that ker$(\alpha)=C_G(H)$. Now apply the first homomorphism theorem.$\square$
Now let's go back to your example. The condition on $H$ ("$ghg^{-1}\in H$ for all $g\in G$ and $h\in H$") is actually equivalent to $H$ being normal in $G$, so $G=N_G(H)$. Further $|H|=p$, so Aut$(H) \cong C_{p-1}$. Now apply the N/C Theorem: $|G/C_G(H)|$ divides $p-1$, but also $|G|$. Since $G$ is a $p$-group, it follows that in fact $G=C_G(H)$, and this is equivalent to $H \subseteq Z(G)$.
Encore - perhaps you want to try this one: let $G$ be a group of odd order and $N \unlhd G$ with $|N|=17$. Prove that $N \subseteq Z(G)$.
