Subgroup of order $p$ is normal

Question

I am trying to show that an arbitrary group $G$ of order $p^n$ has a normal subgroup of order $p$. My first instinct is to say that by Cauchy's Theorem, there is some element $x \in G$ such that the order of $x$ is $p$. Thus, the cyclic group $H = \langle x \rangle$ is a subgroup of $G$ and has order $p$. However, I am not sure how to show that $H$ is normal in $G$. If I could somehow show that $H$ is contained in the center of $G$, this would imply that $H$ is normal in $G$, but I don't know how to show $H$ is in the center of $G$ either. Help!

Answer 1

Any $p$-group $G$ has a non-trivial center $Z(G)$ and hence you can find a subgroup of $Z(G)$ of order $p$ by Cauchy's Theorem applied to the center: there is an element of order $p$, say $z \in Z(G)$ and consider $H=\langle z\rangle$. Now show that this is $H$ in fact a normal subgroup of the whole group.