A group of order 16 has a normal subgroup of order 4

Question

Let $ G$ be a group of order $16$. Show that $G$ must contain a normal subgroup $H$ of order $4$.

I tried the Sylow first theorem, that is $\{e\}\triangleleft H_1\triangleleft H_2\triangleleft H_3\triangleleft G $ where $|H_i|=2^i$. Then I wanted to prove that the normalizer of $H_2$ is $G$. But I couldn't, maybe it's not the correct way to prove it!

Answer 1

If $|Z(G)| \geq 4$ we are done. Otherwise $|Z(G)| = 2$ since $G$ is a $p$-group. Any group of order $8$ has a normal subgroup of order $2$ (so, of index 4), therefore considering the quotient $G/Z(G)$, $G$ must have a normal subgroup of index $4$.

Actually more is true: Normal subgroups of p-groups