If $G$ is a group and $[G:Z(G)]=4$, show that $Z(G)$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$

If $G$ is a group and $[G:Z(G)]=4$, show that $Z(G)$ is isomorphic to
$\mathbb{Z}_2 \times \mathbb{Z}_2$

I want to show that if $G$ is a group and $[G:Z(G)]=4$, then $Z(G)$ is
isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. I know I can do this by
showing that $|Z(G)|=4$. For then $Z(G)$ is isomorphic to either
$\mathbb{Z}_4$ or $\mathbb{Z}_2 \times \mathbb{Z}_2$. The former group is
cyclic, so then $Z(G)$ would have to be cyclic. But if $Z(G)$ is cyclic,
then $G$ is abelian, whence $Z(G)=G$, whence $[G:Z(G)]=1\neq4$. Therefore,
$Z(G)$ must be isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. Any
suggestions on how to show that $|Z(G)|=4$?