We already know that for any odd prime $p$, every non abelian $p$-group contains a subgroup isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$. So can we say something for even prime? I observed that if $G$ is a non abelian group of order $2^n$ with $n \geq 3$, then $G$ does not contain any subgroup isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$ if and only if $G \cong Q_{2^n}$, i.e., generalized quaternion group.
One side is clear, i.e., if $G \cong Q_{2^n}$ then $G$ has no subgroup isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. I am unable to prove the converse part.
A failed attempt: Let $G$ has no subgroup isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. By Sylow's theorem, $G$ must contain a subgroup $H$ of order $8$. Hence $H$ must be isomorphic to either $\mathbb{Z}_8$ or $Q_8$. As $G$ is a $2$-group, it has non trivial center $Z(G)$. If there exists $x \in Z(G) \setminus H$, then we can have a subgroup isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$.
What if $Z(G) < H$? I think in this case $G \cong Q_{2^n}$, but I am unable to prove this. Any hint or comment will be appreciated. Thank you.
