# $2$-Sylow subgroup of nonabelian group of order $56$ must be abelian?

Let $G$ be a nonabelian group of order $56$.

Let $Q$ be the $2$-Sylow subgroup of $G$. Show that $Q$ is isomorphic to $\mathbb Z_8, \mathbb Z_4\times \mathbb Z_2$ or $\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2$.

I figured out that it suffices to prove that $Q$ is abelian, the result will then follow by the fundamental theorem of abelian groups.

I tried considering $Z(Q)$; the center of the group $Q$. Since $Q$ is a $p$-group, its center is nontrivial, hence the possible values are $|Z(Q)|=2,\ 4,\ 8$. If $|Z(Q)|=8$, we are done. If $|Z(Q)|=4$, then $|Q/Z(Q)|=2$ and thus the quotient group is cyclic which implies $Q$ is abelian.

However if $|Z(Q)|=2$, then $|Q/Z(Q)|=4$ and there are two cases, either the cyclic group or the Klein's $4$ group. If it is cyclic group, then we are done. If it is the Klein $4$ group, ..., I am not sure how to proceed.

Thanks for any help!

• If the order is a prime, then the group is cyclic and hence abelian. Sep 25 '15 at 16:04

The claim is false. Let $Q$ be the quaternion group (non-abelian of order $8$) and $G=Q\oplus \mathbb Z/7\mathbb Z$.