Which are the infinite balanced nonabelian groups? This question is motivated by this previous question.

A group $G$ is called a balanced group provided that for all $a,b \in G$, either $ab=ba$ or $a^2 = b^2$.

Following the answers provided by the question referred earlier, are there infinite balanced nonabelian groups other than $Q_8 \times \mathbb{Z}_2^{\omega}$, i.e. a direct product between the quaternion group and a direct product of infinitely many copies of $\mathbb{Z}_2$?
 A: Note that if $ab\neq ba$, then $aab\neq aba$ and so $a$ and $ab$ don't commute either.  The group $\langle a,b : a^2 = b^2 = (ab)^2 \rangle$ is the quaternion group of order $8$.  In particular, two elements of a balanced group either generate an abelian group or a quaternion group of order $8$.
Let $H$ be a cyclic subgroup generated by $a$, and let $b$ be an element of $G$. Then the group $K$ generated by $a$ and $b$ is an abelian or quaternion group and so $H$ is normal in $K$, and so $b$ normalizes $H$.  Since $H$ is normalized by every element of $G$, $H$ is normal in $G$. Since every cyclic subgroup is normal in $G$, every subgroup (being a product of cyclic subgroups, possibly transfinitely) is normal, and $G$ is a Hamiltonian group.
In particular, $G\cong Q_8 \times D\times (\mathbb{Z}/2\mathbb{Z})^{(I)}$ for some index set $I$ and periodic abelian group $D$ in which every element has odd order.
(Correction due to user1729 and Thomas Connor:) Now consider the elements $(i, 1, 1)$ and $(j, d, 1)$ in $G$. Then these elements do not commute, since $i$ and $j$ do not commute, and so one must have $1 = dd$, as the squares must be equal.  Since $d$ has odd order by assumption on $D$, $d = 1$.  In other words, $D = 1$.
Hence $G\cong Q_8 \times (\mathbb{Z}/2\mathbb{Z})^{(I)}$ are the only infinite non-abelian balanced groups.
