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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 $\mathbb{Z}_2$?

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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.

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    Surely you have just proven that every balanced group is Hamiltonian-you haven't proven that every Hamiltonian group is balanced! For example, take $(i, a), (j, a^2) \in Q_8 \times C_3$. Clearly these two elements do not commute, and $(i, a)^2=(-1, a^2)$ but $(j, a^2)^2=(-1, a)$. So this is not balanced!2011-06-06
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    (However, it is easy to see that this contradiction generalises, and so basically every non-abelian balanced group is of the form $Q\times C_2^{(I)}$, $I$ some index set).2011-06-06
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    I think there is a mistake somewhere in your reasonment. Let us consider $G:=Q_8 \times \mathbb{Z}_3$ and take $(i,0,0,\ldots)$ and $(-j,1,1,\ldots)$. Now $(i,0,0,\ldots)(-j,1,1,\ldots)$ $=(-k,1,1,\ldots)$ $\neq (-j,1,1,\ldots)(i,0,0,\ldots)$ $= (k,1,1, \ldots)$. So they do not commute. And $(i,0,0,\ldots)^2$ $=(-1,0,0,\ldots)$ $\neq (-j,1,1,\ldots)^2$ $=(-1,2,2,\ldots)$. So $G$ is not balanced.2011-06-06
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    @Swlabr: thanks, fixed.2011-06-06
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    @Thomas Connor: thanks, fixed.2011-06-06