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Let $G$ be a nonabelian group such that for all $x,y\in G$, $xy\neq yx\implies x^2=y^2$. (Or equivalently, for all $x,y\in G$, either $xy=yx$ or $x^2=y^2$.)

Let $g\in G\setminus Z(G)$. Then $gh\neq hg$ for some $h\in G$. Therefore $gh^{-1}\neq h^{-1}g$ and $g^{-1}h\neq hg^{-1}$. By the assumption, we get $g^2=g^{-2}=h^2=h^{-2}$. This implies that $g^4=1$.

Next let $z\in Z(G)$. Since $(zg^{-1})h\neq h(zg^{-1})$, it follows that $(zg^{-1})^2=h^2$, which implies that $z^2=g^2h^2=1$.

In short, $G$ is a $2$-group of exponent $4$. Such a group exists as the quaternion group $Q_8$ fulfils the property. Are there any characterization on groups with this property? Or are there any other group that fulfils the property?

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2 Answers 2

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We will prove that any such group is the direct product of a quaternion group $Q_8$ and an elementary abelian $2$-group.

As you have shown, $G$ has exponent $4$ (as it cannot have exponent $2$, since such groups are elementary abelian).

Let $g\in G\setminus Z(G)$ have order $2$, and let $h\in G\setminus C_G(g)$. Then $g^2=1$, so $h^2=1$ as well. But $\langle g,h\rangle$ is generated by two involutions and is non-abelian, so must be $D_8$. However, then $gh$ has order $4$ and does not commute with $g$ either, a contradiction.

Thus all involutions are central. Let $g$ and $h$ be non-commuting elements of order $4$, and let $H$ be generated by them. If $x\in C_G(H)$ then $g$ and $hx$ cannot commute either, whence $g^2=(xh)^2=x^2h^2$. Since $g^2=h^2$, we see $x^2=1$, and $C_G(H)$ (and therefore also $Z(G)$) consists of exactly the involutions in $G$.

Notice that $g^4=1$, so $g^{-1}=zg$, where $z=g^2\in Z(G)$. If $[g,h]\neq 1$ then $g^2=h^2=z$, and again $h^{-1}=zh$. Also if $[g,h]\neq 1$ then $[g,gh]\neq 1$, so $(gh)^2=z$. Thus $$h^{-1}gh=h^3gh=hghz=hghgg=zg=g^{-1}.$$ In particular this means that every cyclic subgroup is normal, since $g^h=g^{\pm 1}$ for all $g,h\in G$.

By assumption the two cyclic subgroups do not commute, thus they generate a $Q_8$ subgroup $H$ (noting that $\langle g\rangle\cap \langle h\rangle=\langle g^2\rangle$ as above). Also, if $X=\langle g\rangle$, then $N_G(X)/C_G(X)$ has order $2$, so $|G:C_G(X)|=2$. By symmetry, the same holds for $Y=\langle h\rangle$.

We aim to show that $G=HZ(G)$. But $C_G(H)=C_G(X)\cap C_G(Y)$ has index at most $4$, whereas $H/Z(H)$ has order $4$, so $G=HC_G(H)$. Since $C_G(H)$ is elementary abelian, $C_G(H)=Z(G)$, so $G=HZ(G)$.

Thus $G=HZ(G)$, $Z(G)$ is elementary abelian, and so $G\cong H\times Z$ for some elementary abelian group $Z$.

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  • $\begingroup$ May I know what do you mean by $[H,k]$ and how does $[H,k]=1$ imply that $g$ does not commute with $hk$? $\endgroup$ Sep 22, 2021 at 11:56
  • $\begingroup$ @AlanWang $[H,k]=1$ just means $H$ commutes with $k$, i.e., all elements of $H$ commute with $k$. If $k\in C_G(g)$ and $hk\in C_G(g)$ then $h\in C_G(g)$. But $g$ and $h$ don't commute. $\endgroup$ Sep 22, 2021 at 11:58
  • $\begingroup$ I think it's easy to see that any non-normal subgroup of $G$ would have to have index at most $2$, and so $G$ is one of the groups all of whose subgroups are normal, so your classification is not (at the end of the day) surprising. [Not a criticism of your answer which I +1 ed] $\endgroup$ Sep 22, 2021 at 12:25
  • $\begingroup$ @ancientmathematician I was in the middle of editing with that sort of idea when you commented. Better solution coming! $\endgroup$ Sep 22, 2021 at 12:26
  • $\begingroup$ @AlanWang I have improved it significantly now. And removed the notation that you were not happy with anyway. $\endgroup$ Sep 22, 2021 at 12:47
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The direct product $G$ of the quaternion group and the group $C$ of order two has the property. Let $a$ be the order-two element of the quaternion group, let $b$ be the non-identity element of $C$. The center of $G$ is $Z=\{\,(1,1),(1,b),(a,1),(a,b)\,\}$; the elements of $G$ of order two are the non-identity elements of $Z$; all the elements of $G$ other than those in the center have order four; each element of order four has square $(a,1)$.

So, if two elements of $G$ don't commute, then they are both of order four, and they have the same square.

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