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Is there a finite nonabelian group satisfying all of the following identities? $$ (x^py^p)^2 = (y^px^p)^2, \quad p = 2,3,5,7,11,\ldots (\text{primes}) $$

I thank you all in advance.

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What is the motivation? – Noah Schweber Aug 24 '14 at 18:49
Yes: Extra-special $2$-groups have that property. – Geoff Robinson Aug 24 '14 at 19:02
up vote 14 down vote accepted

Assuming you want that to hold for all $x,y \in G,$ any (non-Abelian) extra-special $2$-group $G$ will have that property. For there are only two squares in $G$: the identity, and the unique element of order $2$ in $Z(G).$ However, note that $(x^{p}y^{p})^{2}$ and $(y^{p}x^{p})^{2}$ are conjugate in $G$, so since they are both central, they are equal. By the way, it is easy to see that any finite group of odd order with the stated property is Abelian.

Later edit following YCor's comment: As remarked after that comment, it appears to me that the condition in the original question is equivalent to $(yx)^{2} = (xy)^{2}$ for all $x,y \in G.$ This is because we may choose a prime $q > |G|,$ and then $g \to g^{q}$ is a bijection from $G$ to itself. Hence, given $x,y \in G,$ there are unique elements $a$ and $b$ in $G$ such that $x = a^{q}$ and $y = b^{q}.$ Hence we have $(xy)^{2} = (yx)^{2}$ by hypothesis. (Late addition: Actually, it's easier than this: Just take a prime $p \equiv 1$ (mod $|G|$) ).

I will prove that a finite group $G$ with $(xy)^{2} = (yx)^{2}$ for all $x,y \in G$ has a normal Sylow $2$-subgroup $S$, and that $G/S$ is Abelian (of odd order). Note that the condition is inherited by homomorphic images. Hence, by induction, we may suppose that $O_{2}(G) = 1.$ The result is clear if $G$ has odd order, as noted above, so suppose that $G$ contains an involution $t.$ Then (by the Baer-Suzki theorem), there is a conjugate $u$ of $t$ such that $\langle t,u \rangle$ (which is a dihedral group) is not a $2$-group. Hence $t$ inverts a non-idetity element $g \in G$ of odd order. Then $v = tg$ is also an involution of $G.$ Also, $tv = g$ has odd order. Also, $vgv = vt$ has odd order as $v = v^{-1}.$ Since $(vt)^{2} = (tv)^{2}$ and both have the same odd order, we have $vt = tv$ ( for if that order is $n,$ we have $vt = ((vt)^{2})^{\frac{n+1}{2}} = ((tv)^{2})^{\frac{n+1}{2}} = tv).$ But this contradicts the fact that $tv = g$ is a non-identity element of odd order, as $t$ and $v$ are involutions.

Perhaps it is possible to say more about the structure of $G$ For example, $G = A_{4}$ does not have the property of the question: if $x,y$ are distinct elements of order $3$ in $A_{4}$ which are conjugate then $(xy)^{2} \neq (yx)^{2}.$

Even later edit: In fact, a finite group $G$ with the stated property is nilpotent with an Abelian Sylow $p$-subgroup for each odd prime $p.$ Given what is proved already, it suffices to prove that the elements of odd order are closed under multiplication. Otherwise, we have $x,y$ of odd order such that $z = xy$ has even order. Then $y = x^{-1}z$ has odd order, so as above, we obtain that $x^{-1}z = zx^{-1}$ since those two elements of odd order have equal squares. Hence $x$ and $xy$ commute, so that $x$ and $y$ commute, a contradiction.

Hence understanding the finite groups $G$ with that stated property reduces to understanding finite $2$-groups $G$ such that $x$ and $yxy$ commute for all $x,y \in G.$

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Thank you very much! – E W H Lee Aug 24 '14 at 19:21
Note: actually they satisfy the identity for $p=1$, which implies the identity for all $p$. This holds for every group satisfying the identity $xy^2=y^2x$. – YCor Aug 24 '14 at 20:29
@YCor: Yes, it is enough that squares are central. Also, I think that there is eventually a prime $q$ which does not divide $|G|$ means that the given condition is equivalent to $(xy)^{2} = (yx)^{2}$ since $g \to g^{q}$ is a bijection from $G$ to $G$ for such a prime. – Geoff Robinson Aug 25 '14 at 9:31

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