When does $(ab)^n = a^n b^n$ imply a group is abelian?

Suppose the identity $(ab)^n = a^n b^n$ holds in a group for some $n\in\mathbb{Z}$. For which $n$ does this necessarily imply the group is abelian? For example, when $n=-1$ or $n=2$, the group must be abelian. Are there any other such $n$, or can we construct a non-abelian group with this property for all $n\neq -1, 2$?

– lhf
Dec 5, 2014 at 16:44
• Also related: math.stackexchange.com/q/103124 Dec 5, 2014 at 16:50
• You mean: suppose that for all $a,b\in G$ the identity ... holds, right? Dec 5, 2014 at 17:13
• @Myself: That's what "identity" means. Dec 5, 2014 at 17:16

Let $n$ be such that $n \neq \pm 1$, $n \neq 2$ (the case $n=1$ is trivial, $-1$ and $2$ imply that the group is abelian).

• If $n$ is a power of $2$ (greater than $2$ in absolute value), then let $G = \mathbb{H}$be the quaternion group. It's nonabelian and has exponent $4$ (which divides $n$), so for all $a,b$, $(ab)^n = e = ee = a^n b^n$.

• If $n$ is not a power of $2$, then let $p > 2$ be a prime number dividing $n$, say $n = pk$. Let $G$ be the group described in this question: it's a nonabelian group that has exponent $p$. Then for all $a,b \in G$, $(ab)^n = ((ab)^p)^k = e = (a^p)^k (b^p)^k = a^n b^n$.

• If $n=-2$, let $G$ be the nonabelian group of exponent $3$ we used before. Then $\forall a \in G, a^{-2} = a$, from which the identity follows. (Thanks to Mikko Korhonen for pointing out the mistake in the first version of this answer).

• Yes, this is a more complete answer. I don't think a separate case for $2^k>4$ is necessary, though; the quaternion group works there too. Dec 5, 2014 at 17:45
• @HenningMakholm Ah yes, I can streamline the argument. Dec 5, 2014 at 17:47
• Is it obvious that the $n=-2$ case implies that the group is abelian? Dec 5, 2014 at 17:51
• @HenningMakholm: No, because it is false. Take $G$ to be a nonabelian group of exponent $3$. Then for any $x \in G$ we have $x^{-2} = x$, so $(xy)^{-2} = xy = x^{-2}y^{-2}$. Dec 6, 2014 at 20:17
• Actually you get all the examples you need from the Heisenberg group (second point in your answer), see my answer. Dec 6, 2014 at 20:40

For any non-abelian group $\;G\;$ of exponent $\;n\;$ we have that $\;1=(ab)^n=a^nb^n\;$ , so for any natural $\;n\;$ for which there exists at least one non abelian group with that exponent we get a counterexample.

• That's not really an answer, you rephrased the question... Dec 5, 2014 at 16:45
• @Najib: It is not a complete answer, but it does give some information not present in the question. For example, this rules out all even numbers $>2$ because of the dihedral group. Dec 5, 2014 at 16:50
• @Najib, I think the above is as complete an answer as your can expect for your question, since it points the very naturals for which one gets that the implication of the group being abelians isn't true. It is nothing close to rephrasing your question. Dec 5, 2014 at 16:55
• @Timbuc: I don't think the answer says that if every group of exponent $n$ is abelian, then $\forall ab[(ab)^n=a^nb^n]$ does guarantee that a group is abelian? Dec 5, 2014 at 16:59
• @Henning, the question was "Are there any other such n , or can we construct a non-abelian group with this property for all n≠−1,2 ?". and I think my answer addresses exactly this. Now, if every group of some exponent $\;n\;$ is abelian, then what do we need the condition $\;(ab)^n=a^nb^n\;$ to determine this for? Dec 5, 2014 at 17:00

Note that if $G$ is a group with exponent dividing $n-1$ or $n$, then $(xy)^n = x^n y^n$ for all $x, y \in G$.

Suppose that $n \neq -1, 2$. By above it would be enough to prove that in this case we can find a nonabelian group with exponent dividing $n-1$ or $n$. By the assumption either $n-1$ or $n$ has an odd prime divisor $p$, so we could use for example the Heisenberg group of order $p^3$, which has exponent $p$.