17

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 $n\neq -1, 2$?

Najib Idrissi
  • 54,185
Sameer Kailasa
  • 8,983

3 Answers3

7

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).

Community
  • 1
Najib Idrissi
  • 54,185
4

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.

Timbuc
  • 34,191
  • 1
    That's not really an answer, you rephrased the question... – Najib Idrissi Dec 05 '14 at 16:45
  • 2
    @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. – hmakholm left over Monica Dec 05 '14 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. – Timbuc Dec 05 '14 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? – hmakholm left over Monica Dec 05 '14 at 16:59
  • 1
    @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? – Timbuc Dec 05 '14 at 17:00
  • In fact every composite number is ruled out. If $p$ and $q$ are (not necessarily different) primes, then the permutation group generated by (a) a product of $p$ disjoint $q$-cycles, and (b) a $p$-cycle containing one element of each of the $q$-cycles, will be non-abelian of exponent $pq$. – hmakholm left over Monica Dec 05 '14 at 17:02
  • I maintain that this answer should have been a comment. You just rephrased the question, in a more tractable way, sure, but compare with my answer... Yours isn't simply incomplete, it's just the first step in the reasoning... – Najib Idrissi Dec 05 '14 at 17:40
  • Comments are not for extended discussion; this conversation has been moved to chat. – user642796 Dec 08 '14 at 12:16
1

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 $p$.

Mikko Korhonen
  • 25,224