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I came across an exercise on groups which states as follows,

If $G$ is a group and there exists $3$ consecutive integers such that $(ab)^i=a^ib^i$ for all $a,b\in G$, then $G$ is abelian. Here $i$ is any of the $3$ integers.

This is not hard to prove as $ab=(ab)^{n+2}(ab)^{-(n+1)}=a^{n+2}ba^{-(n+1)}$ and $ab=(ab)^{n+1}(ab)^{-n}=a^{n+1}ba^{-n}$. Now $a^{n+2}ba^{-(n+1)}= a^{n+1}ba^{-n}$ and $ab=ba$. Here $n$ is the least of the consecutive integers.

If we only have two consecutive integers, then there are counterexamples if $n=0$ (take any non-abelian group). If we assume $n>0$ are there still counterexamples? If there are, how do you go about finding such a group?

Shaun
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spinosarus123
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    If $n=2$ then $G$ is abelian, since $abab=(ab)^2=a^2b^2=aabb$, so $ba=ab$. – David A. Craven Sep 12 '21 at 19:19
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    A group satisfying $(ab)^n=a^nb^n$ for all $a$ and $b$ is said to be $n$-abelian. There is extensive discussion about $n$-abelian groups, and groups that are both $n$-abelian and $m$-abelian for two values of $n,m$, in a number of places. Here is one characterizing the pairs for which every group that is both $n$-abelian and $m$-abelian is necessarily abelian. – Arturo Magidin Sep 12 '21 at 20:25
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    Also, Alperin has described the structure of $n$-abelian groups: you essentially take an abelian group, a group of exponent $n$, and a group of exponent $n-1$, take their product, and then consider subgroups and quotients of subgroups of that group. – Arturo Magidin Sep 12 '21 at 20:26

1 Answers1

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Suppose that $(ab)^n=a^nb^n$ and $(ab)^{n+1}=a^{n+1}b^{n+1}$ for all $a,b\in G$.

If $(ab)^2=a^2b^2$ for all $a,b\in G$, then $G$ is abelian, since $abab=aabb$, so $ab=ba$. Thus if $n=1$ or $n=2$ then $G$ is abelian.

If $n\geq 3$, then let $G$ be a group of exponent dividing $n$, i.e., all elements of $G$ have order dividing $n$. Then $a^n=1$ for all $a\in G$, so the statement $(ab)^n=a^nb^n$ vacuously holds, and $(ab)^{n+1}=a^{n+1}b^{n+1}$ becomes $ab=ab$, and again vacuously holds.

The only issue is to construct a non-abelian group of exponent dividing $n$. If $4\mid n$ then the dihedral group of order $8$ will work. If $2p$ divides $n$ for some odd prime $p$ then the dihedral group of order $2p$ will work. Thus $n$ is odd, and the easiest example here is now the non-abelian group of order $p^3$ and exponent $p$, all upper unitriangular matrices of degree $3$ over $\mathbb F_p$. (Of course this example works for $n$ not a $2$-power as well, but dihedral groups are easier.)

(As an aside, we see why three consecutive numbers break this counterexample, as it becomes the case $n=1$ and $n=2$ again.)

Edit: Let's also show the three consecutive integers in a slightly more conceptual way. The two commutations conditions above yield $$ a^{n+1}b^{n+1}=(ab)^{n+1}=(ab)^n ab=a^nb^nab.$$ Cancelling outer terms from the outer two equal expressions yields $ab^n=b^na$.

Thus for all $a,b\in G$, $a$ commutes with $b^n$. Now, if we also have $(ab)^{n+2}=a^{n+2}b^{n+2}$ then we also have that $a$ and $b^{n+1}$ commute, for all $a,b\in G$. In particular, $a$ commutes with $(b^{-1})^n$ and $b^{n+1}$, so it commutes with their product, $b$.

David A. Craven
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  • For $n\geq 4$, you can also take nonabelian groups of exponent $n-1$ as your basic building blocks, of course. – Arturo Magidin Sep 12 '21 at 20:27
  • @ArturoMagidin If they have exponent $n-1$, don't we get $(ab)^2=a^2b^2$, so they would be abelian? – David A. Craven Sep 12 '21 at 20:29
  • That's why I restricted to $n\geq 4$; for example, a nonabelian group of exponent $3$ satisfies $(ab)^4 = a^4b^4$ and $(ab)^3 = a^3b^3$. – Arturo Magidin Sep 12 '21 at 20:30
  • @ArturoMagidin I think you've misread the first line. I'm assuming $n$ and $n+1$, not $n$ and $n-1$. Nonabelian groups of exponent $n-1$ cannot satisfy $(ab)^{n+1}=a^{n+1}b^{n+1}$. – David A. Craven Sep 12 '21 at 20:36
  • I think I just wasn't clear, my apologies. It amounts to the same thing, depending whether you want "the $n$ I have and the next one" or "two consecutive $n$s, one of them being the $n$ I have." – Arturo Magidin Sep 12 '21 at 20:40
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    @ArturoMagidin Right. Yes, you can use any group of exponent dividing $n$ for the equations with $n$ and $n+1$. I just wanted to provide an example of such a group to prove that the set of non-abelian groups of exponent dividing $n>2$ is non-empty. – David A. Craven Sep 12 '21 at 20:50