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I have the following question I have to prove/disprove:

Let $(G, *)$ be a group. If for every $a, b ∈ G$ we have $(a * b)^6 = a^6 * b^6$ then $G$ is commutative.

I tried:

I know that we have this rule in $G$ that an element 'squared' is the identity. So $(a * b)^2 = e$, then $(a * b)^2 = abab$. Thus I was thinking of , $(a * b)^6 = abababababab$??? I am really not sure if this holds.

We also have $e = e * e = a^2b^2$, thus $e = e * e * e * e * e * e= a^6b^6$,

so we must have: $a^6b^6 = abababababab$,

which gives: $a^{-5}a^6b^6b^{-5} = a^{-5}ababababababb^{-5}$ results in $ab = ba$.

Thus, since $a,b∈G$ were arbitrary, $G$ is commutative.

I hope this approach is going right. I am not sure if it is the right way.

Thanks in advance.

Mathlover
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  • Wait, so in addition to $(ab)^6=a^6b^6$ you are also given $a^2=e$ for all $a,b\in G$? – Teddy38 Sep 29 '20 at 14:42
  • I know $(ab)^2=a^2b^2$ holds, but I am snot sure if it is true what I did. – Mathlover Sep 29 '20 at 14:43
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    It would help if you clearly list all assumptions and conditions at the beginning. – Teddy38 Sep 29 '20 at 14:43
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    "I know that we have this rule in $G$ that an element 'squared' is the identity. So $(a * b)^2 = e$, then $(a * b)^2 = abab$." This is not a rule. What I think you're remembering that if you have a group $G$ and for every $x\in G,$ it is true that $x^2=e$ then $G$ is abelian. But it's very important to know that it is not a rule in general about groups. – Chickenmancer Sep 29 '20 at 15:00
  • Thanks for making that clear! – Mathlover Sep 29 '20 at 15:02
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    One approach to every question such as "Is every group satisfying (X) abelian?" is to check counterexamples, i.e., look at non-abelian groups. Here the smallest counterexample is just the smallest non-abelian group... – YCor Sep 29 '20 at 15:42

2 Answers2

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This is not true for $S_3$. It is not commutative but satisfies $x^6=1$.

markvs
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You need at least another condition to conclude that $G$ is abelian, e.g., that $|G|$ and $n(n-1)$ are coprime, or that $(ab)^m=a^mb^m$ for another $m$ coprime to $n$.

References:

If$(ab)^n=a^nb^n$ & $(|G|, n(n-1))=1$ then $G$ is abelian

A group such that $a^m b^m = b^m a^m$ and $a^n b^n = b^n a^n$ ($m$, $n$ coprime) is abelian?

Dietrich Burde
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