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Let $G$ be a finite group whose order is not divisible by $3$. suppose $(ab)^3 = a^3 b^3$ for all $a,b \in G$. Prove that $G$ must be abelian.

Let$ $G be a finite group of order $n$. As $n$ is not divisible by $3$ ,$3$ does not divide $n$ thus $n$ should be relatively prime to $n$. that is gcd of an $n$ should be $1$. $n = 1 ,2 ,4 ,5 ,7 ,8 ,10 ,11, 13 ,14 ,17,...$ further I know that all groups upto order $5$ are abelian and every group of prime order is cyclic. when it remains to prove the numbers which are greater than $5$ and not prime are abelian.

Am I going the right way? please suggest me the proper way to prove this.

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  • $\begingroup$ This is definitely the wrong appriach. Play around with the identities for a while $\endgroup$ – Pax Kivimae Oct 30 '15 at 16:04
  • $\begingroup$ The only group of prime order $p$ is $\mathbb{Z}_p$, and there are certainly nonabelian groups of order $> 5$. $\endgroup$ – anomaly Oct 30 '15 at 16:04
  • $\begingroup$ No, you're not going the right way. The right way would be to somehow leverage $(ab)^3=a^3b^3$ into $ab=ba$, using algebra. I would assume that you at some point have to use that the order is not divisible by $3$, but it will probably be to show that no third-power (except of the identity) is the identity or something of the sort. Not to specifically exclude groups of order $12$. $\endgroup$ – Arthur Oct 30 '15 at 16:07
  • $\begingroup$ " it remains to prove the numbers which are greater than 5 and not prime are abelian" Wow. Good luck proving that. $\endgroup$ – fleablood Oct 30 '15 at 16:59
  • $\begingroup$ I guess nothing is wrong with you.@fleablood $\endgroup$ – Kavita Oct 30 '15 at 17:02
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First note that given condition says that $ f: G \to G$ defined as $x \to x^3$ is an injective homomorphism of $G$.

Further Note that $$ \forall a,b \in G: \quad ababab = (ab)^{3} = a^{3} b^{3} = aaabbb. $$ Hence, $$ \forall a,b \in G: \quad baba = aabb, \quad \text{or equivalently}, \quad (ba)^{2} = a^{2} b^{2}. $$ Using this fact, we obtain \begin{align} \forall a,b \in G: \quad (ab)^{4} &= [(ab)^{2}]^{2} \\ &= [b^{2} a^{2}]^{2} \\ &= (a^{2})^{2} (b^{2})^{2} \\ &= a^{4} b^{4} \\ &= aaaabbbb. \end{align}

  • On the other hand, \begin{align} \forall a,b \in G: \quad (ab)^{4} &= abababab \\ &= a (ba)^{3} b \\ &= a b^{3} a^{3} b \\ &= abbbaaab. \end{align}

  • Hence, for all $ a,b \in G $, we have $ aaaabbbb = abbbaaab $, which yields $$ f(ab) = a^{3} b^{3} = b^{3} a^{3} = f(ba). $$ As $ f $ is injective, we conclude that $ ab = ba $ for all $ a,b \in G $.Hence $G$ is an abelian group


Added: I think it's worth mentioning that there exist nonabelian group $G$ for which $x \to x^3$ is a group homomorphism.Smallest such example is Heisenberg group of order $27$ which can be thought of as all $3 \times 3$ upper diagonal matrices with $1's$ on the diagonal and other entries in the field of order $3$.As $G$ is of exponent 3 ( i.e. $x^3=1$ for all $x \in G$) hence $f$ is a homomorphism and $G$ is clearly non abelian because for example following two matrices don't commute: $$\left(\begin{array}{ccc} 1 & 1 & 0\\ 0 & 1 & 2\\ 0 & 0 & 1 \end{array}\right),$$ $$\left(\begin{array}{ccc} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{array}\right)$$

In particular this also shows that the condition that $3$ does not divide order($G$) is necessary.

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  • $\begingroup$ Why is $f$ an homomorphism? That is, why $a^3b^3=(ab)^3$? $\endgroup$ – ajotatxe Oct 30 '15 at 16:14
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    $\begingroup$ $x\mapsto x^3$ is injective because the order of the group is not divisible by 3. $\endgroup$ – Laars Helenius Oct 30 '15 at 16:19
  • $\begingroup$ @LaarsHelenius Correct. $\endgroup$ – Arpit Kansal Oct 30 '15 at 16:19
  • $\begingroup$ @ajotatxe probably I don't understand your question.$f$ is clearly a group morphism.It's a given condition $\endgroup$ – Arpit Kansal Oct 30 '15 at 16:21
  • $\begingroup$ @Arpit kansal. Thanks for the "easy to understand" proof, its worth for many upvotes. $\endgroup$ – Kavita Oct 30 '15 at 17:04

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