# To show that group G is abelian if $(ab)^3 = a^3 b^3$ and the order of $G$ is not divisible by 3

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.

LetG 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.

• This is definitely the wrong appriach. Play around with the identities for a while – Pax Kivimae Oct 30 '15 at 16:04
• The only group of prime order $p$ is $\mathbb{Z}_p$, and there are certainly nonabelian groups of order $> 5$. – anomaly Oct 30 '15 at 16:04
• 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 $12$. – Arthur Oct 30 '15 at 16:07
• " it remains to prove the numbers which are greater than 5 and not prime are abelian" Wow. Good luck proving that. – fleablood Oct 30 '15 at 16:59
• I guess nothing is wrong with you.@fleablood – Kavita Oct 30 '15 at 17:02

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.

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