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
Oct 30, 2015 at 16:04
• The only group of prime order $p$ is $\mathbb{Z}_p$, and there are certainly nonabelian groups of order $> 5$. Oct 30, 2015 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 $12$. Oct 30, 2015 at 16:07
• " it remains to prove the numbers which are greater than 5 and not prime are abelian" Wow. Good luck proving that. Oct 30, 2015 at 16:59
• I guess nothing is wrong with you.@fleablood Oct 30, 2015 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 $a^3b^3=(ab)^3$? Oct 30, 2015 at 16:14
• $x\mapsto x^3$ is injective because the order of the group is not divisible by 3. Oct 30, 2015 at 16:19
• @LaarsHelenius Correct. Oct 30, 2015 at 16:19
• @ajotatxe probably I don't understand your question.$f$ is clearly a group morphism.It's a given condition Oct 30, 2015 at 16:21
• @Arpit kansal. Thanks for the "easy to understand" proof, its worth for many upvotes. Oct 30, 2015 at 17:04

Since this has appeared on the top, here is a shorter proof of the result. It starts the same, as it has to: $$(ab)^3=a^3b^3$$ implies $$baba=aabb$$. But now multiply by $$ba$$ on the left: $$bababa=baaabb.$$ But the left-hand side is $$b^3a^3$$, and we cancel off the first $$b$$ to achieve $$b^2a^3=a^3b^2$$, i.e., for all $$a,b$$, $$a^3$$ commutes with $$b^2$$.

Now we use the fact (mentioned elsewhere) that since $$3\nmid |G|$$, $$x\mapsto x^3$$ is a bijection, and so $$b^2$$ in fact commutes with $$a$$, for all $$a,b\in G$$. Consequently, $$b^2$$ commutes with $$a^2$$ as well. But now we go back to our first statement $$baba=aabb$$: we obtain $$baba=aabb=bbaa.$$ Removing the front $$b$$ and the back $$a$$ yields $$ab=ba$$, as needed.

I have followed the similar method as previous answer with a few changes (without defining the $$f$$, thought it would be easier:) )

As $$(ab)^3=a^3b^3$$ for all $$a,b\in G$$ we have \begin{align*} ababab&=aaabbb\\ \Rightarrow baba&=aabb\\ \Rightarrow (ba)^2&=a^2b^2 \end{align*} Consider, \begin{align*}(ab)^4&=((ab)^2)^2\\ &=(b^2a^2)^2\\ &=a^4b^4 \\ &=aaaabbbb\end{align*} Also \begin{align*} (ab)^4&=abababab\\ &=a(ba)^3b \end{align*} Therefore, we get \begin{align}aaaabbbb&=a(ba)^3b\\ \Rightarrow (ba)^3&=(ab)^3\\ \Rightarrow (ab)^{-3}(ba)^3&=e \end{align} Where $$e$$ is identity in $$G$$ $$\Rightarrow [(ab)^{-1}(ba)]^3=e$$ Now for $$x=(ab)^{-1}(ba)$$ ,$$|x|$$ divides $$3$$ by which $$|x|$$ can be $$3$$ or $$1$$. Now $$|x|$$ can not be $$3$$ (as by Lagrange's theorem if $$|x|$$=3 then $$3$$ divides $$|G|$$ which is not true). Thus, \begin{align*} |x|&=1\\ \Rightarrow x&=e\\ \Rightarrow (ab)^{-1}(ba)&=e \end{align*} Multiplying by $$ab$$ from left, $$\Rightarrow ba=ab$$ for all $$a,b\in G$$. Thus, $$G$$ is abelian.

• What is wrong with it?? Sep 27, 2020 at 17:48
• Are there any mistakes?? Sep 27, 2020 at 17:51
• I guess the reason for the downvote is that the question is 5 years old, with an already decent answer. Sep 30, 2020 at 8:30
• I did it to simplify answer Oct 1, 2020 at 9:27
• I don't immediately see why your answer is "simpler". Certainly, it has the same first step as the other answer. Ideally, you should say why your answer is simpler, rather than making people guess. If you just want someone to check your answer then you should post a question saying "I wrote this proof, can someone check if it is correct:..." Oct 1, 2020 at 9:38

I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 24 on p.48 in Herstein's book.
I solved this problem as follows:

$$G\ni x\mapsto x^3\in G$$ is a homomorphism since $$(ab)^3=a^3b^3$$ for all $$a,b\in G$$ by the assumption of this problem.
Assume that there exists $$a\in G$$ such that $$a\ne e$$ and $$a^3=e$$.
If $$a^2=e$$, then $$a^3=a^2\cdot a=e\cdot a=a\ne e$$.
But $$a^3=e$$ by our assumption.
So, $$a^2\ne e$$ must hold.
So, $$o(a)=3$$.
By a famous theorem, $$o(a)\mid\#G$$.
So, $$3\mid\#G$$.
But $$3\not\mid\#G$$ by the assumption of this problem.
So, there does not exist $$a\in G$$ such that $$a\ne e$$ and $$a^3=e$$.
So, if $$a^3=e$$, then $$a=e$$ must hold.
So, the kernel of the homomorphism $$G\ni x\mapsto x^3\in G$$ is $$\{e\}$$.
So, the homomorphism $$G\ni x\mapsto x^3\in G$$ is injective.
Since $$G$$ is finite by the assumption of this problem, the homomorphism $$G\ni x\mapsto x^3\in G$$ is bijective.

Let $$a,b\in G$$.
$$b(abab)a=(ba)^3=b^3a^3=b(bbaa)a$$ by the assumption of this problem.
So, by the left cancellation law and the right cancellation law, we get $$abab=bbaa$$.
So, $$(ab)^2=b^2a^2$$ for any $$a,b\in G$$.
$$(ab)^4=((ab)^2)^2=(b^2a^2)^2$$.
Let $$A:=b^2$$ and $$B:=a^2$$.
Then, $$(b^2a^2)^2=(AB)^2=B^2A^2=a^4b^4$$.
So, $$(ab)^4=a^4b^4$$ for any $$a,b\in G$$.
$$(ab)^4=a^4b^4=a(a^3b^3)b$$.
$$(ab)^4=a(bababa)b=a(ba)^3b$$.
So, $$a(a^3b^3)b=a(ba)^3b$$.
By the left cancellation law and the right cancellation law, we get $$a^3b^3=(ba)^3$$ for any $$a,b\in G$$.
$$(ba)^3=b^3a^3$$ by the assumption of this problem.
So, $$a^3b^3=b^3a^3$$ for any $$a,b\in G$$.

Let $$a,b\in G$$.
Then, there exist $$a^{'},b^{'}\in G$$ such that $$(a^{'})^3=a,(b^{'})^3=b$$ since $$G\ni x\mapsto x^3\in G$$ is surjective.
$$ab=(a^{'})^3(b^{'})^3=(b^{'})^3(a^{'})^3=ba$$.