Let $a, b$ be arbitrary elements of the group $G$. We want to show that $ab=ba$.

By the given relation $(ab)^3=a^3b^3$, we have
\begin{align*}
ababab=a^3b^3.
\end{align*}
Multiplying by $a^{-1}$ on the left and $b^{-1}$ on the right, we obtain
\[baba=a^2b^2,\]
or equivalently we have
\[(ba)^2=a^2b^2 \tag{*}\]
for any $a, b\in G$.

Now we consider $aba^{-1}b^{-1}$ (such an expression is called the commutator of $a, b$).
We have
\begin{align*}
(aba^{-1}b^{-1})^2&=(a^{-1}b^{-1})^2(ab)^2 && \text{by (*)}\\
&=b^{-2}a^{-2}b^2a^2 && \text{by (*)}\\
&=b^{-2}(ba^{-1})^2a^2 && \text{by (*)}\\
&=b^{-2}ba^{-1}ba^{-1}a^2\\
&=b^{-1}a^{-1}ba.
\end{align*}
Hence we have obtained
\[(aba^{-1}b^{-1})^2=b^{-1}a^{-1}ba \tag{**}\]
for any $a, b\in G$.
Taking the square of (**), we obtain
\begin{align*}
(aba^{-1}b^{-1})^4&=(b^{-1}a^{-1}ba)^2\\
&=aba^{-1}b^{-1}. && \text{by (**)}
\end{align*}
It follows that we have
\[(aba^{-1}b^{-1})^3=e,\]
where $e$ is the identity element of $G$.
Since the group $G$ does not have an element of order $3$, this yields that
\[aba^{-1}b^{-1}=e.\]
(Otherwise, the order of the element $aba^{-1}b^{-1}$ would be $3$.)

This is equivalent to
\[ab=ba.\]
Thus, we have obtained $ab=ba$ for any elements $a, b$ in $G$.
Therefore, the group $G$ is abelian.

Related Question.

I came up with this problem when I solved the previous problem:

Problem. Prove that if a group $G$ satisfies $(ab)^2=a^2b^2$ for $a, b \in G$, then $G$ is an abelian group.

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## 1 Response

[…] For a proof of this problem, see the post “Prove a group is abelian if $(ab)^3=a^3b^3$ and no elements of order $3$“. […]