# Prove a Group is Abelian if $(ab)^3=a^3b^3$ and No Elements of Order $3$

## Problem 402

Let $G$ be a group. Suppose that we have
$(ab)^3=a^3b^3$ for any elements $a, b$ in $G$. Also suppose that $G$ has no elements of order $3$.

Then prove that $G$ is an abelian group.

## Proof.

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.

(For a proof of this problem, see the post “Prove a group is abelian if $(ab)^2=a^2b^2$“.)

I wondered what happens if I change $2$ to $3$, and that’s how I made this problem.

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