Prove a Group is Abelian if $(ab)^2=a^2b^2$

Problem 401

Let $G$ be a group. Suppose that
\[(ab)^2=a^2b^2\]
for any elements $a, b$ in $G$. Prove that $G$ is an abelian group.

 
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Proof.

To prove that $G$ is an abelian group, we need
\[ab=ba\]
for any elements $a, b$ in $G$.

By the given relation, we have
\[(ab)^2=a^2b^2.\]
The left hand side is
\[(ab)^2=(ab)(ab),\]
and thus the relation becomes
\[(ab)(ab)=a^2b^2.\]
Equivalently, we can express it as
\[abab=aabb.\]
Multiplying by $a^{-1}$ on the left and $b^{-1}$ on the right, we obtain
\begin{align*}
a^{-1}(abab)b^{-1}=a^{-1}(aabb)b^{-1}.
\end{align*}
Since $a^{-1}a=e, bb^{-1}=e$, where $e$ is the identity element of $G$, we have
\[ebae=eabe.\]
Since $e$ is the identity element, it yields that
\[ba=ab\]
and this implies that $G$ is an abelian group.

Related Question.

I wondered what happens if I change the number $2$ in $(ab)^2=a^2b^2$ into $3$, and created the following problem:

Problem. If $G$ is a group such that $(ab)^3=a^3b^3$ and $G$ does not have an element of order $3$, then $G$ is an abelian group.

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 $3$“.

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