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.

Eckmann–Hilton Argument: Group Operation is a Group Homomorphism
Let $G$ be a group with the identity element $e$ and suppose that we have a group homomorphism $\phi$ from the direct product $G \times G$ to $G$ satisfying
\[\phi(e, g)=g \text{ and } \phi(g, e)=g, \tag{*}\]
for any $g\in G$.
Let $\mu: G\times G \to G$ be a map defined […]

If a Group $G$ Satisfies $abc=cba$ then $G$ is an Abelian Group
Let $G$ be a group with identity element $e$.
Suppose that for any non identity elements $a, b, c$ of $G$ we have
\[abc=cba. \tag{*}\]
Then prove that $G$ is an abelian group.
Proof.
To show that $G$ is an abelian group we need to show that
\[ab=ba\]
for any […]

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

If Every Nonidentity Element of a Group has Order 2, then it’s an Abelian Group
Let $G$ be a group. Suppose that the order of nonidentity element of $G$ is $2$.
Then show that $G$ is an abelian group.
Proof.
Let $x$ and $y$ be elements of $G$. Then we have
\[1=(xy)^2=(xy)(xy).\]
Multiplying the equality by $yx$ from the right, we […]

Fundamental Theorem of Finitely Generated Abelian Groups and its application
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

Quotient Group of Abelian Group is Abelian
Let $G$ be an abelian group and let $N$ be a normal subgroup of $G$.
Then prove that the quotient group $G/N$ is also an abelian group.
Proof.
Each element of $G/N$ is a coset $aN$ for some $a\in G$.
Let $aN, bN$ be arbitrary elements of $G/N$, where $a, b\in […]

Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group
Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order.
(a) Prove that $T(A)$ is a subgroup of $A$.
(The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion […]

Pullback Group of Two Group Homomorphisms into a Group
Let $G_1, G_1$, and $H$ be groups. Let $f_1: G_1 \to H$ and $f_2: G_2 \to H$ be group homomorphisms.
Define the subset $M$ of $G_1 \times G_2$ to be
\[M=\{(a_1, a_2) \in G_1\times G_2 \mid f_1(a_1)=f_2(a_2)\}.\]
Prove that $M$ is a subgroup of $G_1 \times G_2$.
[…]

## 1 Response

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