## 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.

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.

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.

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 by

\[\mu(g, h)=gh.\]
(That is, $\mu$ is the group operation on $G$.)

Then prove that $\phi=\mu$.

Also prove that the group $G$ is abelian.