If a Group $G$ Satisfies $abc=cba$ then $G$ is an Abelian Group

Problem 306

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.

To show that $G$ is an abelian group we need to show that
\[ab=ba\]
for any elements $a, b\in G$.
There are several cases we need to consider. Let us start with an easy case.
If $a=e$ or $b=e$, then we have $ab=ba$.

The next case to consider is $ab=e$. In this case, we have $b=a^{-1}$, and hence $ba=e=ab$.

The last case is $a\neq e, b\neq e, ab\neq e$.
Since $ab\neq e$, the inverse $(ab)^{-1}$ is not the identity as well.
We use the given relation $abc=cba$ with $c=(ab)^{-1}$. We have
\begin{align*}
e&=ab(ab)^{-1}\\
&=(ab)^{-1}ba \qquad \text{ by the relation (*)}\\
\end{align*}
Multiplying this equality by $ab$ on the left we obtain
\[ab=ba.\]

Therefore, for any elements $a, b\in G$ we have proved $ab=ba$, and thus $G$ is an abelian group.

Prove a Group is Abelian if $(ab)^2=a^2b^2$
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.
Proof.
To prove that $G$ is an abelian group, we need
\[ab=ba\]
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\[\phi(e, g)=g \text{ and } \phi(g, e)=g, \tag{*}\]
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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.
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Let $G$ be a finite group. Let $a, b$ be elements of $G$.
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