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.
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Proof.
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.
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