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
\[ab=ba\]
for any elements $a, b$ in $G$.
By the given […]

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 […]

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 […]

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 […]

Abelian Groups and Surjective Group Homomorphism
Let $G, G'$ be groups. Suppose that we have a surjective group homomorphism $f:G\to G'$.
Show that if $G$ is an abelian group, then so is $G'$.
Definitions.
Recall the relevant definitions.
A group homomorphism $f:G\to G'$ is a map from $G$ to $G'$ […]

Non-Abelian Simple Group is Equal to its Commutator Subgroup
Let $G$ be a non-abelian simple group. Let $D(G)=[G,G]$ be the commutator subgroup of $G$. Show that $G=D(G)$.
Definitions/Hint.
We first recall relevant definitions.
A group is called simple if its normal subgroups are either the trivial subgroup or the group […]

Every Cyclic Group is Abelian
Prove that every cyclic group is abelian.
Proof.
Let $G$ be a cyclic group with a generator $g\in G$.
Namely, we have $G=\langle g \rangle$ (every element in $G$ is some power of $g$.)
Let $a$ and $b$ be arbitrary elements in $G$.
Then there exists […]

Abelian Normal Subgroup, Intersection, and Product of Groups
Let $G$ be a group and let $A$ be an abelian subgroup of $G$ with $A \triangleleft G$.
(That is, $A$ is a normal subgroup of $G$.)
If $B$ is any subgroup of $G$, then show that
\[A \cap B \triangleleft AB.\]
Proof.
First of all, since $A \triangleleft G$, the […]