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 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:

For a proof of this problem, see the post “Prove a group is abelian if $(ab)^3=a^3b^3$ and no elements of order $3$“.

The post Prove a Group is Abelian if $(ab)^2=a^2b^2$ first appeared on Problems in Mathematics.

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 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.

The post If a Group $G$ Satisfies $abc=cba$ then $G$ is an Abelian Group first appeared on Problems in Mathematics.

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.

Proof.

$\phi=\mu$

Since $\phi$ is a group homomorphism, for any $g, g’ \in G$ and $h, h’ \in H$, we have
\begin{align*}
\phi( (g,h)(g’,h’))=\phi(g,h)\phi(g’,h’).
\end{align*}
The left hand side is equal to
\[\phi(gg’, hh’),\] and thus we have
\[\phi(gg’, hh’)=\phi(g,h)\phi(g’,h’).\] Setting $g’=h=e$, we have
\begin{align*}
\phi(g, h’)&=\phi(g, e)\phi(e, h’)\\
&= gh’ && \text{ by (*)}\\
&=\mu(g, h’).
\end{align*}
Since this equality holds for any $g \in G$ and $h’\in H$, we obtain
\[\phi=\mu\] as required.

$G$ is an abelian group

Now we prove that $G$ is an abelian group.
Let $g, h \in G$ be any two elements.
Then we have
\begin{align*}
gh&=\phi(e,g)\phi(h,e) && \text{ by (*)}\\
&=\phi(eh, ge) && \text{ since $\phi$ is a homomorphism}\\
&=\phi(h,g)\\
&=\mu(h,g)=hg.
\end{align*}
Thus we have proved that $gh=hg$ for any $g, h \in G$. Thus the group $G$ is abelian.

Combined version

Here is a combined version of the proofs of the two claims at once.
We have for any $g, h\in G$,
\begin{align*}
&\mu(g,h)\\
&=gh=\phi(e,g)\phi(h,e) && \text{ by (*)}\\
&=\phi((e,g)(h,e)) && \text{ since $\phi$ is a homomorphism}\\
&=\phi(eh, ge)\\
&=\phi(h,g) \\
&=\phi(he,eg)=\phi((h,e)(e,g))\\
&=\phi(h,e)\phi(e,g) && \text{ since $\phi$ is a homomorphism}\\
&=hg \qquad \text{ by (*)}\\
&=\mu(h,g)
\end{align*}

From these equalities, we see that $\phi=\mu$ and $G$ is an abelian group.

Remark.

This argument is called the Eckmann–Hilton argument.

The post Eckmann–Hilton Argument: Group Operation is a Group Homomorphism first appeared on Problems in Mathematics.