Exercise 1.6.18 ($g \mapsto g^{2}$ is a homomorphism if and only if $G$ is abelian)

Question

Let $G$ be any group. prove that the map from $G$ to itself defined by $g \mapsto g^{2}$ is a homomorphism if and only if $G$ is abelian.

Richard Ganaye · Accepted Answer

\begin{proof} 
Suppose that $G$ is abelian. Then for all $g, h \in G$,
\begin{align*}
\varphi(gh) &= (gh)^{2}\
&= g (h g)h \
&= g (g h) h & 	ext{ (since $G$ is abelian)}\
&= g^2 h^2\
&= \varphi(g) \varphi(h).
\end{align*}
So $\varphi$ is a homomorphism.

\bigskip
Conversely, suppose that $\varphi$ is a homomorphism. Then for all $g,h \in G$,
\begin{align*}
(h g)^2&=  \varphi(hg) \
&= \varphi(h) \varphi(g))& 	ext{(since $\varphi$ is a homomorphism)}\
&=h^2 g^2.
\end{align*}
So $hghg = hh gg$. Multiplying at the left by $h^{-1}$ and at the right by $g^{-1}$, we obtain $gh = hg$. Therefore $G$ is abelian.

The map from $G$ to itself defined by $g \mapsto g^{2}$ is a homomorphism if and only if $G$ is abelian.

\end{proof}

Exercise 1.6.18 ($g \mapsto g^{2}$ is a homomorphism if and only if $G$ is abelian)

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Exercise 1.6.18 ($g \mapsto g^{2}$ is a homomorphism if and only if $G$ is abelian)

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