Exercise 1.6.17 ($g \mapsto g^{-1}$ 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 $g \mapsto g^{-1}$ is a homomorphism if and only if $G$ is abelian.

Richard Ganaye · Accepted Answer

\begin{proof} We know that in general 
\begin{align}
(gh)^{-1} = h^{-1} g^{-1} 	ext{ for all } g,h \in G.
\end{align}

Suppose that $G$ is abelian. Then for all $g, h \in G$,
\begin{align*}
\varphi(gh) &= (gh)^{-1}\
&= h^{-1} g^{-1}&	ext{(by (1))} \
&= g^{-1} h^{-1} & 	ext{ (since $G$ is abelian)}\
&= \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 &= (h^{-1})^{-1} (g^{-1})^{-1}\
&= \varphi(h^{-1}) \varphi(g^{-1}) \
&= \varphi(h^{-1} g^{-1})& 	ext{(since $\varphi$ is a homomorphism)}\
&= \varphi((gh)^{-1})&(	ext{by (1))}\
&= ((gh)^{-1})^{-1}\
&= gh.
\end{align*}
Therefore $G$ is abelian.

The map from $G$ to itself defined by $g \mapsto g^{-1}$ is a homomorphism if and only if $G$ is abelian.
\end{proof}

Exercise 1.6.17 ($g \mapsto g^{-1}$ is a homomorphism if and only if $G$ is abelian)

Answers

Comments

Exercise 1.6.17 ($g \mapsto g^{-1}$ is a homomorphism if and only if $G$ is abelian)

Answers

Comments

Add answer