Exercise 1.6.22 (If $A$ is abelian, $a \mapsto a^{-1}$ is an isomorphism)

Question

Let $A$ be an abelian group and fix some $k \in \mathbb{Z}$. prove that the map $a \mapsto a^k$ is a homomorphism from $A$ to itself. If $k = -1$ prove that this homomorphism is an isomorphism (i.e., is an automorphism of $A$).

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
Let $k \in \Z$ be some fixed integer and consider the map
$$
\varphi_k
\left\{
\begin{array}{ccl}
A & 	o & A\
a & \mapsto & a^k.
\end{array}
ight.
$$
Since $A$ is abelian, for all $a,\, b \in A$, $(a b)^k = a^k b^k$.

(If $k = 0$, then $(ab)^0 = 1_A = a^0 b^0$. Assume that $(ab)^k = a^k b^k$ for some $k\geq 0$, then $(ab)^{k+1} = (ab)^k ab = a^k b^k a b =a^k a b^k b = a^{k+1} b^{k+1}$, so $(ab)^k = a^k b^k$ is true for every integer $k\geq 0$. Finally, if $k\geq 0$,  $(ab)^{-k} = [(ab)^k]^{-1} = (a^k b^k)^{-1} = [(b^k)]^{-1} [a^k]^{-1} = [a^k]^{-1}[(b^k)]^{-1} = a^{-k} b^{-k}$, so $(ab)^k = a^k b^k$ is true for every integer $k \in \Z$.)

This shows that
$$\varphi(ab) = (ab)^k = a^k b^k = \varphi(a) \varphi(b),$$
so $\varphi$ is a homomorphism.

\bigskip
For all $a \in A$, $\varphi_{-1} (\varphi_{-1}(a)) = (a^{-1})^{-1} = a$, so
$$\varphi_{-1} \circ \varphi_{-1} = \mathrm{Id}_A.$$
This shows that $\varphi_{-1}$ is an involution, therefore $\varphi_{-1}$ is bijective, and $\varphi_{-1}$ is an isomorphism.
\end{proof}

Exercise 1.6.22 (If $A$ is abelian, $a \mapsto a^{-1}$ is an isomorphism)

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Exercise 1.6.22 (If $A$ is abelian, $a \mapsto a^{-1}$ is an isomorphism)

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