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Let $A$ be an abelian group and fix some $k \in \mathbb{Z}$. Prove that the map $\varphi : A \rightarrow A$ given by $a \mapsto a^k$ is a homomorphism. If $k = -1$ show that $\varphi$ is an isomorphism (i.e. an automorphism of $A$).

Solution: $\varphi$ is a homomorphism since $$\varphi(ab) = (ab)^k = a^k b^k = \varphi(a) \varphi(b)$$ by Exercise 1.1.24.

Now fix $k = -1$. $\varphi$ is surjective since for all $a \in A$, $a = \varphi(a^{-1})$. Suppose now that $\varphi(a) = \varphi(b)$; then $a^{-1} = b^{-1}$, so that $a = b$ and $\varphi$ is injective. Thus with $k = -1$, $\varphi$ is an automorphism.