For an abelian group $G=G'$, the endomorphism $x \mapsto x^{-1}$ is both monic and epic

Question

Let $G=G'$ be any abelian group. Define $\phi : G \rightarrow G'$ by $\phi(x) = x^{-1}$. Show that $\phi$ is a monomorphism and an epimorphism.

Answer 1

What you are asked to show is that $\phi$ is a bijective homomorphism, i.e., an isomorphism.

Now, if $\phi(x) = x^{-1}$, it is also the case that $\phi(x^{-1}) = (x^{-1})^{-1} = x$.

We then see that $\phi(\phi(x)) = x$, and hence $\phi = \phi^{-1}$: $\phi$ is its own inverse, and is thereby both injective and surjective.

To convince you further that $\phi$ is surjective, note that we have $G = G'$, and $G$ is a group, and hence closed under inversion; by the definition of a group, the inverse of every element in $G$ is again in $G$. Hence every element in $G' = G$ is the inverse of an element in $G$. That means for every $y \in G'$, there exists an $x \in G$ such that $\phi(x) = y$. That's an explicit satisfaction of surjectivity.

So if you've already shown that $\phi$ satisfies the homomorphism property: $$\phi(xy) = \phi(x)\phi (y)$$ (here we need that $G = G'$ is abelian), you can conclude that $\phi$ is both a monomorphism and an epimorphism.