How to show that $\phi(g)=g^{-1}$ is a group homomorphism iff $G$ is abelian?

Question

For a group $(G,\cdot)$, show that $G$ is abelian if and only if the map $\phi:G\to G$ defined by $$\phi(g)=g^{-1}$$ is a group homomorphism.

Answer 1

Suppose $G$ is abelian. You need to show $\phi(gh) = \phi(g) \phi(h)$ for all $g,h \in G$. If you write down the definitions of $\phi(gh)$, $\phi(g)$, and $\phi(h)$, you should be almost done.

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Suppose $\phi$ is a homomorphism. Given any $g,h \in G$, how can you show $gh = hg$ using the definition of $\phi$ being a homomorphism? (The previous part should give you some intuition.)

Answer 2

$\phi$ is a homomorphism $\iff \phi(g^{-1}h^{-1})=\phi (g^{-1})\phi(h^{-1})\iff (g^{-1}h^{-1})^{-1}=gh\iff hg=gh\forall g,h\in G\iff G$ is abelian.

Here I have used the "socks and shoes property" of inverses: $(ab)^{-1}=b^{-1}a^{-1}$.