Let $\phi:G_1\to G_2$ be a group homomorphism. Show $\phi(g^{-1})=(\phi(g))^{-1}$.

Question

Let $G_1$ and $G_2$ be groups and $\phi:G_1\to G_2$ be a group homomorphism. Show $\phi(g^{-1})=(\phi(g))^{-1}$ $\forall g_1\in G_1$.

So far I have:

$\phi(e)=e$ by definition, where $e$ is the identity element. Then $\phi(g_1)\phi(g_1^{-1})=\phi(g_1g_1^{-1})=\phi(e)=e$ since $\phi$ is a group homomorphism.

I know I can get $\phi(g^{-1})=(\phi(g))^{-1}$ $\forall g_1\in G_1$ from that, but I don't know HOW.

Any help/hints would be welcomed. ^_^

Answer 1

You're very nearly there. You have $\phi(g_1)\phi(g_1^{-1}) = e_{G_2}$ but what you're missing is that $G_2$ is a group and hence if two elements multiply to give the identity, they are inverses of each other. Therefore $\phi(g_1^{-1}) = \phi(g_1)^{-1}$.

Answer 2

You have it by definition! $ba=ab=e$ means the same as $a^{-1}=b$; $b^{-1}=a$. That is, you have proved precisely that $\phi(g)\phi(g^{-1})=\phi(g^{-1})\phi(g)=e$, so you're done.

Answer 3

The two answers already present nail it, but I’ll add a mechanical proof.

You have $\phi(g) \, \phi(g^{-1}) = e_2$, now multiply it on the left by $\phi(g)^{-1}$ (the inverse in $G_2$) and you get $\phi(g)^{-1} (\phi(g) \, \phi(g^{-1})) = \phi(g)^{-1} e_2$, which by applying associativity makes $e_2\, \phi(g^{-1}) = \phi(g)^{-1} e_2$. Now just absorb all $e_2$ and you’re there!