Group Homomorphism Preserves Inverses

Theorem
Let $$\phi: \left({G, \circ}\right) \to \left({H, *}\right)$$ be a group homomorphism.

Let:
 * $$e_G$$ be the identity of $$G$$;
 * $$e_H$$ be the identity of $$H$$.

Then:
 * $$\forall x \in G: \phi \left({x^{-1}}\right) = \left({\phi \left({x}\right)}\right)^{-1}$$.

Proof 1
The result follows directly from the morphism property of $$\circ$$ under $$\phi$$.

Let:
 * $$e_G$$ be the identity of $$G$$;
 * $$e_H$$ be the identity of $$H$$.

Let $$x \in G$$.

Then:

$$ $$ $$

So by definition of inverse, $$\phi \left({x^{-1}}\right)$$ is the inverse of $$\phi \left({x}\right)$$.

Proof 2
A direct application of Homomorphism to Group Preserves Identity and Inverses.

Proof 3
From Homomorphism of Product with Inverse, we have:
 * $$\forall x, y \in G: \phi \left({x \circ y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}$$

Putting $$x = e_G$$ and $$y = x$$ we have:

$$ $$ $$ $$