Group Homomorphism of Product with Inverse

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


 * 1) $$\forall x, y \in G: \phi \left({x \circ y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}$$
 * 2) $$\forall x, y \in G: \phi \left({y^{-1} \circ x}\right) = \left({\phi \left({y}\right)}\right)^{-1} * \phi \left({x}\right)$$

Proof
1. $$\phi \left({x \circ y^{-1}}\right) * \phi \left({y}\right) = \phi \left({\left({x \circ y^{-1}}\right) \circ y}\right) = \phi \left({x}\right)$$

Multiplying on the right by $$\left({\phi \left({y}\right)}\right)^{-1}$$ we get

$$\phi \left({x \circ y^{-1}}\right) * id_H =\phi \left({x}\right)*\left({\phi \left({y}\right)}\right)^{-1}$$

2. $$\phi \left({y}\right) * \phi \left({y^{-1} \circ x}\right) = \phi \left({y \circ \left({y^{-1} \circ x}\right)}\right) = \phi \left({x}\right)$$

Multiplying on the left by $$\left({\phi \left({y}\right)}\right)^{-1}$$ we get

$$e_H * \phi \left({y^{-1} \circ x}\right)= \left({\phi \left({y}\right)}\right)^{-1}*\phi \left({x}\right)$$