Group Homomorphism of Product with Inverse

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Theorem

Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.


Then:

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


Proof

Result $(1)$:

\(\displaystyle \map \phi {x \circ y^{-1} } * \map \phi y\) \(=\) \(\displaystyle \map \phi {\paren {x \circ y^{-1} } \circ y}\) Definition of Group Homomorphism
\(\displaystyle \) \(=\) \(\displaystyle \map \phi x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \phi {x \circ y^{-1} } * e_H\) \(=\) \(\displaystyle \map \phi x * \paren {\map \phi y}^{-1}\) multiplying on the right by $\paren {\map \phi y}^{-1}$


Result $(2)$:

\(\displaystyle \map \phi y * \map \phi {y^{-1} \circ x}\) \(=\) \(\displaystyle \map \phi {y \circ \paren {y^{-1} \circ x} }\) Definition of Group Homomorphism
\(\displaystyle \) \(=\) \(\displaystyle \map \phi x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle e_H * \map \phi {y^{-1} \circ x}\) \(=\) \(\displaystyle \paren {\map \phi y}^{-1} * \map \phi x\) multiplying on the left by $\paren {\map \phi y}^{-1}$

$\blacksquare$


Sources