Group Homomorphism Preserves Inverses/Proof 1

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Theorem

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.

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

Let:

$e_G$ be the identity of $G$
$e_H$ be the identity of $H$


Then:

$\forall x \in G: \map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$


Proof

Let $x \in G$.


Then:

\(\displaystyle \map \phi x * \map \phi {x^{-1} }\) \(=\) \(\displaystyle \map \phi {x \circ x^{-1} }\) Definition of Group Homomorphism
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {e_G}\) Definition of Inverse Element
\(\displaystyle \) \(=\) \(\displaystyle e_H\) Group Homomorphism Preserves Identity

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

$\Box$


Similarly:

\(\displaystyle \map \phi {x^{-1} } * \map \phi x\) \(=\) \(\displaystyle \map \phi {x^{-1} \circ x}\) Definition of Group Homomorphism
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {e_G}\) Definition of Inverse Element
\(\displaystyle \) \(=\) \(\displaystyle e_H\) Group Homomorphism Preserves Identity

So, again by definition, $\map \phi {x^{-1} }$ is the left inverse of $\map \phi x$.

$\Box$


Finally, as $\map \phi {x^{-1} }$ is both:

a left inverse of $\map \phi x$

and:

a right inverse of $\map \phi x$

it is by definition an inverse.

From Inverse in Group is Unique, $\map \phi {x^{-1} }$ is the only such element.

Hence the result.

$\blacksquare$


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