Group Isomorphism Preserves Inverses/Proof 2

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Theorem

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

Let:

$e_G$ be the identity of $\struct {G, \circ}$
$e_H$ be the identity of $\struct {H, *}$.


Then:

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


Proof

Let $g \in G$.

\(\ds \map \phi g * \map \phi {g^{-1} }\) \(=\) \(\ds \map \phi {g \circ g^{-1} }\) Definition of Group Isomorphism
\(\ds \) \(=\) \(\ds \map \phi {e_G}\) Definition of Inverse Element
\(\ds \) \(=\) \(\ds e_H\) Group Isomorphism Preserves Identity

It follows from Inverse in Group is Unique that $\map \phi {g^{-1} }$ is the unique inverse element of $\map \phi g$ in $\struct {H, *}$.

That is:

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

$\blacksquare$


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