Group Isomorphism Preserves Inverses/Proof 2
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Theorem
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group isomorphism.
Let:
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:
- $\forall g \in G: \map \phi {g^{-1} } = \paren {\map \phi g}^{-1}$
$\blacksquare$
Sources
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.5$: Theorem $4$