Group Isomorphism Preserves Identity/Proof 2

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:

$\map \phi {e_G} = e_H$


Proof

\(\ds \map \phi {e_G}\) \(=\) \(\ds \map \phi {e_G \circ e_G}\) Definition of Identity Element
\(\ds \) \(=\) \(\ds \map \phi {e_G} * \map \phi {e_G}\) Definition of Group Isomorphism
\(\ds \) \(=\) \(\ds \map \phi {e_G} * \map \phi {e_G}\) Definition of Group Isomorphism

It follows from Identity is only Idempotent Element in Group that $\map \phi {e_G}$ is the identity of $H$.

That is:

$\map \phi {e_G} = e_H$

$\blacksquare$


Sources

  • 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.5$: Theorem $4$
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