Group Isomorphism Preserves Identity

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 1

An group isomorphism is by definition a group epimorphism.

The result follows from Epimorphism Preserves Identity.

$\blacksquare$

Proof 2

\(\ds \map \phi {e_G}\)

\(=\)

\(\ds \map \phi {e_G \circ e_G}\)

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

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 28 \gamma$