Group Homomorphism Preserves Identity/Proof 1

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:

$\map \phi {e_G} = e_H$

$\ds \map \phi {e_G}$

$=$

$\ds \map \phi {e_G \circ e_G}$

Definition of Identity Element

$\ds $

$=$

$\ds \map \phi {e_G} * \map \phi {e_G}$

Definition of Morphism Property

That is:

$\ds \map \phi {e_G} * e_H$

$=$

$\ds \map \phi {e_G} * \map \phi {e_G}$

Definition of Identity Element

$\ds \leadsto \ \ $

$\ds e_H$

$=$

$\ds \map \phi {e_G}$

$\blacksquare$

1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Morphisms
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 47.1$ Homomorphisms and their elementary properties
1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms