Group Homomorphism Preserves Identity/Proof 1

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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:

$\map \phi {e_G} = e_H$


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 Morphism Property


That is:

\(\ds \map \phi {e_G} * e_H\) \(=\) \(\ds \map \phi {e_G} * \map \phi {e_G}\) Definition of Identity Element
\(\ds \leadsto \ \ \) \(\ds e_H\) \(=\) \(\ds \map \phi {e_G}\) Cancellation Laws

$\blacksquare$


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