Group Homomorphism Preserves Identity

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Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.

Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.


$e_G$ be the identity of $G$
$e_H$ be the identity of $H$.


$\map \phi {e_G} = e_H$

Proof 1

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


Proof 2

A direct application of Homomorphism to Group Preserves Identity.


Proof 3

From Group Homomorphism of Product with Inverse, we have:

$\forall x, y \in G: \map \phi {x \circ y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$

Putting $x = y$ we have:

\(\ds \map \phi {e_G}\) \(=\) \(\ds \map \phi {x \circ x^{-1} }\)
\(\ds \) \(=\) \(\ds \map \phi x * \paren {\map \phi x}^{-1}\)
\(\ds \) \(=\) \(\ds e_H\)