Group Homomorphism Preserves Identity
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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:
Then:
- $\map \phi {e_G} = e_H$
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 |
$\blacksquare$
Proof 2
A direct application of Homomorphism to Group Preserves Identity.
$\blacksquare$
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:
\(\ds \map \phi {e_G}\) | \(=\) | \(\ds \map \phi {x \circ x^{-1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x * \paren {\map \phi x}^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e_H\) |
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 60 \alpha$