Group Homomorphism Preserves Identity/Proof 3
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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
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
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Corollary $8.7 \ \text{(i)}$