Homomorphism of Generated Group
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Theorem
Let $\struct {G, \circ}$ and $\struct {H, \circ}$ be groups.
Let $\phi: G \to H$ and $\psi: G \to H$ be homomorphisms.
Let $\gen S = G$ be the group generated by $S$.
Let:
- $\forall x \in S: \map \phi x = \map \psi x$
Then:
- $\phi = \psi$
Proof
Let $H = \set {x \in G: \map \phi x = \map \psi x}$.
From Elements of Group with Equal Images under Homomorphisms form Subgroup, $H$ is a subgroup of $G$.
But from the definition of the group generated by $S$, the smallest subgroup that contains $S$ is $G$ itself.
Thus:
- $G = \set {x \in G: \map \phi x = \map \psi x}$
and so:
- $\phi = \psi$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.8$