# Homomorphism of Generated Group

## 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:

$\phi = \psi$

$\blacksquare$