Elements of Group with Equal Images under Homomorphisms form Subgroup

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

Let $f: G \to H$ and $g: G \to H$ be group homomorphisms.

Then the set:

$S = \set {x \in G: \map f x = \map g x}$

is a subgroup of $G$.


Let the identities of $\struct {G, \circ}$ and $\struct {H, *}$ be $e_G$ and $e_H$ respectively.

By Homomorphism to Group Preserves Identity:

$\map f {e_G} = \map g {e_G} = e_H$

Thus $e_G \in S$, and so $S \ne \O$.

Similarly, from Homomorphism to Group Preserves Inverses, $x \in S \implies x^{-1} \in S$.

Let $x, y \in S$. Then:

\(\ds \map f {x \circ y}\) \(=\) \(\ds \map f x * \map f y\) Morphism Property
\(\ds \) \(=\) \(\ds \map g x * \map g y\) Definition of $S$
\(\ds \) \(=\) \(\ds \map g {x \circ y}\) Morphism Property

Thus $x \circ y \in S$.

So, by the Two-Step Subgroup Test:

$S \le G$