Elements of Group with Equal Images under Homomorphisms form Subgroup

Theorem
Let $\left({G, \circ}\right)$ and $\left({H, *}\right)$ be groups.

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

Then the set:


 * $S = \left\{{x \in G: f \left({x}\right) = g \left({x}\right)}\right\}$

is a subgroup of $G$.

Proof

 * Let the identities of $\left({G, \circ}\right)$ and $\left({H, *}\right)$ be $e_G$ and $e_H$ respectively.

By Homomorphism to Group Preserves Identity, $f \left({e_G}\right) = g \left({e_G}\right) = e_H$.

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


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


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

Thus $x \circ y \in S$.

So, by the Two-step Subgroup Test, $S \le G$.