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 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 and Inverses, $$f \left({e_G}\right) = g \left({e_G}\right) = e_H$$.

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


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


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

$$ $$ $$

Thus $$x \circ y \in S$$.

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