Kernel of Group Homomorphism is Subgroup

Theorem
The kernel of a group homomorphism is a subgroup of its domain:

$$\mathrm{ker} \left({\phi}\right) \le \mathrm{Dom} \left({\phi}\right)$$

Proof
Let $$\phi: \left({G, \circ}\right) \to \left({H, *}\right)$$ be a group homomorphism.


 * From Homomorphism to Group Preserves Identity and Inverses, $$\phi \left({e_G}\right) = e_H$$, so $$e_G \in \mathrm{ker} \left({\phi}\right)$$.

Therefore $$\mathrm{ker} \left({\phi}\right) \ne \varnothing$$.


 * Let $$x, y \in \mathrm{ker} \left({\phi}\right)$$, so that $$\phi \left({x}\right) = \phi \left({y}\right) = e_H$$. Then:

So $$x^{-1} \circ y \in \mathrm{ker} \left({\phi}\right)$$, and from the One-step Subgroup Test, $$\mathrm{ker} \left({\phi}\right) \le S$$.