Kernel of Group Homomorphism is Subgroup

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


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

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

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

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

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

Then:

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

Also see

 * Kernel is Normal Subgroup of Domain