Kernel of Group Homomorphism is Subgroup

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


 * $\map \ker \phi \le \Dom \phi$

Proof
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.

From Homomorphism to Group Preserves Identity, $\map \phi {e_G} = e_H$, so $e_G \in \map \ker \phi$.

Therefore $\map \ker \phi \ne \O$.

Let $x, y \in \map \ker \phi$, so that $\map \phi x = \map \phi y = e_H$.

Then:

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

Also see

 * Kernel is Normal Subgroup of Domain