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:

 $\ds \map \phi {x^{-1} \circ y}$ $=$ $\ds \map \phi {x^{-1} } * \map \phi y$ Definition of Morphism Property $\ds$ $=$ $\ds \paren {\map \phi x}^{-1} * \map \phi y$ Homomorphism with Identity Preserves Inverses $\ds$ $=$ $\ds e_T^{-1} * e_T$ as $x, y \in \map \ker \phi$ $\ds$ $=$ $\ds e_T$ Definition of Identity Element

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

$\blacksquare$