Kernel of Group Homomorphism is Subgroup

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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, $\map \ker \phi \le S$.

$\blacksquare$


Also see


Sources

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