Kernel 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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \phi \left({x^{-1} \circ y}\right)\)	\(=\)	\(\displaystyle \phi \left({x^{-1} }\right) * \phi \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Morphism Property
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\phi \left({x}\right)}\right)^{-1} * \phi \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Homomorphism with Identity Preserves Inverses
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e_T^{-1} * e_T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $x, y \in \ker \left({\phi}\right)$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e_T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Behaviour of identity

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

$\blacksquare$

Also see

• Kernel is Normal Subgroup of Domain

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 7.4$
• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.10$: Theorem $22$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 47.7$