Kernel of Group Homomorphism Corresponds with Normal Subgroup of Domain

Theorem
Let $\left({G, \circ}\right)$ and $\left({H, *}\right)$ be groups.

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

Let $\ker \left({\phi}\right)$ be the kernel of $\phi$.

Then there exists $N \triangleleft G$, a normal subgroup of $G$ such that:
 * $N = \ker \left({\phi}\right)$

Conversely, let $N \triangleleft G$ be normal subgroup of $G$.

Then there exists $\phi: \left({G, \circ}\right) \to \left({H, *}\right)$, a group homomorphism, whose kernel $\ker \left({\phi}\right)$ is such that:
 * $\ker \left({\phi}\right) = N$

Proof
The first statement is Kernel is Normal Subgroup of Domain:
 * The kernel of $\phi$ is a normal subgroup of the domain of $\phi$:


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

The second statement is Natural Epimorphism to Quotient Group:
 * The mapping $\phi: G \to G / N$, defined as:
 * $\phi: G \to G / N: \phi \left({x}\right) = x N$
 * is a group epimorphism, and its kernel is $N$.