Kernel of Group Homomorphism Corresponds with Normal Subgroup of Domain

Theorem

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.

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

Let $\map \ker \phi$ be the kernel of $\phi$.

Then there exists $N \lhd G$, a normal subgroup of $G$ such that:

$N = \map \ker \phi$

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

Then there exists $\phi: \struct {G, \circ} \to \struct {H, *}$, a group homomorphism, whose kernel $\map \ker \phi$ is such that:

$\map \ker \phi = N$

The kernel of $\phi$ is a normal subgroup of the domain of $\phi$:

$\map \ker \phi \lhd \Dom \phi$

The mapping $\phi: G \to G / N$, defined as:

$\phi: G \to G / N: \map \phi x = x N$

$\blacksquare$