# Universal Property of Quotient Group

## Theorem

Let $G$ and $H$ be groups.

Let $N \trianglelefteq G$ be an normal subgroup.

Let $\pi: G \to G / N$ be the quotient epimorphism.

Let $f: G \to H$ be a group homomorphism with $N \subseteq \ker f$.

Then there exists a unique group homomorphism $\overline f: G / N \to H$ such that $f = \overline f \circ \pi$.

$\xymatrix{ G \ar[d]^\pi \ar[r]^{\forall f} & H \\ G/N \ar[ru]_{\exists ! \bar f} }$

## Proof

### Existence

Let $\sim$ denote (left) congruence modulo $N$.

From Congruence Modulo Subgroup is Equivalence Relation, $\sim$ is an equivalence relation on $X$.

For all $g \in G$, let $\eqclass g \sim$ denote the equivalence class of $g$ under $\sim$.

In particular, since $N \subseteq \ker f$, we have that $f$ is $\sim$-invariant.

Hence, by Universal Property of Quotient Set, there exists a unique such mapping $\overline f: G / \sim{} \to H$, which is given by:

$\forall g \in G: \map {\overline f} {\eqclass g \sim} = \map f g$

From Universal Property of Quotient Set, $f$ is well-defined.

By definition of (left) coset space, $G / N = G / \sim$.

for all $g \in G$ the equivalence class $\eqclass g \sim$ is the (left) coset of $N$ containing $g$, which is $g N$.

Thus $\overline f$ is given by:

$\forall g \in G: \map {\overline f} {g N} = \map f g$

We have that $\overline f$ is a group homomorphism.

Indeed, for all $x, y \in G$ we have:

 $\ds \map {\overline f} {\paren {x N} \paren {y N} }$ $=$ $\ds \map {\overline f} {\paren {x y} N}$ Definition of Quotient Group $\ds$ $=$ $\ds \map f {x y}$ Definition of $\overline f$ $\ds$ $=$ $\ds \map f x \map f y$ Definition of Group Homomorphism $\ds$ $=$ $\ds \map {\overline f} {x N} \map {\overline f} {y N}$ Definition of $\overline f$

Thus we have shown the existence of such a group homomorphism.

$\Box$

### Uniqueness

By Universal Property of Quotient Set, there exists a unique such mapping $\overline f$.

A fortiori, there exists at most one such group homomorphism.

$\blacksquare$