theorem 1   Let $G,H$ be groups, $f\colon G \to H$ a homomorphism, and let $N$ be a normal subgroup of $G$ contained in $\ker(f)$ . Then there exists a unique homomorphism $h\colon G/N \to H$ so that $h \circ \varphi=f$ , where $\varphi$ denotes the canonical homomorphism from $G$ to $G/N$ .

Furthermore, if $f$ is onto, then so is $h$ ; and if $\ker(f)=N$ , then $h$ is injective.

Proof. We'll first show the uniqueness. Let $h_1, h_2\colon G/N \to H$ functions such that $h_1 \circ \varphi=h_2 \circ \varphi$ . For an element $y$ in $G/N$ there exists an element $x$ in $G$ such that $\varphi(x)=y$ , so we have $$ h_1(y)=(h_1 \circ \varphi)(x)=(h_2 \circ \varphi)(x)=h_2(y $$ for all $y \in G/N$ , thus $h_1=h_2$ .

Now we define $h: G/N \to H,\; h(gN)=f(g)\;\forall\;g \in G$ . We must check that the definition is independent of the given representative; so let $gN=kN$ , or $k \in gN$ . Since $N$ is a subset of $\ker(f)$ , $g^{-1}k \in N$ implies $g^{-1}k \in \ker(f)$ , hence $f(g)=f(k)$ . Clearly $h \circ \varphi=f$ .

Since $x \in \ker(f)$ if and only if $h(xN)=1_H$ , we have $$ \ker(h)=\{xN \mid x \in \ker(f)\}=\ker(f)/N $$