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# fundamental homomorphism theorem

The following theorem is also true for rings (with ideals instead of normal subgroups) or modules (with submodules instead of normal subgroups).

###### 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

$\ker(h)=\{xN\mid x\in\ker(f)\}=\ker(f)/N.$ |

∎

A consequence of this is: If $f\colon G\to H$ is onto with $\ker(f)=N$, then $G/N$ and $H$ are isomorphic.

## Mathematics Subject Classification

20A05*no label found*

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