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 $G\mathrm{,}H$ be groups, $f\mathrm{:}G\mathrm{\to }H$ a homomorphism, and let $N$ be a normal subgroup of $G$ contained in $\ker \mathit{}\mathrm{(}f\mathrm{)}$. Then there exists a unique homomorphism $h\mathrm{:}G\mathrm{/}N\mathrm{\to }H$ so that $h\mathrm{\circ }\phi \mathrm{=}f$, where $\phi$ denotes the canonical homomorphism from $G$ to $G\mathrm{/}N$.

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

Proof.

We’ll first show the uniqueness. Let $h_1,h_2:G/N\to H$ functions such that $h_1\circ \phi =h_2\circ \phi$. For an element $y$ in $G/N$ there exists an element $x$ in $G$ such that $\phi (x)=y$, so we have

$$h_1(y)=(h_1\circ \phi )(x)=(h_2\circ \phi )(x)=h_2(y)$$

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 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 $h\circ \phi =f$.

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 $f:G\to H$ is onto with $\ker (f)=N$, then $G/N$ and $H$ are isomorphic.

Title	fundamental homomorphism theorem
Canonical name	FundamentalHomomorphismTheorem
Date of creation	2013-03-22 15:35:06
Last modified on	2013-03-22 15:35:06
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	9
Author	yark (2760)
Entry type	Theorem
Classification	msc 20A05