PlanetMath: fundamental homomorphism theorem

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (232)
Orphanage
Unclass'd
Unproven (519)
Corrections (21)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

fundamental homomorphism theorem

(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 be groups, $f\colon G \to H$ a homomorphism, and let be a normal subgroup of 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 to .

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

there exists an element

such that $\varphi(x)=y$ , so we have

$\displaystyle h_1(y)=(h_1 \circ \varphi)(x)=(h_2 \circ \varphi)(x)=h_2(y)$

for all $y \in G/N$ , thus

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 , or $k \in gN$ . Since is a subset of $\ker(f)$ , $g^{-1}k \in N$ implies $g^{-1}k \in \ker(f)$ , hence . Clearly $h \circ \varphi=f$ .