correspondence between normal subgroups and homomorphic images

Assume, that $G$ and $H$ are groups. If $f:G\to H$ is a group homomorphism, then the first isomorphism theorem states, that the function $F:G/\ker (f)\to \mathrm{im}(f)$ defined by $F(g\ker (f))=f(g)$ is a well-defined group isomorphism. Note that $\ker (f)$ is always normal in $G$.

This leads to the following question: is there a correspondence between normal subgroups of $G$ and homomorphic images of $G$? We will try to answer this question, but before that, let us introduce some notion.

First of all, homomorphic image $\mathrm{im}(f)$ is not only a subgroup of $H$. Actually homomorphic image contains also some data about homomorphism. This observation leads to the following definition:

Definition. Let $G$ be a group. Pair $(H,f)$ is called a homomorphic image of $G$ iff $H$ is a group and $f:G\to H$ is a surjective group homomorphism. We will say that two homomorphic images $(H,f)$ and $(H^{\prime },f^{\prime })$ of $G$ are isomorphic (or equivalent), if there exists a group isomorphism $F:H\to H^{\prime }$ such that $F\circ f=f^{\prime }$.

It is easy to see, that this isomorphism relation is actually an equivalence relation and thus we may speak about isomorphism classes of homomorphic images (which will be denoted by $[H,f]$ for homomorphic image $(H,f)$). Furthermore, if $N\subset G$ is a normal subgroup, then $(G/N,{\pi }_N)$ is a homomorphic image, where ${\pi }_N:G\to G/N$ is a projection, i.e. ${\pi }_N(g)=gN$. Let

$$\mathrm{norm}(G)=\{N\subseteq G|N\text{ is normal subgroup}\};$$

$$\mathrm{h}.\mathrm{im}(G)=\{[H,f]|(H,f)\text{ is a homomorphic image of }G\}.$$

Proposition. Function $T:\mathrm{norm}(G)\to \mathrm{h}.\mathrm{im}(G)$ defined by $T(N)=[G/N,{\pi }_N]$ is a bijection.

Proof. First, we will show, that $T$ is onto. Let $(H,f)$ be a homomorphic image of $G$. Let $N=\ker (f)$. Then (due to the first isomorphism theorem), there exists a group isomorphism $F:G/N\to H$ defined by $F(gN)=f(g)$. This shows, that

$$f(g)=F(gN)=F({\pi }_N(g))=(F\circ {\pi }_N)(g)$$

and thus $(G/N,{\pi }_N)$ is isomorphic to $(H,f)$. Therefore

$$T(N)=[G/N,{\pi }_N]=[H,f],$$

which completes this part.

Now assume, that $T(N)=T(N^{\prime })$ for some normal subgroups $N,N^{\prime }\in \mathrm{norm}(G)$. This means, that $(G/N,{\pi }_N)$ and $(G/N^{\prime },{\pi }_{N^{\prime }})$ are isomorphic, i.e. there exists a group isomorphism $F:G/N\to G/N^{\prime }$ such that $F\circ {\pi }_N={\pi }_{N^{\prime }}$. Let $x\in N^{\prime }=\ker ({\pi }_{N^{\prime }})$ and denote by $e\in G/N^{\prime }$ the neutral element. Then, we have

$$e={\pi }_{N^{\prime }}(x)=F({\pi }_N(x))$$

and (since $F$ is an isomorphism) this is if and only if $x\in \ker ({\pi }_N)=N$. Thus, we’ve shown that $N^{\prime }\subseteq N$. Analogously (after considering $F^{-1}$) we have that $N\subseteq N^{\prime }$. Therefore $N=N^{\prime }$, which shows, that $T$ is injective. This completes the proof. $\mathrm{\square }$

Title	correspondence between normal subgroups and homomorphic images
Canonical name	CorrespondenceBetweenNormalSubgroupsAndHomomorphicImages
Date of creation	2013-03-22 19:07:11
Last modified on	2013-03-22 19:07:11
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	4
Author	joking (16130)
Entry type	Theorem
Classification	msc 20A05
Classification	msc 13A15
Related topic	HomomorphicImageOfGroup