# 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 $F:G/\mathrm{ker}(f)\to\mathrm{im}(f)$ defined by $F(g\mathrm{ker}(f))=f(g)$ is a well-defined group isomorphism. Note that $\mathrm{ker}(f)$ is always normal in $G$.

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

 $\mathrm{norm}(G)=\{N\subseteq G\ |\ N\mbox{ is normal subgroup}\};$
 $\mathrm{h.im}(G)=\{[H,f]\ |\ (H,f)\mbox{ is a homomorphic image of }G\}.$

Function $T:\mathrm{norm}(G)\to\mathrm{h.im}(G)$ 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 $N=\mathrm{ker}(f)$. 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

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

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 $x\in N^{\prime}=\mathrm{ker}(\pi_{N^{\prime}})$ 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. $\square$

Title correspondence between normal subgroups and homomorphic images CorrespondenceBetweenNormalSubgroupsAndHomomorphicImages 2013-03-22 19:07:11 2013-03-22 19:07:11 joking (16130) joking (16130) 4 joking (16130) Theorem msc 20A05 msc 13A15 HomomorphicImageOfGroup