The isomorphism theorems

1. If $f:G\rightarrow H$ is a homomorphism, then ${ G\over {ker(f)}}\cong im(f).$

This is easy to see by identifying the image f(g) with the coset gker(f). The map is a well defined isomorphism.

[In other words, we identify $h\in im(f)$ with its inverse image $f^{-1}(h)=\{g\in G:f(g)=h\};$ the latter is simply gker(f), where f(g)=h. In the case of nonlinear maps the preimage f^-1(h) is usually called the level surface corresponding to h.]

The first isomorphism theorem tells us that the factor groups of G over its various normal subgroups are, up to isomorphism, precisely the homomorphic images of G.

2. If $H \leq G$ and $N\unlhd G$, then ${ H\over {H\cap N}}\cong{{HN}\over N}$.

To see this, consider the restriction to H of the natural homomorphism $G\rightarrow G/N$. Its kernel is $H\cap N$. Its image consists of all cosets of N having representatives in H, which is precisely the group (HN)/N. The first isomorphism theorem allows us now to conclude that $H/(H\cap N)$ is isomorphic to (HN)/N.

3. If $H\unlhd G,$ $K\unlhd G,$ and $K\leq G,$ then ${{G/K}\over {H/K}}\cong{G\over H}.$

Define $f:G/K\rightarrow G/H$ by f(xK)=xH. This is a well defined homomorphism whose kernel is H/K and whose image is G/H. The result follows now from the first isomorphism theorem.

4. If $H\unlhd G$, then the map $S\rightarrow S/H$ is a bijection between the set of all subgroups containing H and the set of all subgroups of G/H. Normal subgroups correspond to normal subgroups under this bijection.

This is an important theorem that is self evident. The statement on normality follows from the third isomorphism theorem.