Group extension problem

Statement

The group extension problem for two groups $N$ and $H$, is the problem of finding all groups $G$ with $N$ as a normal subgroup of $G$, and the quotient group $G/N$ isomorphic to $H$.

Congruence classes formulation

In this formulation, we're thinking of $N$ and $H$ as specific groups, and looking at short exact sequences:

$1\to N\to G\to H\to 1$

where two short exact sequences:

$1\to N\to G_1\to H\to 1$

and:

$1\to N\to G_2\to H\to 1$

are equivalent if there is an isomorphism from $G_1$ to $G_2$ that induces the identity map on $N$ and on $H$.

Formulation upto automorphisms

This is a more general formulation, where we declare two short exact sequences:

$1\to N\to G_1\to H\to 1$

and:

$1\to N\to G_2\to H\to 1$

are equivalent if there is an isomorphism from $G_1$ to $G_2$ that induces an automorphism on $N$ and an automorphism on $H$.

The set of group extensions upto automorphisms of $N$ and $H$ is a quotient of the set of all group extensions, by an equivalence relation.

Classifying group extensions for an abelian normal subgroup

If $N$ is an abelian group, then there is a procedure to classify all group extensions with normal subgroup $N$.

Finding a list of possible actions

The first step is to note that the quotient group acts on the normal subgroup. In other words, given any group $G$ with a specified normal subgroup $N$ and a quotient group $H$, there is a homomorphism:

$\phi :H\to \mathrm{A}\mathrm{u}\mathrm{t}(N)$

This automorphism is determined by the congruence class of the extensions, so we can partition all congruence classes of extensions as a disjoint union of congruence classes of extensions corresponding to elements of $\mathrm{H}\mathrm{o}\mathrm{m}(H,\mathrm{A}\mathrm{u}\mathrm{t}(N))$.

If we're looking at extensions modulo equivalence upto automorphisms, then an extension doesn't define a unique map $\phi :H\to \mathrm{A}\mathrm{u}\mathrm{t}(N)$. Rather, the map is defined uniquely up to pre-composition with automorphisms of $H$ and conjugation by automorphisms of $N$. Thus, when classifying equivalence classes of extensions in this fashion, it suffices to consider equivalence classes of elements in $\mathrm{H}\mathrm{o}\mathrm{m}(H,\mathrm{A}\mathrm{u}\mathrm{t}(N))$.

Finding all congruence classes for a given action

Having fixed a homomorphism:

$\phi :H\to \mathrm{A}\mathrm{u}\mathrm{t}(N)$

the next step is to find all congruence classes of extensions that give rise to this homomorphism. This is, in fact, described using the second cohomology group $H^2(H;N)$. There are a number of shortcuts to computing this group when $H$ or $N$ have special structure (for instance second cohomology group for trivial group action of finite cyclic group on finite cyclic group).

Multiple congruence classes of extensions may again be equivalent upto automorphisms, so the set of equivalence classes of extensions is not necessarily a group.