Group extension problem

Statement

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

Congruence classes formulation

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

$1\to N\to G\to Q\to 1$

where two short exact sequences:

$1\to N\to G_1\to Q\to 1$

and:

$1\to N\to G_2\to Q\to 1$

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

Group extensions that are equivalent in this fashion are called congruent group extensions and the equivalence classes are called congruence classes of group extensions.

Formulation upto automorphisms, or "pseudo-congruence"

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

$1\to N\to G_1\to Q\to 1$

and:

$1\to N\to G_2\to Q\to 1$

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

Group extensions that are equivalent in this fashion are called pseudo-congruent group extensions and the equivalence classes are called pseudo-congruence classes of group extensions.

Classifying group extensions for an abelian normal subgroup

If $N$ is an abelian group, then the following 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 $Q$, there is a homomorphism:

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

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 $\mathrm{H}\mathrm{o}\mathrm{m}(Q,\mathrm{A}\mathrm{u}\mathrm{t}(N))$.

Identifying homomorphism types for pseudo-congruence

If we're looking at extensions modulo equivalence up to automorphisms (i.e., pseudo-congruence), then an extension doesn't define a unique map $\phi :Q\to \mathrm{A}\mathrm{u}\mathrm{t}(N)$. Rather, the map is defined uniquely up to pre-composition with automorphisms of $Q$ 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 $\mathrm{H}\mathrm{o}\mathrm{m}(Q,\mathrm{A}\mathrm{u}\mathrm{t}(N))$ under these actions, or rather, under the combined action of $\mathrm{A}\mathrm{u}\mathrm{t}(N)\times \mathrm{A}\mathrm{u}\mathrm{t}(Q)$.

Finding all congruence classes for a given action

What we're trying to do: For a fixed homomorphism

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

, find all the possible congruence classes of group extensions that would induce that homomorphism.

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 $H_{\phi }^2(Q;N)$. There are a number of shortcuts to computing this group when $Q$ or $N$ have special structure (for instance second cohomology group for trivial group action of finite cyclic group on finite cyclic group).

Finding all pseudo-congruence classes for a given action

Multiple congruence classes of extensions may again be equivalent up to automorphisms (i.e., they may be pseudo-congruent). The set of pseudo-congruence classes of extensions for a given action can be obtained from the set of congruence classes as follows: take the equivalence classes under the natural action of the following subgroup of $\mathrm{A}\mathrm{u}\mathrm{t}(N)\times \mathrm{A}\mathrm{u}\mathrm{t}(Q)$: $C_{\mathrm{A}\mathrm{u}\mathrm{t}(N)}(\phi (Q))\times W$ where $W$ is the subgroup of $\mathrm{A}\mathrm{u}\mathrm{t}(Q)$ comprising those automorphisms that do not affect the image under $\phi$, i.e., $W=\{\sigma \in \mathrm{A}\mathrm{u}\mathrm{t}(Q)\mid \phi (\sigma (q))=\phi (q)\mathrm{\forall }q\in Q\}$.

Note in particular that if $\phi$ is trivial, then the acting group is the entire group $\mathrm{A}\mathrm{u}\mathrm{t}(N)\times \mathrm{A}\mathrm{u}\mathrm{t}(Q)$.

Classifying group extensions for an arbitrary normal subgroup

As before, we denote the normal subgroup as $N$ and the quotient group as $Q$.

Finding a list of outer actions

The first step is to note that quotient group maps to outer automorphism group of normal subgroup. Thus, for any group extension with normal subgroup identified with $N$ and quotient group identified with $Q$, we can construct a homomorphism:

$\alpha :Q\to \mathrm{O}\mathrm{u}\mathrm{t}(N)$

where the group on the right is the outer automorphism group of $N$. In the case that $N$ is abelian, this reduces to the situation discussed earlier, because in that case the outer automorphism group is identified with the automorphism group.

Thus, a first step to classifying all the group extensions with normal subgroup $N$ and quotient group $Q$ is to determine the set $\mathrm{H}\mathrm{o}\mathrm{m}(Q,\mathrm{O}\mathrm{u}\mathrm{t}(N))$ of all possible group homomorphisms from $Q$ to $\mathrm{O}\mathrm{u}\mathrm{t}(N)$. Then, for each such homomorphism, we will determine all the possible group extensions whose corresponding homomorphism is that homomorphism. The overall set of congruence classes of group extensions is thus a disjoint union over $\mathrm{H}\mathrm{o}\mathrm{m}(Q,\mathrm{O}\mathrm{u}\mathrm{t}(N))$ of the set of group extensions for each element therein.

Finding all congruence classes for a given outer action

What we're trying to do: Given a homomorphism

$\alpha :Q\to \mathrm{O}\mathrm{u}\mathrm{t}(N)$

, find all the possible congruence classes of group extensions with normal subgroup

$N$

, quotient group

$Q$

, and induced outer action

$\alpha$

.

There is a canonical homomorphism associated with the group $N$ as an abstract group (see outer automorphism group maps to automorphism group of center):

$\pi :\mathrm{O}\mathrm{u}\mathrm{t}(N)\to \mathrm{A}\mathrm{u}\mathrm{t}(Z(N))$

where $Z(N)$ denotes the center of $N$.

We define the composite homomorphism $\phi =\pi \circ \alpha$. Thus, $\phi \in \mathrm{H}\mathrm{o}\mathrm{m}(Q,\mathrm{A}\mathrm{u}\mathrm{t}(Z(N))$. Now, consider the group $H_{\phi }^2(Q,Z(N))$.

First, we note that the group $H_{\phi }^2(Q,Z(N))$ acts canonically on the set of group extensions with normal subgroup $N$, quotient group $Q$, and outer action $\alpha$.

Our claim is that there are two possibilities:

1. There are no group extensions with normal subgroup $N$, quotient group $Q$, and outer action $\alpha$.
2. The canonical action of $H_{\phi }^2(Q,Z(N))$ on the set of group extensions with normal subgroup $N$, quotient group $Q$, and outer action $\alpha$ is equivalent to the regular group action, i.e., it is transitive with a single orbit.

The third cohomology group element criterion

To figure out whether case (1) or case (2) is operative, we need to compute a particular element of the third cohomology group $H_{\phi }^3(Q,Z(N))$. If this element is a non-identity element of the group, then case (1) holds. If this element is the identity element of the group, then case (2) holds. Note in particular that if the third cohomology group is trivial, then case (2) must hold, and we do not need to compute the element explicitly in that case.

Here is how the element is computed. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.

Finding all pseudo-congruence classes for a given outer action

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.