The format of this post will be a little unorganized. I start off with the definition of group cohomology without giving much motivation. Then, I properly define a bar resolution for groups in complete detail. With this setup, I properly motivate group cohomology, work out some extremely interesting examples to see why one should care about this particular free resolution. I’ve also decided to make another post later on a generalization of the bar resolution to the case of monads and a further generalization to algebras. I am currently taking a course on Homotopy Type Theory and I’ve found it blissfully interesting so there might be a few posts in the future on that too. I’ve only recently noticed that I’ve never actually posted anything on either algebraic topology or homotopy theory, which is actually my main interest so I might update that soon enough.

A representation of a group over a field is just a module. It is a common philosophy that one can study the structure of a group by studying its representations. In some sense, group cohomology relates to topology but I’ll get to this later on in the post.

Consider as a where acts trivially and let be a representation,i.e module. Extend this to a . The group cohomology with coefficients in is defined by

. One can immediately see that . A map in satisfies . But since the action on is trivial,. This implies that . Sending to any element of corresponds exactly to the fixed points . So, corresponds to the fixed points .

# Bar resolution

So, let’s describe these things in terms of the bar resolution. The calculation of the functor entails finding a projective resolution $latex F_{\bullet} \mapsto X \to 0$ and applying the contravariant functor. Taking ‘homology’ yields . The bar resolution actually gives such a resolution, in fact, a free resolution.

Let be the free on , that is, the set of symbols for and where all those expressions are simply formal symbols for the generated free module. We have the following free resolution, whose maps will be described soon.

[Insert Image]

where

the augmentation map is given by on the basis elements.

is given by:

where the maps is defined on the basis of the free module which extends to the entire group.

,

for . Keep note of the dimension above. Finally,

.

There is topological motivation for this seemingly bizarre construction.Let us say that the aim is to somehow construct a simplicial object from . For every ordered tuple, one can insert a simplex and can act on these faces by diagonal action:

If any element of the tuple is , the simplex degenerates into lower dimension, for example from the differential maps we saw above, if . The differnetial maps match with the this interpretation of degeneracy and face maps of the simiplicial object.You can alternatively also define a normalized version of the bar resolution where the maps are replaced by tuples where the element are non-identity. It is easy to check that the sequence is a chain complex though this involves some annoying calculations. Also, the sequence is exact as we’ll show in the following lemma.

## Lemma 1:

The sequence above is a chain complex that is for and . It is also a -free resolution i.e it is exact.

### Proof of Lemma 1:

### Step 1:Proving that it is a chain complex

Pick a basis element . .

I encourage the reader to work out the case for . Now, for , we can do a little trick to simplify the calculations. Define a new set of modules by

for with the action given by a diagonal action on the tuples as follows

.

I will omit the details but essentially, what happens is that where we have a bijection of those tuples. So, we just hacked our module to be a free module on one lower dimension. Verify that it is indeed a bijection and you get an isomorphism for given by

.

Isomorphism follows trivially from the fact that left multiplication by is a bijection on a group. There may be other ways to map the basis elements too.

Define by

.

We have that by the usual calculation one encounters in chain complexes. This will obviously be useful soon. We shall prove that from which it follows that iff since is an isomorphism.

Now apply to this expression and check that it is equal to . It is a slightly annoying calculation but much less painful than dealing with the original equation. Now, it remains to deal with the case for (we’ve already done j=1), which I’ve already left as an exercise to the reader. Anyways, moving onto the next step

### Step 2:Complex is chain-homotopic to the identity

We define the chain homotopy for (taking the -1 case as )on the basis elements by

, simply adding the element to the symbol. We simply use the isomorphism and transfer the entire problem to the ‘s to get

so we get that

. Now I leave it to the reader to verify through simple calculations that

for . Lower cases can be dealt with separately.

## Group cohomology again and a few applciations

Ok, now that we’ve constructed this module resolution. By definition,

Now, it’s great that we’ve got a free resolution but obviously the hope is that one gets some meaningful results with the bar resolution. Let’s observe a few key ideas.

on the level of sets. Basically, assigning the values at the basis elements gives all maps from to by extending linearly.

# Motivation

Let’s look at some low-dimensional cases to see what exactly is going on.

I describe in the first section of the post.

Just like as described in the first section, even , the collection of maps since specifying the values on the basis elements determines the entire homomorphism. Under this representation, the boundary maps can be written more explicitly as:

We can call the elements of and the elements of as , the cycles and boundaries respectively or cocycles and coboundaries, if one wishes to be more precise. It is a chain complex and it is exact though I won’t bother much with the details. The upshot is that know we have the following description of the cohomology group

.

Great! Now we have just one final simple thing to put in place before we look at some applications.

## Theorem 1

Suppose is a short exact sequence of modules, then there is right-derived long exact sequence:

I covered the proof of this theorem for the general case of derived functors in my other post.

I guess, as a simple consequence of the proof of dimension shifting I completed in the other post(or you could prove easily on your own), we have the following theorem which is useful in the context of group cohomology.

## Theorem 2(Dimension shifting):

If is a short exact sequence of modules such that is trivial, then we have the following natural(wrt to chian complex morphisms) isomorphisms

for .

The result essentially allows us to compute homology by computing homology of with respect to some other representation in a lower dimension. Now, let us move onto some examples.

# The Extension Problem

This is probably most the classical application of group cohomology which I am sure the reader is already quite familiar with so I shall discuss it only little detail.

A classical problem in group theory was to find all possible extensions of a group by a normal subgroup. If we have two groups , it is natural to ask what groups are extensions of by , that is, fit into the exact sequence

An obvious example is a the direct product . One could also add the extra condition that it be a split exact sequence which is equivalent to a semidirect-product . Two extensions are said to be equivalent if the following diagrams commute:

[Insert Image]

Let’s think of a few extensions of groups. For the groups , we have the trivial extension and the well known extension where acts by inversion on .This is the dihedral group of order 6.(Sorry if I have messed up the order of placement of the normal subgroup, I never remember how it works. I think I may have even messed up the semidirect product symbol too. Who cares, I always hated that symbol anyways and abhor the idea of trying to fix it).

We’re going to prove that the elements of the cohomology group are in one-to-one correspondence to extensions up to equivalence. We begin by first defining a section on the level of sets as follows: If there is an extension of the form

[Insert Image]

We would like to define an action on from the exact sequence. Since the sequence is exact, there exists some such that $latex \pi(e_{g})=g$. A section is basically a choice of such ‘s. We’ll explain the representative notation soon. acts on by conjugation . Note that and also that conjugation by this element since is abelian. So, the action is independent of the choice of .

Let the section be . Each represents an equivalence class . Both and are in the same equivalence class for , so by exactness, there is a unique satisfying

Define a function by , the value determined above. Note that every element in can be written uniquely as by exactness. Let’s see how multiplication works in :

. Since ,. Since is abelian

where is the action of defined above.

Associativity of multiplication in (I leave it to the reader to check this) gives the equality:

which is exactly the condition for to be in . Finally, let be another section of the extension and let be its factor set. We’ll show next that and differ by a coboundary, an element of which implies that every extension has a well-defined cohomology class in .

Obviously, lie in the same coset since they both map to under . This implies that there is a unique such that . Define by .

Also,

Equating the two expressions yields,

giving the needed result.

Now, we have to prove the opposite assertion , that, an element of yields a unique(upto equivalence) extension of by . I encourage the reader to find the remaining details in a textbook like Dummit and Foote or Aluffi(I am nor sure if he covers it though).

The element in corresponds to a split extension because then the section is a genuine homomorphism and so .

## Schur-Zassenhaus Theorem

Consider the exact sequence of groups:

[Insert Image]

If then the sequence is split, equivalently,

If the order of , a normal subgroup of , is coprime to the order of the index of in , then there is a semidirect product such that for some complement .

We’ll first deal with the abelian case and that’ll turn out to be useful in the general proof. Towards, that we have the following Lemma.

## Lemma 2:

Let be a finite group of order and let be an abelian group with , then are modules, that is, they are annihilated on multiplication by .

### Proof of Lemma 2:

Let be the bar resolution of and let be the endomorphism of given by multiplication by on and multiplication by for for . We show that is nullhomotopic. Consider the chain homotopy given by the formula on the basis elements

I leave it as an exercise to the reader to show that it is nullhomotopic, a trivial calculation.

## Corollary 2.1

Let be a finite group of order and let be either a vector space over or a module then for .

### Proof of Corollary 2.1

Multiplication by is an isomorphism on in either case. Applying to the bar resolution , consider a class .

which is only if is . So, .

This resolves the Schur-Zassenhaus theorem for the abelian case below.

## Corollary 2.2(Abelian Case)

If is an abelian group, then so every extension of by is split.

With this, we reduce the proof of the general case to this specific case through some techniques. Before that though, there’s just a random group theory fact we’re going to use.

## Lemma 3:

Let be a group and let by a Sylow-p subgroup. Then, is a subgroup of .

### Proof of Lemma 3:

### Proof of Schur Zassenhaus Theorem

We proceed by induction on , the order of , the first element in the sequence. Let be a prime that divides $n=ord(A)$ and consider , a Sylow-p subgroup of . We know that by standard group theory via. the class equation, for example. Consider , the normalizer of in . Using Frattinis’s argument,

.

So, the index divides and is a subgroup of by Lemma 3. So, . So, we get an extension

by the Second Isomorphism Theorem and using Frattini’s argument. If , using the induction hypothesis, this sequence splits so there is a subgroup of which is isomorphic to . On the other hand, if , we get that and we get the extension

by the Third Isomorphism theorem. This splits by the induction hypothesis. Let be the subgroup of isomorphic to . Pulling back, let where is the natural quotient map . From this, we get the extension,

. Using Corollary 2.2(since is abelian), there is subgroup of isomorphic to so there is a subgroup of isomorphic to which, as established, is isomorphic to . Q.E.D