Group cohomology of cyclic group:Z2

Classifying space and corresponding chain complex
The classifying space of the cyclic group of order two is $$\R\mathbb{P}^\infty$$, viz., countable-dimensional real projective space (read more about this space as a topological space on the Topology Wiki).

A chain complex that can be used to compute the homology for the classifying space and hence also for the group is:

$$\dots \stackrel{\cdot 0}{\to} \mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z} \stackrel{\cdot 0}{\to} \mathbb{Z} \to \dots \stackrel{\cdot 2}{\to} \mathbb{Z} \stackrel{\cdot 0}{\to} \mathbb{Z}$$

where the subscript for the last written entry is $$0$$, and hence the multiplication by 2 maps arise from even to odd subscripts and the multiplication by zero maps arise from odd to even subscripts.

Over the integers
The homology groups with coefficients in the ring of integers $$\mathbb{Z}$$ are given as follows:

$$H_p(\mathbb{Z}/2\mathbb{Z};\mathbb{Z}) = \left\lbrace\begin{array}{rl}\mathbb{Z}/2\mathbb{Z}, &p = 1,3,5,\dots\\0, & p = 2,4,6, \dots \\ \mathbb{Z},& p = 0\\\end{array}\right.$$

The first few cohomology groups are given below:

Over an abelian group
The homology groups with coefficients in an abelian group $$M$$ (which we may treat as a module over a ring $$R$$) are given by:

$$H_p(\mathbb{Z}/2\mathbb{Z};M) = \left\lbrace\begin{array}{rl} M/2M, & p=1,3,5,\dots\\ \operatorname{Ann}_M(2), & p = 2,4,6, \dots \\ M, & p = 0\\\end{array}\right.$$

where $$\operatorname{Ann}_M(2)$$ is the 2-torsion submodule of $$M$$, i.e., the submodule of $$M$$ comprising elements whose double is zero.

These homology groups can be computed directly from the chain complex, and they can also be computed using the homology groups over the integers along with the universal coefficients theorem for group homology.

We apply the "finitely generated abelian groups" case of the universal coefficients theorem. The case $$p = 0$$ is easy. For $$p \ge 1$$, $$H_p(\mathbb{Z}/2\mathbb{Z};\mathbb{Z})$$ is always a finite group, so in the notation of the theorem, $$r_p = 0$$. Thus, the term $$M^{r_p}$$ vanishes in all cases.

For odd $$p$$, we have $$T_{p-1} = 0$$, so $$\operatorname{Tor}(T_{p-1},M)$$ vanishes, and we are left with:

$$H_p(\mathbb{Z}/2\mathbb{Z};M) \cong T_p \otimes M \cong M/2M$$

For even $$p > 0$$, we have $$T_p = 0$$, so $$T_p \otimes M$$ vanishes, and we are left with:

$$H_p(\mathbb{Z}/2\mathbb{Z};M) \cong \operatorname{Tor}(T_{p-1},M) \cong \operatorname{Ann}_M(2)$$

The above reasoning is summarized in the table below:

Important case types for abelian groups
Note that the third case, where $$M$$ is 2-divisible but not necessarily uniquely so, cannot arise if $$M = R$$ and it is a unital ring. So when taking coefficients over a unital ring, there is no need to distinguish between 2-divisibility and unique 2-divisibility.

Over the integers
The cohomology groups with coefficients in the ring of integers are given as below:

$$H^p(\mathbb{Z}/2\mathbb{Z};\mathbb{Z}) = \left\lbrace\begin{array}{rl}0, &p = 1,3,5,\dots\\\mathbb{Z}/2\mathbb{Z}, & p = 2,4,6, \dots \\ \mathbb{Z},& p = 0\\\end{array}\right.$$

Basically, the even/odd role gets interchanged. This is because in the cochain complex, the arrows are all pointing in the reverse direction.

Over an abelian group
The cohomology groups with coefficients in an abelian group $$M$$ (which we may treat as a module over a ring $$R$$) are given by:

$$H^p(\mathbb{Z}/2\mathbb{Z};M) = \left\lbrace\begin{array}{rl} \operatorname{Ann}_M(2), & p=1,3,5,\dots\\ M/2M, & p = 2,4,6, \dots \\ M, & p = 0\\\end{array}\right.$$

where $$\operatorname{Ann}_M(2)$$ is the 2-torsion submodule of $$M$$, i.e., the submodule of $$M$$ comprising elements whose double is zero.

We can deduce this directly from the cochain complex, but can also deduce this from the homology groups using the dual universal coefficients theorem for group cohomology.

We apply the "finitely generated abelian groups" case of the dual universal coefficients theorem. The case $$p = 0$$ is easy. For $$p \ge 1$$, $$H_p(\mathbb{Z}/2\mathbb{Z};\mathbb{Z})$$ is always a finite group, so in the notation of the theorem, $$r_p = 0$$. Thus, the term $$M^{r_p}$$ vanishes in all cases.

For odd $$p$$, we have $$T_{p-1} = 0$$, so $$\operatorname{Ext}(T_{p-1},M)$$ vanishes, and we are left with:

$$H^p(\mathbb{Z}/2\mathbb{Z};M) \cong \operatorname{Hom}(T_p,M) \cong \operatorname{Ann}_M(2)$$

For even $$p > 0$$, we have $$T_p = 0$$, so $$\operatorname{Hom}(T_p,M)$$ vanishes, and we are left with:

$$H^p(\mathbb{Z}/2\mathbb{Z};M) \cong \operatorname{Ext}(T_{p-1},M) \cong M/2M$$

The above reasoning is summarized in the table below:

Over the integers
The cohomology groups over the integers come with a ring structure. The structure of the ring is $$\mathbb{Z}[x]/\langle 2x \rangle$$. It is almost the same as $$(\mathbb{Z}/2\mathbb{Z})[x]$$ but the key difference is that the constant terms can vary over all of $$\mathbb{Z}$$. The identification is as follows: $$x^k$$ is the unique nonzero element in $$H^{2k}(\mathbb{Z}/2\mathbb{Z};\mathbb{Z})$$.

Over a 2-divisible ring
If $$R$$ is a 2-divisible unital ring, then it is also uniquely 2-divisible. In this case, all cohomology groups in positive degree vanish, and $$H^*(\mathbb{Z}/2\mathbb{Z};R)$$ is isomorphic to $$R$$, occuring in the $$H^0$$ part.

In particular, this includes the case of $$R$$ a field of any characteristic other than two, as well as the case of any ring (not necessarily a field) of finite positive characteristic.

Over the integers
The Tate cohomology groups with coefficients in the ring of integers are given as below:

$$\hat{H}^p(\mathbb{Z}/2\mathbb{Z};\mathbb{Z}) = \left\lbrace\begin{array}{rl}0, &p \qquad \operatorname{odd} \\\mathbb{Z}/2\mathbb{Z}, & p \qquad \operatorname{even}\\\end{array}\right.$$

Over an abelian group
The cohomology groups with coefficients in an abelian group $$M$$ (which we may treat as a module over a ring $$R$$) are given by:

$$\hat{H}^p(\mathbb{Z}/2\mathbb{Z};M) = \left\lbrace\begin{array}{rl} T, & p \quad \operatorname{odd}\\ M/2M, & p \quad \operatorname{even} \\\end{array}\right.$$

where $$T$$ is the 2-torsion submodule of $$M$$, i.e., the submodule of $$M$$ comprising elements whose double is zero.

In particular, we see the following cases:

Schur multiplier
The Schur multiplier, defined as the second cohomology group for trivial group action $$H^2(G,\mathbb{C}^\ast)$$ and also as the second homology group $$H_2(G,\mathbb{Z})$$, is the trivial group.

In other words, cyclic group:Z2 is a Schur-trivial group. See also cyclic implies Schur-trivial.

Schur covering groups
Cyclic group:Z2 is its own Schur covering group, because the Schur multiplier is trivial.

Second cohomology groups for trivial group action
As noted above, if $$M$$ is a finite abelian group, all the cohomology groups (for trivial group action) $$H^p(\mathbb{Z}/2\mathbb{Z};M)$$ are isomorphic to $$(\mathbb{Z}/2\mathbb{Z})^r$$ where $$r$$ is the rank (i.e., minimum size of generating set) for the 2-Sylow subgroup of $$M$$. In particular, this is also true for the second cohomology group for trivial group action.

The corresponding extensions to the elements of this second cohomology group are all abelian group extensions. We list some cases below:

Second cohomology groups for inverse map action
For any abelian group $$M$$, consider the action of $$\mathbb{Z}/2\mathbb{Z}$$ on $$M$$ via the inverse map: the non-identity element of $$\mathbb{Z}/2\mathbb{Z}$$ sends every element of $$M$$ to its negative (i.e., its inverse).

The second cohomology group $$H^2(\mathbb{Z}/2\mathbb{Z};M)$$ is isomorphic to the subgroup $$T$$ of $$M$$ where $$T$$ is the 2-torsion subgroup of $$M$$.

In particular, if $$M$$ is a finite abelian group whose 2-Sylow subgroup has rank $$r$$, then this second cohomology group is isomorphic to $$(\mathbb{Z}/2\mathbb{Z})^r$$.

Note that for finite abelian groups $$M$$, the second cohomology group for the inverse map action is isomorphic to the second cohomology group for the trivial group action. However, this need not be true for infinite abelian groups, because the former is $$T$$ whereas the latter is $$M/2M$$. In particular, for $$M = \mathbb{Z}$$, the second cohomology group for the inverse map action is the trivial group and the second cohomology group for the trivial group action is isomorphic to cyclic group:Z2.

The zero element of the second cohomology group corresponds to the extension arising as an external semidirect product $$M \rtimes \mathbb{Z}/2\mathbb{Z}$$ where the latter acts by the inverse map. This is called the generalized dihedral group for $$M$$.

Note that if $$M$$ is a group of exponent two (i.e., an elementary abelian 2-group), then the inverse map action on $$M$$ coincides with the trivial group action.

Second cohomology groups for other actions
Suppose $$M$$ is an abelian group and $$\tau:M \to M$$ is an automorphism of order two. We can define an action of $$\mathbb{Z}/2\mathbb{Z}$$ on $$M$$ where the non-identity element acts by the automorphism $$\tau$$. The second cohomology group for this action is given as the quotient $$M^{\tau}/((1 + \tau)M)$$ where:

$$\! M^{\tau} := \{ m \in M \mid \tau(m) = m \}$$

$$\! (1 + \tau)M := \{ \tau(m) + m \mid m \in M \}$$

Note that the second cohomology group for trivial group action is a special case where $$M^{\tau} = M$$ and $$(1 + \tau)M = 2M$$, yielding $$M/2M$$. The second cohomology group for the inverse map action is another special case where $$M^\tau = T$$ is the 2-torsion submodule and $$(1 + \tau)M$$ is trivial.