Group cohomology of cyclic group:Z2

This article gives specific information, namely, group cohomology, about a particular group, namely: cyclic group:Z2.
View group cohomology of particular groups | View other specific information about cyclic group:Z2

Classifying space and corresponding chain complex

The classifying space of the cyclic group of order two is $\mathbb {R} \mathbb {P} ^{\infty }$, 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.

Homology groups for trivial group action

To look at the same material from a topological/algebraic topology perspective, check out the homology of countable-dimensional real projective space at the Topology Wiki

FACTS TO CHECK AGAINST (homology group for trivial group action):
First homology group: first homology group for trivial group action equals tensor product with abelianization
Second homology group: formula for second homology group for trivial group action in terms of Schur multiplier and abelianization|Hopf's formula for Schur multiplier
General: universal coefficients theorem for group homology|homology group for trivial group action commutes with direct product in second coordinate|Kunneth formula for group homology

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:

$\!p$	$\!0$	$\!1$	$\!2$	$\!3$	$\!4$	$\!5$	$\!6$	$\!7$
$\!H_{p}$	$\mathbb {Z}$	$\mathbb {Z} /2\mathbb {Z}$	$\!0$	$\mathbb {Z} /2\mathbb {Z}$	$\!0$	$\mathbb {Z} /2\mathbb {Z}$	$\!0$	$\mathbb {Z} /2\mathbb {Z}$

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. [SHOW MORE]

We apply the "finitely generated abelian groups" case of the universal coefficients theorem. The case $p=0$ is easy. For $p\geq 1$, $H_{p}(\mathbb {Z} /2\mathbb {Z} ;\mathbb {Z} )$ 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:

Case on $p$	$H_{p}(\mathbb {Z} /2\mathbb {Z} ;\mathbb {Z} )$	$H_{p-1}(\mathbb {Z} /2\mathbb {Z} ;\mathbb {Z} )$	$M^{r_{p}}$ part	$T_{p}\otimes M$ part where $T_{p}$ is the torsion part of $H_{p}$ with coefficients in $\mathbb {Z}$	$\operatorname {Tor} (T_{p-1},M)$ where $T_{p-1}$ is the torsion part of $H_{p-1}$ with coefficients in $\mathbb {Z}$	$H_{p}(G;M)$ (direct sum of preceding three columns)
0	$\mathbb {Z}$	0	$M$ because $r_{p}=1$	0 because $T_{p}=0$	0 because $T_{p-1}=0$	$M$
odd positive	$\mathbb {Z} /2\mathbb {Z}$	either $\mathbb {Z}$ or 0	0 because $r_{p}=0$	$M/2M$ because $T_{p}=\mathbb {Z} /2\mathbb {Z}$	0 because $T_{p-1}=0$	$M/2M$
even positive	0	$\mathbb {Z} /2\mathbb {Z}$	0 because $r_{p}=0$	0 because $T_{p}=0$	$\operatorname {Ann} _{M}(2)$ because $T_{p-1}=\mathbb {Z} /2\mathbb {Z}$	$\operatorname {Ann} _{M}(2)$

Important case types for abelian groups

Case on $R$ or $M$	Conclusion about odd-indexed homology groups, i.e., $H_{p},p=1,3,5,\dots$	Conclusion about even-indexed homology groups, i.e., $H_{p},p=2,4,6,\dots$
$M$ is uniquely 2-divisible, i.e., every element of $M$ has a unique half. This includes the case that $M$ is a field of characteristic not 2.	all zero groups	all zero groups
$M$ is 2-torsion-free, i.e., no nonzero element of $M$ doubles to zero	$M/2M$ (need more information about $M$ to compute this)	all zero groups
$M$ is 2-divisible, but not necessarily uniquely so, e.g., $M=\mathbb {Q} /\mathbb {Z}$	all zero groups	$\operatorname {Ann} _{M}(2)$ (need more information about $M$ to compute this)
$M=\mathbb {Z} /2^{n}\mathbb {Z}$, $n$ any natural number	all isomorphic to $\mathbb {Z} /2\mathbb {Z}$	all isomorphic to $\mathbb {Z} /2\mathbb {Z}$
$M$ is a finite abelian group	all isomorphic to $(\mathbb {Z} /2\mathbb {Z} )^{r}$ where $r$ is the rank (i.e., minimum number of generators) for the 2-Sylow subgroup of $M$	all isomorphic to $(\mathbb {Z} /2\mathbb {Z} )^{r}$ where $r$ is the rank (i.e., minimum number of generators) for the 2-Sylow subgroup of $M$
$M$ is a finitely generated abelian group	all isomorphic to $(\mathbb {Z} /2\mathbb {Z} )^{r+s}$ where $r$ is the rank for the 2-Sylow subgroup of the torsion part of $M$ and $s$ is the free rank (i.e., the rank as a free abelian group of the torsion-free part) of $M$	all isomorphic to $(\mathbb {Z} /2\mathbb {Z} )^{r}$ where $r$ is the rank for the 2-Sylow subgroup of $M$ and $s$ is the free rank (i.e., the rank as a free abelian group of the torsion-free part) of $M$

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.

Cohomology groups for trivial group action

To look at the same material from a topological/algebraic topology perspective, check out the cohomology of countable-dimensional real projective space at the Topology Wiki

FACTS TO CHECK AGAINST (cohomology group for trivial group action):
First cohomology group: first cohomology group for trivial group action is naturally isomorphic to group of homomorphisms
Second cohomology group: formula for second cohomology group for trivial group action in terms of Schur multiplier and abelianization
In general: dual universal coefficients theorem for group cohomology relating cohomology with arbitrary coefficientsto homology with coefficients in the integers. |Cohomology group for trivial group action commutes with direct product in second coordinate | Kunneth formula for group cohomology

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.[SHOW MORE]

We apply the "finitely generated abelian groups" case of the dual universal coefficients theorem. The case $p=0$ is easy. For $p\geq 1$, $H_{p}(\mathbb {Z} /2\mathbb {Z} ;\mathbb {Z} )$ 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:

Case on $p$	$H_{p}(\mathbb {Z} /2\mathbb {Z} ;\mathbb {Z} )$	$H_{p-1}(\mathbb {Z} /2\mathbb {Z} ;\mathbb {Z} )$	$M^{r_{p}}$ part	$\operatorname {Hom} (T_{p},M)$ part where $T_{p}$ is the torsion part of $H_{p}$ with coefficients in $\mathbb {Z}$	$\operatorname {Ext} (T_{p-1},M)$ where $T_{p-1}$ is the torsion part of $H_{p-1}$ with coefficients in $\mathbb {Z}$	$H^{p}(G;M)$ (direct sum of preceding three columns)
0	$\mathbb {Z}$	0	$M$ because $r_{p}=1$	0 because $T_{p}=0$	0 because $T_{p-1}=0$	$M$
odd positive	$\mathbb {Z} /2\mathbb {Z}$	either $\mathbb {Z}$ or 0	0 because $r_{p}=0$	$\operatorname {Ann} _{M}(2)$ because $T_{p}=\mathbb {Z} /2\mathbb {Z}$	0 because $T_{p-1}=0$	$\operatorname {Ann} _{M}(2)$
even positive	0	$\mathbb {Z} /2\mathbb {Z}$	0 because $r_{p}=0$	0 because $T_{p}=0$	$M/2M$ because $T_{p-1}=\mathbb {Z} /2\mathbb {Z}$	$M/2M$

Important case types for abelian groups

Case on $M$	Conclusion about odd-indexed cohomology groups, i.e., $H^{p},p=1,3,5,\dots$	Conclusion about even-indexed cohomology groups, i.e., $H^{p},p=2,4,6,\dots$
$M$ is uniquely 2-divisible, i.e., every element of $M$ has a unique half	all zero groups	all zero groups
$M$ is 2-torsion-free, i.e., no nonzero element of $M$ doubles to zero	all zero groups	$M/2M$ (need more information about $M$ to compute this)
$M$ is 2-divisible, but not necessarily uniquely so, e.g., $M=\mathbb {Q} /\mathbb {Z}$	$\operatorname {Ann} _{2}(M)$ (need more information about $M$ to compute this)	all zero groups
$M=\mathbb {Z} /2^{n}\mathbb {Z}$	all isomorphic to $\mathbb {Z} /2\mathbb {Z}$	all isomorphic to $\mathbb {Z} /2\mathbb {Z}$
$M$ is a finite abelian group	all isomorphic to $(\mathbb {Z} /2\mathbb {Z} )^{r}$ where $r$ is the rank (i.e., minimum number of generators) for the 2-Sylow subgroup of $M$	all isomorphic to $(\mathbb {Z} /2\mathbb {Z} )^{r}$ where $r$ is the rank (i.e., minimum number of generators) for the 2-Sylow subgroup of $M$
$M$ is a finitely generated abelian group	all isomorphic to $(\mathbb {Z} /2\mathbb {Z} )^{r}$ where $r$ is the rank for the 2-Sylow subgroup of the torsion part $M$	all isomorphic to $(\mathbb {Z} /2\mathbb {Z} )^{r+s}$ where $r$ is the rank (i.e., minimum number of generators) for the 2-Sylow subgroup of $M$ and $s$ is the rank as a free abelian group of the torsion-free part of $M$

Cohomology ring

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 characteristic two

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Tate cohomology groups for trivial group action

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:

Case on $R$ or $M$	Conclusion about odd-indexed Tate cohomology groups, i.e., ${\hat {H}}^{p},p=\dots ,-5,-3,-1,1,3,5,\dots$	Conclusion about even-indexed Tate cohomology groups, i.e., ${\hat {H}}^{p},p=\dots ,-4,-2,0,2,4,6,\dots$
$M$ is uniquely 2-divisible, i.e., every element of $M$ has a unique half	all zero groups	all zero groups
$M$ is 2-torsion-free, i.e., no nonzero element of $M$ doubles to zero	all zero groups	unclear
$M$ is 2-divisible, but not necessarily uniquely so, e.g., $M=\mathbb {Q} /\mathbb {Z}$	unclear	all zero groups
$M=\mathbb {Z} /2^{n}\mathbb {Z}$	all isomorphic to $\mathbb {Z} /2\mathbb {Z}$	all isomorphic to $\mathbb {Z} /2\mathbb {Z}$
$M$ is a finite abelian group	all isomorphic to $(\mathbb {Z} /2\mathbb {Z} )^{r}$ where $r$ is the rank (i.e., minimum number of generators) for the 2-Sylow subgroup of $M$	all isomorphic to $(\mathbb {Z} /2\mathbb {Z} )^{r}$ where $r$ is the rank (i.e., minimum number of generators) for the 2-Sylow subgroup of $M$

Second cohomology group and extensions

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:

Case for $M$	Isomorphism class of second cohomology group	Link to page	Short description of extensions
$M$ has odd order	trivial group	--	The only extension is a direct product of $M$ and the cyclic group of order two.
$M=\mathbb {Z} /2\mathbb {Z}$, i.e., cyclic group:Z2	cyclic group:Z2	second cohomology group for trivial group action of Z2 on Z2	The extensions are Klein four-group (for zero element of cohomology group) and cyclic group:Z4 (for nonzero element)
$M=\mathbb {Z} /4\mathbb {Z}$, i.e., cyclic group:Z4	cyclic group:Z2	second cohomology group for trivial group action of Z2 on Z4	The extensions are direct product of Z4 and Z2 (for zero element of cohomology group) and cyclic group:Z8 (for nonzero element)
$M=\mathbb {Z} /2\mathbb {Z} \times \mathbb {Z} /2\mathbb {Z}$, i.e., Klein four-group	Klein four-group	second cohomology group for trivial group action of Z2 on V4	The extensions are elementary abelian group:E8 (for zero element of cohomology group) and direct product of Z4 and Z2 (three copies; different versions for each of the nonzero elements)
$M=\mathbb {Z}$, i.e., group of integers	cyclic group:Z2	--	direct product of Z and Z2 (zero element of cohomology group) and group of integers (nonzero element of cohomology group).

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.

Case for $M$	Isomorphism class of second cohomology group	Link to page	Short description of extensions
$M$ has odd order	trivial group	--	The only extension is the generalized dihedral group corresponding to $M$.
$M=\mathbb {Z} /2\mathbb {Z}$, i.e., cyclic group:Z2	cyclic group:Z2	second cohomology group for trivial group action of Z2 on Z2	The extensions are Klein four-group (for zero element of cohomology group) and cyclic group:Z4 (for nonzero element)
$M=\mathbb {Z} /4\mathbb {Z}$, i.e., cyclic group:Z4	cyclic group:Z2	second cohomology group for nontrivial group action of Z2 on Z4	The extensions are dihedral group:D8 (for zero element of cohomology group) and quaternion group (for nonzero element of cohomology group)
$M=\mathbb {Z} /2\mathbb {Z} \times \mathbb {Z} /2\mathbb {Z}$, i.e., Klein four-group	Klein four-group	second cohomology group for trivial group action of Z2 on V4	The extensions are elementary abelian group:E8 (for zero element of cohomology group) and direct product of Z4 and Z2 (three copies; different versions for each of the nonzero elements)
$M=\mathbb {Z}$, i.e., group of integers	trivial group		The infinite dihedral group is the only group.

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.