# Group cohomology of elementary abelian group of prime-square order

This article gives specific information, namely, group cohomology, about a family of groups, namely: elementary abelian group of prime-square order.
View group cohomology of group families | View other specific information about elementary abelian group of prime-square order

Suppose $p$ is a prime number. We are interested in the elementary abelian group of prime-square order $E_{p^2} = (\mathbb{Z}/p\mathbb{Z})^2 = \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}$.

## Particular cases

Value of prime $p$ elementary abelian group of prime-square order cohomology information
2 Klein four-group group cohomology of Klein four-group
3 elementary abelian group:E9 group cohomology of elementary abelian group:E9
5 elementary abelian group:E25 group cohomology of elementary abelian group:E25

## Homology groups for trivial group action

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

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

The even and odd cases can be combined giving the following alternative description:

$H_q(\mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z};\mathbb{Z}) = \left\lbrace \begin{array}{rl} (\mathbb{Z}/p\mathbb{Z})^{q/2 + 3(1 - (-1)^q)/4}, & \qquad q > 0 \\ \mathbb{Z}, & \qquad q = 0 \\\end{array}\right.$ The first few homology groups are given below:

$q$ $0$ $1$ $2$ $3$ $4$ $5$
$H_q$ $\mathbb{Z}$ $(\mathbb{Z}/p\mathbb{Z})^2 = E_{p^2}$ $\mathbb{Z}/p\mathbb{Z}$ $(\mathbb{Z}/p\mathbb{Z})^3 = E_{p^3}$ $(\mathbb{Z}/p\mathbb{Z})^2 = E_{p^2}$ $(\mathbb{Z}/p\mathbb{Z})^4 = E_{p^4}$
rank of $H_q$ as an elementary abelian $p$-group -- 2 1 3 2 4

### Over an abelian group

The homology groups with coefficients in an abelian group $M$ are given as follows:

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

Here, $M/pM$ is the quotient of $M$ by $pM = \{ px \mid x \in M \}$ and $\operatorname{Ann}_M(p) = \{ x \in M \mid px = 0 \}$.

These homology groups can be computed in terms of the homology groups over integers using the universal coefficients theorem for group homology.

### Important case types for abelian groups

Case on $M$ Conclusion about odd-indexed homology groups, i.e., $H_q, q = 1,3,5,\dots$ Conclusion about even-indexed homology groups, i.e., $H_q, q = 2,4,6,\dots$
$M$ is uniquely $p$-divisible, i.e., every element of $M$ can be divided uniquely by $p$. This includes the case that $M$ is a field of characteristic not $p$. all zero groups all zero groups
$M$ is $p$-torsion-free, i.e., no nonzero element of $M$ multiplies by $p$ to give zero. $(M/pM)^{(q+3)/2}$ $(M/pM)^{q/2}$
$M$ is $p$-divisible, but not necessarily uniquely so, e.g., $M = \mathbb{Q}/\mathbb{Z}$ $(\operatorname{Ann}_M(p))^{(q-1)/2}$ $(\operatorname{Ann}_M(p))^{(q+2)/2}$
$M = \mathbb{Z}/p^n\mathbb{Z}$, $n$ any natural number $(\mathbb{Z}/p\mathbb{Z})^{q+1}$ $(\mathbb{Z}/p\mathbb{Z})^{q+1}$
$M$ is a finite abelian group isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{r(q + 1)}$ where $r$ is the rank (i.e., minimum number of generators) for the $p$-Sylow subgroup of $M$ isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{r(q + 1)}$ where $r$ is the rank (i.e., minimum number of generators) for the $p$-Sylow subgroup of $M$
$M$ is a finitely generated abelian group all isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{r(q + 1) + s(q + 3)/2}$ where $r$ is the rank for the $p$-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 $(\mathbb{Z}/p\mathbb{Z})^{r(q + 1) + sq/2}$ where $r$ is the rank for the $p$-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 $M$

## Cohomology groups for trivial group action

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 integers are given as below:

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

The odd and even cases can be combined as follows:

$H^q(\mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z};\mathbb{Z}) = \left\lbrace \begin{array}{rl} (\mathbb{Z}/p\mathbb{Z})^{q/2 + 1/4 + 3(-1)^q/4}, & q > 0\\ \mathbb{Z}, & q = 0 \\\end{array}\right.$

The first few cohomology groups are given below:

$q$ $\! 0$ $\! 1$ $\! 2$ $\! 3$ $\! 4$ $\! 5$
$H^q$ $\mathbb{Z}$ 0 $(\mathbb{Z}/p\mathbb{Z})^2 = E_{p^2}$ $\mathbb{Z}/p\mathbb{Z}$ $(\mathbb{Z}/p\mathbb{Z})^3 = E_{p^3}$ $(\mathbb{Z}/p\mathbb{Z})^2 = E_{p^2}$
rank of $H^q$ as an elementary abelian $p$-group -- 0 2 1 3 2

### Over an abelian group

The cohomology groups with coefficients in an abelian group $M$ are given as follows:

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

Here, $M/pM$ is the quotient of $M$ by $pM = \{ px \mid x \in M \}$ and $\operatorname{Ann}_M(p) = \{ x \in M \mid px = 0 \}$.

These can be deduced from the homology groups with coefficients in the integers using the dual universal coefficients theorem for group cohomology.

### Important case types for abelian groups

Case on $M$ Conclusion about odd-indexed cohomology groups, i.e., $H^q, q = 1,3,5,\dots$ Conclusion about even-indexed homology groups, i.e., $H^q, q = 2,4,6,\dots$
$M$ is uniquely $p$-divisible, i.e., every element of $M$ can be divided by $p$ uniquely. This includes the case that $M$ is a field of characteristic not 2. all zero groups all zero groups
$M$ is $p$-torsion-free, i.e., no nonzero element of $M$ multiplies by $p$ to give zero. $(M/pM)^{(q-1)/2}$ $(M/pM)^{(q+2)/2}$
$M$ is $p$-divisible, but not necessarily uniquely so, e.g., $M = \mathbb{Q}/\mathbb{Z}$ $(\operatorname{Ann}_M(p))^{(q+3)/2}$ $(\operatorname{Ann}_M(p))^{q/2}$
$M = \mathbb{Z}/p^n\mathbb{Z}$, $n$ any natural number $(\mathbb{Z}/p\mathbb{Z})^{q+1}$ $(\mathbb{Z}/p\mathbb{Z})^{q+1}$
$M$ is a finite abelian group isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{r(q + 1)}$ where $r$ is the rank (i.e., minimum number of generators) for the $p$-Sylow subgroup of $M$ isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{r(p + 1)}$ where $r$ is the rank (i.e., minimum number of generators) for the $p$-Sylow subgroup of $M$
$M$ is a finitely generated abelian group all isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{r(q + 1) + s(q - 1)/2}$ where $r$ is the rank for the $p$-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 $(\mathbb{Z}/p\mathbb{Z})^{r(q + 1) + s(q + 3)/2}$ where $r$ is the rank for the $p$-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 $M$

## Tate cohomology

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## Growth of ranks of cohomology groups

### Over the integers

With the exception of the zeroth homology group and cohomology group, the homology groups and cohomology groups over the integers are all elementary abelian $p$-groups.

For the homology groups, the rank (i.e., dimension as a vector space over the field of $p$ elements) is a function of $q$ that is a sum of a linear function (of slope 1/2) and a periodic function (of period 2). The same is true for the cohomology groups, although the precise description of the periodic function differs.

• For homology groups, choosing the periodic function so as to have mean zero, we get that the linear function is $q \mapsto q/2 + 3/4$ and the periodic function is $3(-1)^{q+1}/4$.
• For cohomology groups, choosing the periodic function so as to have mean zero, we get that the linear function is $q \mapsto q/2 + 1/4$ and the periodic function is $3(-1)^q/4$.

Note that:

• The intercept for the cohomology groups is 1/4, as opposed to the intercept of 3/4 for the homology groups. This is explained by the somewhat slower start of cohomology groups on account of $H^1$ being torsion-free.
• The periodic parts for homology groups and cohomology groups are negatives of each other, indicating an opposing pattern that is explained by looking at the dual universal coefficients theorem for group cohomology.

### Over the prime field

If we take coefficients in the prime field $\mathbb{F}_p$, then the ranks of the homology and cohomology groups both grow as linear functions of $q$. The linear function in both cases is $q \mapsto q + 1$. Note that in this case, the homology groups and cohomology groups are vector spaces over $\mathbb{F}_p$ and the cohomology group is the vector space dual of the homology group.

Note that there is no periodic part when we are working over the prime field.