# Group cohomology of 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 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 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. $(M/pM)^{(q-3)/2}$ $(M/pM)^{(q+2)/2}$
$M$ 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 $M$