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

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

Suppose $p$ is a prime number. We are interested in the elementary abelian group of prime-cube order

$E_{p^3} = (\mathbb{Z}/p\mathbb{Z})^3 = \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}$

## Particular cases

Value of prime $p$ elementary abelian group of prime-cube order cohomology information
2 elementary abelian group:E8 group cohomology of elementary abelian group:E8
3 elementary abelian group:E27 group cohomology of elementary abelian group:E27
5 elementary abelian group:E125 group cohomology of elementary abelian group:E125

## 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

The homology groups over the integers are given as follows:

$\! H_q(E_{p^3};\mathbb{Z}) = \left\lbrace\begin{array}{rl} (\mathbb{Z}/p\mathbb{Z})^{\frac{q^2 + 4q + 7}{4}}, & \qquad q = 1,3,5,\dots \\ (\mathbb{Z}/p\mathbb{Z})^{\frac{q^2+4q}{4}}, & \qquad q = 2,4,6,\dots \\ \mathbb{Z}, & \qquad q = 0 \\\end{array}\right.$

The first few homology groups are given as follows:

$q$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$H_q$ $\mathbb{Z}$ $(\mathbb{Z}/p\mathbb{Z})^3$ $(\mathbb{Z}/p\mathbb{Z})^3$ $(\mathbb{Z}/p\mathbb{Z})^7$ ($\mathbb{Z}/p\mathbb{Z})^8$ $(\mathbb{Z}/p\mathbb{Z})^{13}$ $(\mathbb{Z}/p\mathbb{Z})^{15}$ $(\mathbb{Z}/p\mathbb{Z})^{21}$ $(\mathbb{Z}/p\mathbb{Z})^{24}$
rank of $H_q$ as an elementary abelian $p$-group -- 3 3 7 8 13 15 21 24

### Over an abelian group

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

$\! H_q(E_{p^3};M) = \left\lbrace\begin{array}{rl} (M/pM)^{\frac{q^2 + 4q + 7}{4}} \oplus (\operatorname{Ann}_M(p))^{\frac{q^2 + 2q - 3}{4}}, & \qquad q = 1,3,5,\dots \\ (M/pM)^{\frac{q^2 + 4q}{4}} \oplus (\operatorname{Ann}_M(p))^{\frac{q^2 + 2q + 4}{4}}, & \qquad q = 2,4,6,\dots \\ M, & \qquad q = 0\\\end{array}\right.$

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

### Important case types for abelian groups

Case on $M$ Conclusion for odd-indexed homology groups $H_q$, $q = 1,3,5,\dots$ Conclusion for even-indexed homology groups $H_q$, $q = 2,4,6,\dots$
$M$ is uniquely $p$-divisible, i.e., every element of $M$ can be uniquely divided 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. $(M/pM)^{\frac{q^2 + 4q + 7}{4}}$ $(M/pM)^{\frac{q^2 + 4q}{4}}$
$M$ is $p$-divisible, but not necessarily uniquely so, e.g., $M = \mathbb{Q}/\mathbb{Z}$. $(\operatorname{Ann}_M(p))^{\frac{q^2 + 2q - 3}{4}}$ $(\operatorname{Ann}_M(p))^{\frac{q^2 + 2q + 4}{4}}$
$M = \mathbb{Z}/p^n\mathbb{Z}$, $n$ any natural number $(\mathbb{Z}/p\mathbb{Z})^{\frac{(q+1)(q+2)}{2}}$ $(\mathbb{Z}/p\mathbb{Z})^{\frac{(q+1)(q+2)}{2}}$
$M$ is a finite abelian group All isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\frac{r(q+1)(q+2)}{2}}$ where $r$ is the rank of the $p$-Sylow subgroup of $M$ $(\mathbb{Z}/p\mathbb{Z})^{\frac{r(q+1)(q+2)}{2}}$ where $r$ is the rank of the $p$-Sylow subgroup of $M$
$M$ is a finitely generated abelian group All isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\frac{r(q+1)(q+2)}{2} + \frac{s(q^2 + 4q + 7)}{4}}$ where $r$ is the rank of the $p$-Sylow subgroup of the torsion part of $M$ and $s$ is the rank of the free part of $M$. All isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\frac{r(q+1)(q+2)}{2} + \frac{s(q^2 + 4q)}{4}}$ where $r$ is the rank of the $p$-Sylow subgroup of the torsion part of $M$ and $s$ is the rank of the 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 as follows:

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

The first few cohomology groups are given below:

$q$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$H^q$ $\mathbb{Z}$ 0 $(\mathbb{Z}/p\mathbb{Z})^3$ $(\mathbb{Z}/p\mathbb{Z})^3$ $(\mathbb{Z}/p\mathbb{Z})^7$ $(\mathbb{Z}/p\mathbb{Z})^8$ $(\mathbb{Z}/p\mathbb{Z})^{13}$ $(\mathbb{Z}/p\mathbb{Z})^{15}$ $(\mathbb{Z}/p\mathbb{Z})^{21}$
rank of $H^q$ as an elementary abelian $p$-group -- 0 3 3 7 8 13 15 21

### Over an abelian group

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

$H^q(E_{p^3};M) = \left\lbrace \begin{array}{rl} (M/pM)^{\frac{q^2 + 2q - 3}{4}} \oplus (\operatorname{Ann}_M(p))^{\frac{q^2 + 4q + 7}{4}}, & \qquad q = 1,3,5,\dots \\ (M/pM)^{\frac{q^2 + 2q + 4}{4}} \oplus (\operatorname{Ann}_M(p))^{\frac{q^2 + 4q}{4}}, & \qquad q = 2,4,6,\dots \\ M, & \qquad q = 0 \\\end{array}\right.$

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

### Important case types for abelian groups

Case on $M$ Conclusion for odd-indexed cohomology groups $H^q$, $q = 1,3,5,\dots$ Conclusion for even-indexed cohomology groups $H^q$, $q = 2,4,6,\dots$
$M$ is uniquely $p$-divisible, i.e., every element of $M$ can be uniquely divided 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. $(M/pM)^{\frac{q^2 +2q - 3}{4}}$ $(M/pM)^{\frac{q^2 + 2q + 4}{4}}$
$M$ is $p$-divisible, but not necessarily uniquely so, e.g., $M = \mathbb{Q}/\mathbb{Z}$. $(\operatorname{Ann}_M(p))^{\frac{q^2 + 4q + 7}{4}}$ $(\operatorname{Ann}_M(p))^{\frac{q^2 + 4q}{4}}$
$M = \mathbb{Z}/p^n\mathbb{Z}$, $n$ any natural number $(\mathbb{Z}/p\mathbb{Z})^{\frac{(q+1)(q+2)}{2}}$ $(\mathbb{Z}/p\mathbb{Z})^{\frac{(q+1)(q+2)}{2}}$
$M$ is a finite abelian group All isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\frac{r(q+1)(q+2)}{2}}$ where $r$ is the rank of the $p$-Sylow subgroup of $M$ $(\mathbb{Z}/p\mathbb{Z})^{\frac{r(q+1)(q+2)}{2}}$ where $r$ is the rank of the $p$-Sylow subgroup of $M$
$M$ is a finitely generated abelian group All isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\frac{r(q+1)(q+2)}{2} + \frac{s(q^2 + 2q - 3)}{4}}$ where $r$ is the rank of the $p$-Sylow subgroup of the torsion part of $M$ and $s$ is the rank of the free part of $M$. All isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\frac{r(q+1)(q+2)}{2} + \frac{s(q^2 + 2q + 4)}{4}}$ where $r$ is the rank of the $p$-Sylow subgroup of the torsion part of $M$ and $s$ is the rank of the free part of $M$.

## 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 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 quadratic function and a periodic function with 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 quadratic function is $q \mapsto (q^2 + 4q)/4 + (7/8)$, and the periodic function is $7(-1)^{q + 1}/8$.
• For cohomology groups, choosing the periodic function so as to have mean zero, we get that the quadratic function is $q \mapsto (q^2 + 2q)/4 + (1/8)$, and the periodic function is $7(-1)^q/8$.

Note that:

• The cohomology groups grow slightly slower than the homology groups. Basically, the $q^{tH}$ cohomology group is the $(q-1)^{th}$ homology group, so the cohomology groups lag behind slightly.
• The periodic function for the cohomology groups is opposite that for the homology groups. This follows from the dual universal coefficients theorem for group cohomology.

### Over the prime field

If we take coefficients in the prime field $\mathbb{F}_p$, the ranks of the homology groups and cohomology groups both grow as quadratic functions of $q$. The quadratic function in both cases is $q \mapsto (q + 1)(q + 2)/2$. Note that in this case, the homology groups and cohomology groups are both 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.