Group cohomology of elementary abelian group of prime-cube order

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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.

Tate cohomology groups for trivial group action

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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 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.