# Second cohomology group for trivial group action of free abelian group of rank two on group of integers

This article gives information about the second cohomology group for trivial group action (i.e., the second cohomology group with trivial action) of the group free abelian group of rank two on group of integers. The elements of this classify the group extensions with group of integers in the center and free abelian group of rank two the corresponding quotient group. Specifically, these are precisely the central extensions with the given base group and acting group.
The value of this cohomology group is group of integers.
Get more specific information about free abelian group of rank two |Get more specific information about group of integers|View other constructions whose value is group of integers

This article describes the second cohomology group $H^2(\mathbb{Z}^2;\mathbb{Z})$ corresponding to the trivial group action of the free abelian group of rank two ($\mathbb{Z}^2 = \mathbb{Z} \times \mathbb{Z}$) on the group of integers ($\mathbb{Z}$). The second cohomology group is itself isomorphic to $\mathbb{Z}$, the group of integers.

## Computation in terms of group cohomology

The cohomology group can be computed as an abstract group using the group cohomology of free abelian groups. The general formula is:

$H^q(\mathbb{Z}^n;\mathbb{Z}) = \mathbb{Z}^{\binom{n}{q}}$

In our case, $n = q = 2$, so $\binom{n}{q} = 1$, and we obtain that the cohomology group is $\mathbb{Z}$.

## Elements

### Summary

FACTS TO CHECK AGAINST (second cohomology group for trivial group action):
Background reading on relationship with extension groups: Group extension problem
Arithmetic functions of extension group:
order (thus all extension groups have the same order): order of extension group is product of order of normal subgroup and quotient group
nilpotency class: nilpotency class of extension group is between nilpotency class of quotient group and one more for central extension
derived length: derived length of extension group is bounded by sum of derived length of normal subgroup and quotient group
minimum size of generating set: minimum size of generating set of extension group is bounded by sum of minimum size of generating set of normal subgroup and quotient group|minimum size of generating set of quotient group is at most minimum size of generating set of group
WHAT'S THE TABLE BELOW?: Recall that there is a correspondence:
Elements of the group $H^2(G;A)$ for the trivial group action $\leftrightarrow$ congruence classes of central extensions with the specified subgroup $A$ and quotient group $G$.
This descends to a correspondence:
Orbits for the group action of $\operatorname{Aut}(G) \times \operatorname{Aut}(A)$ on $H^2(G;A)$ $\leftrightarrow$ pseudo-congruence classes of central extensions.
The table below breaks down the second cohomology group as a union of these orbits, with (as a general rule) each row describing one orbit, i.e., one "cohomology class type", aka one "pseudo-congruence class" of central extensions. The number of rows is the number of pseudo-congruence classes of central extensions.

By a priori considerations, all extension groups must have nilpotency class either one or two, and must have Hirsch length as well as polycyclic breadth equal to 3. The minimum size of generating set must be either 2 or 3.

Cohomology class type (i.e., as element of second cohomology group) Number of cohomology classes Corresponding group extension Stem extension? Base characteristic in whole group? Nilpotency class of whole group Derived length of whole group Minimum size of generating set of whole group (must be at least 2, at most 3)
zero 1 free abelian group of rank three No No 1 1 3
generator (1 or -1 if we identify the cohomology group with $\mathbb{Z}$) 2 unitriangular matrix group:UT(3,Z) Yes Yes 2 2 2
any other element (one type for each positive integer $n > 1$) 2 (for each positive integer greater than 1) central product of unitriangular matrix group:UT(3,Z) and group of integers where the center of the former is identified with the subgroup $n\mathbb{Z}$ in $\mathbb{Z}$ No Yes 2 2 3