Second cohomology group for trivial group action of UT(3,p) on Zp

Description of the group
This article discusses the second cohomology group for trivial group action:

$$\! H^2(G;A)$$

where $$\! G = UT(3,p)$$ is unitriangular matrix group:UT(3,p), i.e., it is the unitriangular matrix group of degree three over the field of $$p$$ elements and $$A = \mathbb{Z}_p$$ is the group of prime order.

For $$p \ge 5$$, the group is isomorphic to an elementary abelian group of prime-fourth order.

The cases $$p = 2$$ and $$p = 3$$ differ somewhat from the general case. See:


 * Second cohomology group for trivial group action of D8 on Z2 for the $$p = 2$$ case
 * Second cohomology group for trivial group action of UT(3,3) on Z3 for the $$p = 3$$ case.

Elements
As noted above, some of the details differ for the $$p = 2$$ and $$p = 3$$ cases.

Note that all these extensions are central extensions with the base normal subgroup isomorphic to group of prime order and the quotient group isomorphic to prime-cube order group:U(3,p). Due to the fact that order of extension group is product of order of normal subgroup and quotient group, the order of each of the extension groups is $$p \times p^3 = p^4$$.

Some, but not all, of the extensions are stem extensions. These are not the Schur covering groups, but rather are quotients of the Schur covering group, which has order $$p^5$$.

The minimum size of generating set of the extension group is at least equal to 2 (which is the minimum size of generating set of the quotient group) and at most equal to 3 (which is the sum of the minimum size of generating set of the normal subgroup and the quotient group).

The nilpotency class is at least 2 and at most 3 in all cases. It is at least 2 because the quotient prime-cube order group:U(3,p) has nilpotency class two. It is at most 3 because the sum of the nilpotency class of the normal subgroup and quotient group is 3, and the extension is a central extension. The derived length is always exactly 2 because nilpotency class 2 or 3 forces derived length exactly 2, using derived length is logarithmically bounded by nilpotency class.

Under the action of the automorphism group of the acting group
Each of the cohomology class types mentioned above forms one orbit under the action of the automorphism group of $$U(3,p)$$.

General background
We know from the general theory that there is a natural short exact sequence:

$$0 \to \operatorname{Ext}^1_{\mathbb{Z}}(G^{\operatorname{ab}},A) \to H^2(G;A) \stackrel{\operatorname{Skew}}{\to} \operatorname{Hom}(H_2(G;\mathbb{Z}),A) \to 0$$

where $$G^{\operatorname{ab}}$$ is the abelianization of $$G$$ and its image comprises those extensions where the restricted extension of the derived subgroup $$[G,G]$$ on $$A$$ is trivial and the corresponding extension of the quotient group is abelian. Also, $$H_2(G;\mathbb{Z})$$ is the Schur multiplier of $$G$$.

We also know, again from the general theory, that the short exact sequence above splits, i.e., the image of $$\operatorname{Ext}^1_{\mathbb{Z}}(G^{\operatorname{ab}},A)$$ in $$H^2(G;A)$$ has a complement inside $$H^2(G;A)$$. However, there need not in general be a natural or even an automorphism-invariant choice of splitting.

In this case
For this choice of $$G$$ and $$A$$, $$G^{\operatorname{ab}}$$ is isomorphic to the elementary abelian group of prime-square order. The corresponding group $$\operatorname{Ext}^1_{\mathbb{Z}}(G^{\operatorname{ab}},A)$$ is also isomorphic to the elementary abelian group of prime-square order.

The Schur multiplier $$H_2(G;\mathbb{Z})$$ is the elementary abelian group of prime-square order, hence $$\operatorname{Hom}(H_2(G;\mathbb{Z}),A)$$ is also isomorphic to elementary abelian group of prime-square order.

The image of $$\operatorname{Ext}^1_{\mathbb{Z}}(G^{\operatorname{ab}},A)$$ in $$H^2(G;A)$$ comprises the $$p^2$$ non-stem extensions, which give groups of overall nilpotency class two. It has $$p^2$$ cosets in the whole cohomology group, and all the $$p^2 - 1$$ non-identity cosets are stem extensions giving groups of overall nilpotency class three.

To split the short exact sequence in an automorphism-invariant fashion, we could pick as a complement the subgroup generated by images of cyclicity-preserving 2-cocycles: comprising the identity element (extension direct product of U(3,p) and Zp) and the $$p^2 - 1$$ cyclicity-preserving extensions that give SmallGroup(p^4,7).

A quick schematic of the picture is:

The actual picture has $$p^2 - 1$$ distinct columns in place of the second column and $$p^2 - 1$$ distinct rows in place of the second row. Each row except the first contains, in its columns other than the first column, $$p - 1$$ copies of SmallGroup(p^4,8) and $$(p^2 - p)/2$$ copies each of SmallGroup(p^4,9) and SmallGroup(p^4,10). Each column except the first contains, in its rows other than the first row, $$p - 1$$ copies of SmallGroup(p^4,8) and $$(p^2 - p)/2$$ copies each of SmallGroup(p^4,9) and SmallGroup(p^4,10).