Elementary abelian subgroups of prime-square order in unitriangular matrix group:UT(3,p)

This article is about a particular subgroup in a group parametrized by a prime (so we get one group for each prime and a corresponding subgroup for it), up to equivalence of subgroups (i.e., an isomorphism of groups that induces the corresponding isomorphism of subgroups). The subgroup is (generically) elementary abelian group of prime-square order and the group is (generically) prime-cube order group:U(3,p) (see subgroup structure of prime-cube order group:U(3,p)).
The subgroup is a normal subgroup and the quotient group is (generically) group of prime order.
VIEW: Group-subgroup pairs with the same subgroup part | Group-subgroup pairs with the same group part| Group-subgroup pairs with the same quotient part | All pages on prime-parametrized particular subgroups in groups

Let $p$ be an odd prime (The case $p=2$, where we get Klein four-subgroups of dihedral group:D8 and cyclic maximal subgroup of dihedral group:D8, is somewhat different, though many things are similar).

The group $G=U(3,p)$ is the group of unipotent upper-triangular matrices over the prime field of $p$ elements, and is also the unique (up to isomorphism) non-abelian group of order $p^{3}$ and exponent $p$. It is defined by the presentation:

$G:=\langle g,h,k\mid g^{p}=h^{p}=k^{p}=e,gh=hg,gk=kg,khk^{-1}=g^{-1}h\rangle$.

In the matrix description, each matrix $(a_{ij})$ can be described by the three entries $a_{12},a_{13},a_{23}$. The matrix looks like:

${\begin{pmatrix}1&a_{12}&a_{13}\\0&1&a_{23}\\0&0&1\end{pmatrix}}$

The multiplication of matrices $A=(a_{ij})$ and $B=(b_{ij})$ gives the matrix $C=(c_{ij})$ where:

• $c_{12}=a_{12}+b_{12}$
• $c_{13}=a_{13}+b_{13}+a_{12}b_{23}$
• $c_{23}=a_{23}+b_{23}$

With the matrix description, we can set $g$ as the matrix with $a_{13}=1$ and the other two entries zero, $h$ as the matrix with $a_{12}=1$ and the other two entries zero, and $k$ as the matrix with $a_{23}=1$ and the other two entries zero.

Now define the subgroup:

$H_{0}:=\{{\begin{pmatrix}1&a_{12}&a_{13}\\0&1&0\\0&0&1\end{pmatrix}}:a_{12},a_{13}\in \mathbb {F} _{p}\}$

In addition to this, consider the following $p$ subgroups, parametrized by $\{1,2,3,\dots ,p\}$:

$H_{i}:=\{{\begin{pmatrix}1&ia_{23}&a_{13}\\0&1&a_{23}\\0&0&1\end{pmatrix}}:a_{13},a_{23}\in \mathbb {F} _{p}\}$

All the $p+1$ subgroups $H_{0},H_{1},\dots ,H_{p}$ are elementary abelian subgroups of prime-square order and they are all automorphic subgroups, i.e., they are all related by automorphisms of the whole group.

Note that $H_{0}$ and $H_{p}$ in particular are pattern subgroups with respect to the given choice of basis.

Arithmetic functions

Function	Value	Explanation
order of whole group	$p^{3}$
order of subgroup	$p^{2}$
index of subgroup	$p$
size of conjugacy class of subgroup	1
number of conjugacy classes in automorphism class of subgroup	$p+1$
size of automorphism class of subgroup	$p+1$