Center of 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) group of prime 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) elementary abelian group of prime-square order.
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Definition

Let $p$ be an odd prime (The case $p=2$, where we get center 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:

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

In other words, this is the subgroup given by $a_{12}=a_{23}=0$.

Cosets

The subgroup has $p^{2}$ cosets each of size $p$, where each coset is parametrized by fixed values of $a_{12}and<math>a_{23}$. In other words, for any fixed $a_{12}\in \mathbb {F} _{p}$ and $a_{23}\in \mathbb {F} _{p}$, the corresponding coset of $H$ in $G$ is the set of all matrices with those values of $a_{12}$ and $a_{23}$, and with $a_{13}$ varying over all of $\mathbb {F} _{p}$.

Complements

The subgroup has no permutable complements and also no lattice complements.

Properties related to complementation

Property	Meaning	Satisfied?	Explanation
complemented normal subgroup	normal subgroup with a complement	No	nilpotent and non-abelian implies center is not complemented
permutably complemented subgroup	subgroup with a permutable complement.	No	nilpotent and non-abelian implies center is not complemented
lattice-complemented subgroup	subgroup with a lattice complement.	No	nilpotent and non-abelian implies center is not complemented

Effect of subgroup operators

Function	Value as subgroup (descriptive)	Value as subgroup (link)	Value as group
normalizer	the whole group	--	prime-cube order group:U(3,p)
centralizer	the whole group	current page	prime-cube order group:U(3,p)
normal core	the subgroup itself	current page	group of prime order
normal closure	the subgroup itself	current page	group of prime order
characteristic core	the subgroup itself	current page	group of prime order
characteristic closure	the subgroup itself	current page	group of prime order
commutator with whole group	the trivial subgroup	current page	trivial group

Subgroup-defining functions

The subgroup is a characteristic subgroup of the whole group and arises as a result of many subgroup-defining functions on the whole group. Some of these are given below.

Subgroup-defining function	Meaning in general	Why it takes this value
center	set of elements that commute with every element	To see that every element of $H$ is in the center, simply use the formula for multiplication and simplify algebraically. The same algebraic approach can be used to show that nothing else is in the center.
derived subgroup	subgroup generated by commutators of all pairs of elements in the group, smallest subgroup with abelian quotient	The quotient (called the abelianization) is $G/H$, which is isomorphic to elementary abelian group of prime-square order. No smaller subgroup works, because the quotient by the trivial subgroup is isomorphic to $G$, which is non-abelian.
socle	subgroup generated by all the minimal normal subgroups	It is the unique minimal normal subgroup (hence is also a monolith). In general, for a finite $p$-group, the socle is $\mho ^{1}(Z(G))$.
Frattini subgroup	intersection of all the maximal subgroups	$H$ is the intersection of maximal subgroups obtained by taking $H$ along with any single element outside it.
ZJ-subgroup	center of the join of abelian subgroups of maximum order	The join of abelian subgroups of maximum order (the Thompson subgroup) is the whole group, so its center is $H$.

Characteristicity and related properties satisfied

Property	Meaning	Satisfied?	Explanation
normal subgroup	invariant under inner automorphisms	Yes	center is normal
characteristic subgroup	invariant under all automorphisms	Yes	center is characteristic, derived subgroup is characteristic
fully invariant subgroup	invariant under all endomorphisms	Yes	derived subgroup is fully invariant
verbal subgroup	generated by set of words	Yes	derived subgroup is verbal
normal-isomorph-free subgroup	no other isomorphic normal subgroup	Yes
isomorph-free subgroup, isomorph-containing subgroup	No other isomorphic subgroups	No	There are other subgroups of order $p$. In fact, every element of the group has order $p$.
isomorph-normal subgroup	Every isomorphic subgroup is normal	No	There are other subgroups of order two that are not normal: $\langle x\rangle$ etc.
homomorph-containing subgroup	contains all homomorphic images	No	There are other subgroups of order $p$.