$p$-Power conjugacy classes in $U(n,q)$ and $T(n,q)$

TL;DR

This paper investigates the behavior of p-power maps on the unitriangular and triangular groups over finite fields, revealing large conjugacy classes in the images and providing recursive formulas for counting elements.

Contribution

It introduces new results on the structure of p-power maps in $U(n,q)$ and $T(n,q)$, including size estimates and recursive counting methods.

Findings

The image of p-power maps contains large conjugacy classes.

Recursive formulas are provided for counting elements in the image.

The results extend understanding of p-power maps in algebraic groups.

Abstract

Let $q$ be a $p$-power where $p$ is a fixed prime. In this paper, we look at the $p$-power maps on unitriangular group $U(n,q)$ and triangular group $T(n,q)$. In the spirit of Borel dominance theorem for algebraic groups, we show that the image of this map contains large size conjugacy classes. For the triangular group we give a recursive formula to count the image size.

Equations

$$|T^p|=\sum _{(a_1,\dots ,a_k)⊢n,k<q}\left(\frac{(q-1)\cdots (q-k)n!}{{\prod }_{b=1}^n{m}_b!(b!{)}^{m_b}}\right)\left(\prod _{i=1}^k|U(a_i,q{)}^p|\right)q^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)-{\sum }_{i=1}^k\left(\genfrac{}{}{0ex}{}{a_i}{2}\right)}$$

$$\frac{|T^p|}{|T|}\ge \frac{2^{n-2}}{9(q-1{)}^{n-2}{q}^{(p-1)(n-p)}}.$$

$$(n,n-1)<(n-1,n-2)<(n,n-2)<(n-2,n-3)<\cdots <(n-1,1)<(n,1).$$

$$\mu_{(r,s)}(A):=\big{(}\mu(a_{ij})\big{)}_{(i,j)\leq(r,s)},$$

$${\mathcal{G}}_{(r,s)}:=\{A=(a_{ij})\in U(n,q)\mid a_{ij}=0\text{mbox}forall(i,j)\le (r,s)\}.$$

$${\mathcal{N}}_{(r,s)}:={\mathcal{G}}_{(r,s{)}^{\ast }}/{\mathcal{G}}_{(r,s)},$$

$$U_l(n,q):=\left\{A=(a_{ij})\in U(n,q)\mid a_{ij}=0,\text{mbox}foralli-j\le l\right\}$$

$$U(n,q)=U_0(n,q)\supset U_1(n,q)\supset \cdots \supset U_l(n,q)\supset \cdots \supset U_{n-1}(n,q)=\{1\}$$

$$A(a_1,a_2,\dots ,a_{n-l-1})=I+\sum _{i=1}^{n-l-1}{a}_i{e}_{l+1+i,i}\in U_l(n,q),$$

$${\mathcal{A}}_m=\{A(a_1,a_2,\dots ,a_{n-l-1})\mid a_i=0\text{mbox}fori\le m,a_i\in F_q^{\times }\text{mbox}fori>m\}$$

$${\mathcal{B}}_m=\{A(a_1,a_2,\dots ,a_{n-l-1})\mid a_i\in F_q^{\times }\text{mbox}fori<m,a_i=0\text{mbox}fori\ge m\}.$$

$$G_m=\{B=(b_{ij})\in U_{l+1}(n,q)\mid b_{ij}=0\text{mbox}foralll+2<i\le l+m+1,1\le j<m\}.$$

$$[A,U(n,q)]=\{[A,C]\mid C\in U(n,q)\}.$$

$$b_{ij}$$

$$b_{i,j}=\sum _{r_1=j+1}^{i-1}\sum _{r_2=r_1+1}^{i-1}\cdots \sum _{r_{p-1}=r_{p-2}+1}^{i-1}{a}_{i,r_{p-1}}{a}_{r_{p-1},r_{p-2}}\cdots a_{r_1,j}$$

$$b_{i,j}=\sum _{r_1=j+1}^{i-1}\sum _{r_2=r_1+1}^{i-1}\cdots \sum _{r_{p-1}=r_{p-2}+1}^{i-1}{a}_{i,r_{p-1}}{a}_{r_{p-1},r_{p-2}}\cdots a_{r_1,j}.$$

$$b_{p+i+1,p+i}:=(b_{i+2,i+1}\cdots b_{p+i,p+i-1}{)}^{-1}{a}_{i+1},$$

$$\displaystyle|U(n,q)^{p}|\geq q^{\frac{(n-p)(n-p-1)}{2}}\big{(}(q-1)^{n-p}\big{)}+\displaystyle\sum_{m=1}^{\lfloor{\frac{n-p}{2}}\rfloor}q^{\frac{(n-p)(n-p-1)}{2}-\frac{m(m-1)}{2}}\big{(}2(q-1)^{n-p-m}\big{)}$$

$$|U(n,q{)}^p|$$

$$|U(n,q{)}^p|>q^{\frac{(n-p)(n-p-1)}{2}}\left((q-1{)}^{n-p}+2(q-1{)}^{n-p-1}\right).$$

$$\frac{|U(n,q{)}^p|}{|U_{p-1}(n,q)|}>\frac{q^{\frac{(n-p)(n-p-1)}{2}}\left((q-1{)}^{n-p}+2(q-1{)}^{n-p-1}\right)}{q^{\frac{(n-p)(n-p+1)}{2}}},$$

$$\frac{|U(n,q{)}^p|}{|U_{p-1}(n,q)|}>{\left(1-\frac{1}{q}\right)}^{n-p-1}\left(1+\frac{1}{q}\right).$$

$$\frac{|U(n,q{)}^p|}{|U_{p-1}(n,q)|}>{\left(1-\frac{1}{q}\right)}^{n-p-1}\left(1+\frac{1}{q}\right)\ge {\left(1-\frac{1}{q}\right)}^q\left(1+\frac{1}{q}\right)>\frac{1}{3}.$$

$$|D_{\delta }(n,q)|=\frac{(q-1)!n!}{(q-k-1)!{\prod }_{i=1}^n\left((i!{)}^{m_i}\cdot m_i!\right)}.$$

$$d=(\underset{a_1}{\underbrace{d_1,\dots ,d_1}},\dots ,\underset{a_k}{\underbrace{d_k,\dots ,d_k}}),$$

$$|T^p|=\sum _{\delta ⊢n}\mid D_{\delta }\mid [U:C_{\delta }]\mid C_{\delta }^p\mid$$

$$T^p=\bigcup _{d\in D,v\in C_U(d{)}^p}(dv{)}^U$$

$$\mid \bigcup _{v\in C^p}(dv{)}^U\mid =[U:C]\sum _{i=1}^{m_d}\mid v_{d,i}^C\mid =[U:C]\mid C^p\mid .$$

$$\mid T^p\mid =\sum _{d\in D}[U:C_U(d)]|C_U(d{)}^p|=\sum _{\delta ⊢n,l(\delta )<q}\mid D_{\delta }\mid [U:C_{\delta }]\mid C_{\delta }^p\mid$$

Full text

The first named author gratefully acknowledges the hospitality of Harish-Chandra Research Institute (Allahabad) and IISER Pune. The second named author would like to acknowledge support of SERB-MATRICS grant and the hospitality of the Department of Mathematics, University of Florence, Italy during his visits.

1. Introduction

Let $G$ be a finite group and $w$ be a word. The word $w$ defines a map into $G$ called a word map. It has been a subject of intensive investigation whether these maps are surjective on finite simple and quasi-simple groups; we refer to the article by Shalev [Sh] for a survey on this subject. A more general problem is to determine the image $w(G)$ of a word map and, in particular, its size. In this paper we investigate power maps, that is, maps corresponding to the word $w=X^{p}$, for the lower-triangular matrix group $T(n,q)$ and lower unitriangular matrix group $U(n,q)$ over finite fields $\mathbb{F}_{q}$, where $q$ is a $p$-power for a fixed prime $p$. Results concerning the verbal subgroup, that is the group generated by the image of the power map, for triangular and unitriangular group can be found in [Bi] and [So].

Motivated by the Borel’s dominance theorem for algebraic groups, Gordeev, Kunyavskiĭ and Plotkin started investigating the image of a non-surjective word map more closely (see [GKP, GKP1, GKP2, GKP3]). In the spirit of questions raised in [GKP3, Section 4] for algebraic groups, we address, for the groups $T(n,q)$ and $U(n,q)$, the question: Which semisimple, and unipotent elements lie in the image of the power maps and whether it contains ‘large’ conjugacy classes?

One of the motivations for our interest in the triangular and unitriangular groups lies in the fact that $T(n,q)$ is a Borel subgroup of $GL(n,q)$ and $U(n,q)$ is a Sylow $p$-subgroup of $GL(n,q)$. In the finite groups of Lie type, the regular semisimple elements play an important role as they are dense (see [GKSV]). Considering the image of a word map on maximal tori has turned out to be useful in getting asymptotic results. Thus, we aim at considering the large size conjugacy classes in $U(n,q)$, described in [VA2], and try to understand if they are in the image under the power map $w=X^{p}$. (Note that, clearly, raising to a power coprime to $p$ gives a bijection of $U(n,q)$). In what follows, we use the notation $G^{p}$ for the image $w(G)$ of a group $G$ under the word map given by $w=X^{p}$ (we call it power map). So, $G^{p}=\{g^{p}\mid g\in G\}$ is the set consisting of the $p$-powers of the elements of $G$. We remark that the verbal width with respect to power maps, that is, the smallest number $k$ such that the product of $k$-copies of $G^{p}$, coincides with the verbal subgroup $\langle G^{p}\rangle$, has already been determined: see [Bi, Theorem 5] for $G=U(n,q)$ and [So, Theorem 1] for $G=T(n,q)$.

It is known (see [Bi, Theorem 3] or Proposition 3.4) that $U(n,q)^{p}$ is contained in the subgroup $U_{p-1}(n,q)=\left\{(a_{ij})\in U(n,q)\mid a_{ij}=0,\ \forall\ i-j\leq p-1\right\}$ consisting of the lower triangular matrices with the first $p-1$ sub-diagonals having zero entries. Moreover, $U(n,q)^{p}=1$ if and only if $n\leq p$, and $U(n,q)^{p}=U_{p-1}(n,q)$ if and only if $n=p+1$ and $p+2$. Our first result, for $n\geq p+3$, is the following estimate on the set of pth powers in $U(n,q)$.

Theorem A**.**

Let $q$ be a power for a prime $p$ and $n$ an integer such that $n\geq p+3$. Then, the set $U(n,q)^{p}$ is a proper generating subset of $U_{p-1}(n,q)$ and $|U(n,q)^{p}|>\frac{1}{3}|U_{p-1}(n,q)|$ when $q\geq n-p-1$.

Next we prove the following result, which reduces the counting of $p$-powers for $T(n,q)$ to that of unitriangular groups of smaller size.

Theorem B**.**

Let $q$ be a $p$-power and suppose $q>2$. Then for the group $T=T(n,q)$ we have,

[TABLE]

where the $m_{b}$’s are obtained by writing the partition $(a_{1},\ldots,a_{k})$ in power notation as $1^{m_{1}}\ldots n^{m_{n}}$.

Using the estimate in Theorem A, we hence get

Corollary C**.**

Let $q$ be a power of a prime $p$ such that $q>n-p-1$. Then for the group $T=T(n,q)$ we have,

[TABLE]

We conclude the section with a quick layout. Theorem A is proved in Section 3 and Theorem B and Corollary C in Section 4. All groups considered in what follows are tacitly assumed to be finite.

2. Conjugacy classes in $U(n,q)$

The conjugacy classes of the unitriangular group $U(n,q)$, considered as the group of upper unitriangular matrices, have been studied in a series of papers by Arregi and Vera-López; we will use, in particular, the results in [VA1, VA2]. For the convenience of the reader, we reproduce some notations and results from [VA1] in the setting of lower unitriangular matrices, i.e., swapping the notation by taking transpose.

Let us order the index set $\mathcal{I}=\{(i,j)\mid 1\leq j\leq i\leq n\}$ in the following manner:

[TABLE]

To every $A=(a_{ij})\in U(n,q)$ and $(r,s)\in\mathcal{I}$, one associates a vector $\mu_{(r,s)}(A)$ (the $(r,s)$-weight of $A$) as follows:

[TABLE]

where $\mu(a_{ij})=0$ if $a_{ij}=0$ and $\mu(a_{ij})=1$ if $a_{ij}\neq 0$. $\mu_{(n,1)}(A)$ is called the weight of $A$ and is simply denoted by $\mu(A)$. So, $\{\mu(A)\mid A\in U(n,q)\}=\{0,1\}^{\frac{n(n-1)}{2}}$ and we totally order this set of weights by lexicographical order (considering $0<1$). For a given index $(r,s)\in\mathcal{I}$, we order $\mu_{(r,s)}(A)$ in the same manner. We remark that in [VA1], the word ‘type’ is used in place of ‘weight’. But we will use ‘weight’ as we use ‘type’ for some other purpose.

For $(r,s)\in\mathcal{I}$, define

[TABLE]

It is a routine check to see that $\mathcal{G}_{(r,s)}$ is a normal subgroup of $U(n,q)$ having order $q^{ns-r-\frac{s(s-1)}{2}}$. As proved in [VA1, Theorem 3.2], every conjugacy class in $U(n,q)/\mathcal{G}_{(r,s)}$ contains a unique element of minimum $(r,s)$-weight. A matrix $A\in U(n,q)$ is said to be canonical if $A\mathcal{G}_{(r,s)}$ is the unique element of its conjugacy class in $U(n,q)/\mathcal{G}_{(r,s)}$ having minimal $(r,s)$-weight for all $(r,s)\in\mathcal{I}$.

For each $(r,s)\in\mathcal{I}$, let us define

[TABLE]

where $(r,s)^{*}$ denotes the preceding pair of $(r,s)$ in the ordering of $\mathcal{I}$ defined above. It follows from [VA1, Lemma 3.4] that for every $A\in U(n,q)$ and $(r,s)\in\mathcal{I}$ the number of conjugacy classes in $U(n,q)/\mathcal{G}_{(r,s)}$ which intersect with $\bar{A}\mathcal{N}_{(r,s)}$ is either $1$ or $q$, where $\bar{A}=A\mathcal{G}_{(r,s)}$. We say that $(r,s)\in\mathcal{I}$ is an inert point of $A\in U(n,q)$ if the number in the preceding statement is $1$.

The following two results are restatements of [VA1, Lemma 3.7, Lemma 3.8] for lower unitriangular matrices.

Lemma 2.1**.**

Let $A\in U(n,q)$ be a canonical matrix such that $a_{rs}\neq 0$ and $a_{js}=0$ for all $j$ such that $s<j<r$. Then the pairs $(r,s^{\prime})$, with $s^{\prime}<s$, are inert points of $A$.

Lemma 2.2**.**

Let $A\in U(n,q)$ be a canonical matrix such that $a_{rs}\neq 0$ and $a_{ri}=0$ for all $i$, $s<i<r$. Then the pair $(r^{\prime},s)$ for any $r^{\prime}>r$ is an inert point of $A$ if $a_{jr^{\prime}}=0$ for all $j>r^{\prime}$.

We set the following notation. Given $k\in\{0,1,\ldots,n-1\}$, we say that the array of entries $(a_{k+1,1},a_{k+2,2},\ldots,a_{n,n-k})$ is the $k^{th}$-sub-diagonal of the matrix $A=(a_{i,j})$. For $l$ such that $0\leq l\leq n-1$, define

[TABLE]

consisting of lower unitriangular matrices whose first $l$ sub-diagonals have all zero entries. We remark that

[TABLE]

is the lower central series of $U(n,q)$, with U_{l}(n,q)=\gamma_{l+1}\big{(}U(n,q)\big{)}, and that the $U_{l}(n,q)$ are the only fully invariant subgroups of $U(n,q)$ ([Bi, Theorem 1]).

Having fixed a dimension $n$, in $GL(n,q)$ we denote by $I$ the identity matrix and by $e_{rs}$ the elementary matrix with $1$ at $(r,s)^{th}$ place and [math] elsewhere. We now turn our attention to some relevant elements of the subgroups $U_{l}(n,q)$.

For $0\leq l\leq n-2$, set

[TABLE]

where $a_{1},a_{2},\ldots,a_{n-l-1}\in\mathbb{F}_{q}$.

We have the following important property of the elements defined in (2.1)

Lemma 2.3**.**

For every choice of $0\leq l\leq n-2$ and $a_{1},a_{2},\ldots,a_{n-l-1}\in\mathbb{F}_{q}$, the element $A(a_{1},a_{2},\ldots,a_{n-l-1})$ is a canonical element of $U(n,q)$.

Proof.

In order to show that $A=A(a_{1},a_{2},\ldots,a_{n-l-1})\in U_{l}(n,q)$ is a canonical element of $U(n,q)$, we need to prove that each non-zero entry on the $l+1$-th subdiagonal of $A\mathcal{G}_{(r,s)}$ will continue to be non-zero in every $U(n,q)/\mathcal{G}_{(k,l)}$-conjugate of $A\mathcal{G}_{(r,s)}$ for all $(r,s)\in\mathcal{I}$. More generally, we observe that if $A=(a_{i,j})\in U_{l}(n,q)$ and $B=(b_{i,j})\in U(n,q)$, then the $(l+1)$-th subdiagonal of $A$ and $B^{-1}AB$ are identical modulo $\mathcal{G}_{(r,s)}$ for all pairs $(r,s)\in\mathcal{I}$. In fact, it is readily checked that the element in the $(l+1+k,k)$-th place, for $k=1,\ldots,n-l-1$, of the $(l+1)$-th subdiagonal of both $AB$ and $BA$, is simply $a_{l+1+k,k}+b_{l+1+k,k}$ modulo $\mathcal{G}_{(r,s)}$. This shows that $A$ is canonical in $U(n,q)$. ∎

We conclude this section with the following result in which we single out conjugacy classes of $U_{l}(n,q)$ of considerably large orders, including the largest ones.

Proposition 2.4**.**

Let $0\leq l\leq n-2$. For $0\leq m\leq\lfloor{\frac{n-l-1}{2}}\rfloor+1$, set

[TABLE]

and for $\lfloor{\frac{n-l-1}{2}}\rfloor<m\leq n-l$, set

[TABLE]

Then, the elements in $\mathcal{A}_{m}$ are representatives of distinct $U(n,q)$-conjugacy classes of size $q^{\frac{(n-l-1)(n-l-2)}{2}-\frac{m(m-1)}{2}}$ and the elements in $\mathcal{B}_{m}$ are representatives of distinct $U(n,q)$-conjugacy classes of size $q^{\frac{(n-l-1)(n-l-2)}{2}-\frac{(n-l-m)(n-l-m-1)}{2}}$.

Proof.

Since, by Lemma 2.3, the elements in $\mathcal{A}_{m}$ and $\mathcal{B}_{m}$ are canonical elements of $U(n,q)$, it follows by [VA1, Corollary 3.3] that these are pair-wise non-conjugate in $U(n,q)$.

Let $A\in\mathcal{A}_{m}$. Then for each $t$, $m+1\leq t\leq n-l-1$, it follows from Lemma 2.1 that there are $t-1$ inert points of $A$ corresponding to $a_{t}$. So the number of inert points of $A$ is at least $\frac{(n-l-1)(n-l-2)}{2}-\frac{m(m-1)}{2}$. Thus, by [VA1, Theorem 3.5], the conjugacy class of $A$ in $U(n,q)$ has size at least $q^{\frac{(n-l-1)(n-l-2)}{2}-\frac{m(m-1)}{2}}$. We claim that it can not be bigger than this. Let $G_{m}$ denote the subset of $U_{l+1}(n,q)$ defined as

[TABLE]

It is not difficult to see that $G_{m}$ is a normal subgroup of $U(n,q)$ having order

$q^{\frac{(n-l-1)(n-l-2)}{2}-\frac{m(m-1)}{2}}$. Notice that $[A,U(n,q)]\subseteq G_{m}$, where

[TABLE]

This shows that the size of the conjugacy class of $A$ in $U(n,q)$ is at the most $|G_{m}|$, as claimed. Hence the assertion for the elements of $\mathcal{A}_{m}$ holds.

Assertion for the elements in $\mathcal{B}_{m}$ holds on the same lines using Lemma 2.2, which completes the proof. ∎

3. Unitriangular matrix group

We look at the power map $w=X^{p}$ on the unitriangular group $U(n,q)$. We begin by stating the following results from [Bi] to improve readability of this section.

Lemma 3.1**.**

Let $A$ be a lower unitriangular matrix in $U(n,q)$ such that $A-I=(a_{ij})$. Then the matrix $A^{m}=I+(b_{ij})$ is given by:

[TABLE]

We use Lemma 3.1 to prove the following result for $p$th powers.

Corollary 3.2**.**

Let $A\in U(n,q)$ be such that $A-I=(a_{i,j})$ and $A^{p}=I+(b_{i,j})$. Then, $b_{i,j}=0$ for all $i-j<p$ and

[TABLE]

otherwise. In particular, if $n\leq p$, then $A^{p}=I$, and if $n>p$, then $U(n,q)^{p}\subseteq U_{p-1}(n,q)$.

Proof.

Since the binomial coefficients appearing in the formula of Lemma 3.1 for $m=p$ are all zero modulo $p$, except possibly the last one, we get

[TABLE]

If $i-j<p$, this is an empty sum, that is, it’s [math]. This happens for all pairs $(i,j)$ if $n<p$; giving $A^{p}=I$. If $i-j\geq p$, which actually implies that $n\geq p$, then $a_{i,j}$’s are given by the expression as stated, and obviously fall in $U_{p-1}(n,q)$. ∎

As an immediate consequence, we have the following result.

Proposition 3.3**.**

For $n>p$ and $l=p-1$, every element of $\mathcal{A}_{m}$ and $\mathcal{B}_{m}$ (defined in Proposition 2.4) is a pth power in $U(n,q)$.

Proof.

We first show that the elements $A:=A(a_{1},a_{2},\ldots,a_{n-l-1})$ defined in (2.1) for $a_{1},a_{2},\ldots,$ $a_{n-l-1}\in\mathbb{F}_{q}^{\times}$ are $p$th powers. Let $b_{2,1}=\cdots=b_{p,p-1}=1$. Then iteratively define

[TABLE]

for $0\leq i<n-p$. Now consider the lower unitriangular matrix $C:=(c_{i,j})$, where $c_{i,i-1}=b_{i,i-1}$ for $2\leq i\leq n$ and $c_{i,j}=0$ for $i-j>1$. Using Corollary 3.2, it is a routine computation to show that $C^{p}=A$.

Now let $A:=A(a_{1},a_{2},\ldots,a_{n-l-1})\in\mathcal{B}_{m}$. Then, by the definition, $a_{i}\in F_{q}^{\times}~{}\mbox{for}~{}i<m,a_{i}=0~{}\mbox{for}~{}i\geq m$. Thus, in the above procedure, $b_{i,i-1}=0$ for $p+m\leq i\leq n$. Considering $C:=(c_{i,j})$, where $c_{i,i-1}=b_{i,i-1}$ for $2\leq i\leq n$ and $c_{i,j}=0$ for $i-j>1$, we see, again using Corollary 3.2, that $C^{p}=A$.

For $A:=A(a_{1},a_{2},\ldots,a_{n-l-1})\in\mathcal{A}_{m}$, let $b_{i,i-1}=0$ for $2\leq i\leq m+1$, $b_{i,i-1}=1$ for $m+2\leq i\leq m+p$ and then iteratively define

[TABLE]

for $m\leq i<n-p$. Again considering $C:=(c_{i,j})$, where $c_{i,i-1}=b_{i,i-1}$ for $2\leq i\leq n$ and $c_{i,j}=0$ for $i-j>1$, it follows that $C^{p}=A$, which completes the proof. ∎

The following proposition, which follows from the above formulas, is proved in [Bi, Theorem 2, Theorem 3] (also see [Hu, III, Satz 16.5]).

Proposition 3.4**.**

Let $q$ be a $p$-power. Then,

1. (1)

for $n\leq p$, $U(n,q)^{p}=1$; 2. (2)

for $n=p+1$ and $n=p+2$, $U(n,q)^{p}=U_{p-1}(n,q)$; 3. (3)

for $n\geq p+3$, $U(n,q)^{p}\subset U_{p-1}(n,q)$ and $\langle U(n,q)^{p}\rangle=U_{p-1}(n,q)$.

We now provide a lower bound on $|U(n,q)^{p}|$.

Proposition 3.5**.**

Let $q$ be a power of $p$ and $n$ an integer such that $n\geq p+3$. Then, if $n-p$ is even,

[TABLE]

and if $n-p$ is odd,

[TABLE]

where $r=\lfloor{\frac{n-p}{2}}\rfloor+1$.

Proof.

It follows from Proposition 3.3 that every element of $\mathcal{A}_{m}$ as well as of $\mathcal{B}_{m}$ is a $p$th power in $U(n,q)$. The result now follows by considering the sizes of all distinct conjugacy classes of elements of $\mathcal{A}_{m}$ and $\mathcal{B}_{m}$ obtained in Proposition 2.4. ∎

We are now ready to prove Theorem A.

Proof of Theorem A.

The first assertion follows from Proposition 3.4. For the second assertion, by Proposition 3.5, we have

[TABLE]

Hence,

[TABLE]

which implies

[TABLE]

Thus, if we take $q\geq n-p-1$, then we get

[TABLE]

This completes the proof. ∎

We conclude this section with some computations using MAGMA [MAGMA], which are as follows.

[TABLE]

In view of the values in the last row of this table, we remark that the condition on $q$ in Theorem A can not be completely dropped.

4. Triangular matrix group

In this section we consider the group of triangular matrices $T(n,q)$, where $q$ is a power of a prime $p$, aiming at computing the size $|T(n,q)^{p}|$ of the set of its $p$-powers. Since the group $T(n,2)=U(n,2)$ we assume $q>2$, now onwards. We begin with setting up some notation. We denote by $D(n,q)$ the subgroup of $T(n,q)$ consisting of the diagonal matrices. The elements of $D(n,q)$ can be grouped in “types” in such a way that all elements of each type have the isomorphic centralizers in $T(n,q)$. We recall that a partition of a positive integer $n$ is a sequence of positive integers $\delta=(a_{1},\ldots,a_{k})$ such that $a_{1}\geq a_{2}\geq\ldots\geq a_{k}>0$ and $\displaystyle{\sum_{j=1}^{k}}a_{j}=n$. One can also write the partition $\delta$ in power notation $1^{m_{1}}2^{m_{2}}\ldots n^{m_{n}}$ where $m_{i}=|\{a_{j}\mid a_{j}=i\}|$ is the number of parts $a_{j}$’s equal to $i$, for $1\leq i\leq n$; so, $m_{i}\geq 0$ and $\displaystyle{\sum_{i=1}^{n}}im_{i}=n$.

Let $\Pi=\{X_{1},X_{2},\ldots,X_{k}\}$ be a set-partition of $I_{n}=\{1,2,\ldots,n\}$, i.e., a family of non-empty and pairwise disjoint subsets of $I_{n}$, whose union is $I_{n}$. Setting $a_{i}=|X_{i}|$ and assuming, as we may, that $a_{i}\geq a_{j}$ for $i\leq j$, the tuple $\delta=(a_{1},a_{2},\ldots,a_{k})$ is a partition of the number $n$; we say that $\delta$ is the type of $\Pi$.

A diagonal matrix $d\in D(n,q)$, seen as a map from the set $I_{n}=\{1,2,\ldots,n\}$ to the set $F_{q}^{\times}$ of the non-zero elements of the field $F_{q}$, determines in a natural way a set-partition $\Pi_{d}$ of $I_{n}$, namely the family of the non-empty fibers of the map $d$. We set $\delta=\tau(\Pi_{d})$ as the type of $d$ and we write $\delta=\tau(d)$.

We denote the number $k$ of parts in $\delta$ by $l(\delta)$, the length of $\delta$, and we observe that there exist elements of type $\delta$ in $D$ if and only if $l(\delta)<q$. (Thus not all partitions of $n$ may appear as type of an element in $D(n,q)$, when $q\leq n$).

Given a partition $\delta$ of $n$ with $l(\delta)<q$, we denote by $D_{\delta}(n,q)=\{d\in D\mid\tau(d)=\delta\}$ the set of all elements in $D(n,q)$ of the given type $\delta$.

Lemma 4.1**.**

Let $n$ be a positive integer and let $\delta=(a_{1},a_{2},\ldots,a_{k})$ be a partition of $n$. Write $\delta$ in power notation as $1^{m_{1}}\ldots n^{m_{n}}$ and assume that $k=l(\delta)<q$. Then

[TABLE]

Proof.

Let $\Delta$ be the set consisting of the set-partitions of $I_{n}$ having type $\delta$. As above, we associate to a diagonal element $d\in D_{\delta}$ a set-partition $\Pi_{d}\in\Delta$, and observe that all the fibers of the map $\pi:D_{\delta}\rightarrow\Delta$ defined by $\pi(d)=\Pi_{d}$, have the same size $\frac{(q-1)!}{(q-k-1)!}$ (the number of injective maps from a set of $k=l(\delta)$ elements into a set of $q-1$-elements). On the other hand, the cardinality $|\Delta|$ is easily determined by looking at the natural transitive action of the symmetric group $S_{n}$ on $\Delta$ and observing that the stabilizer in $S_{n}$ of a partition of type $\delta=1^{m_{1}}\ldots n^{m_{n}}$ has size $\prod_{i=1}^{n}\left((i!)^{m_{i}}\cdot m_{i}!\right)$. ∎

Lemma 4.2**.**

For any partition $\delta$ of $n$, with $l(\delta)<q$, and for any element $d\in D_{\delta}(n,q)$, all centralizers $C_{U(n,q)}(d)$ belong to the same isomorphism class. Moreover, if $\delta=(a_{1},a_{2},\ldots,a_{k})$, then $\mid C_{U(n,q)}(d)\mid=q^{\sum_{i=1}^{k}{\binom{a_{i}}{2}}}.$

Proof.

Let $\delta=(a_{1},a_{2},\ldots,a_{k})$ be a given partition of $n$, with $k<q$, and let

[TABLE]

where $d_{1},\ldots,d_{k}$ are distinct elements of $F_{q}$, be a ’standard-form’ element in $D_{\delta}$. Write $G=GL_{n}(q)$, $U=U(n,q)$. It is well known that $C_{G}(d)=\prod_{i=1}^{k}GL_{a_{i}}(q)$, the subgroup of $\delta$-block matrices.

Let $b=b_{\pi}\in G$ be a permutation matrix, where $\pi\in S_{n}$. We will show that $C_{U}(d^{b})$ is isomorphic to $C_{U}(d)$. In order to do this, it is enough to show that the two subgroups have the same order, since $C_{U}(d)=C_{G}(d)\cap U$ is a Sylow $p$-subgroup of $C_{G}(d)$ and $C_{U}(d^{b})=C_{G}(d)^{b}\cap U\cong C_{G}(d)\cap U^{b^{-1}}$ is a $p$-subgroup. We denote by $M=\prod_{i=1}^{k}M_{a_{i}}(q)$, the $\delta$-blocks matrix algebra and write $C=C_{G}(d)$. We observe that $M\cap U=C\cap U$, $M^{b}\cap U=C^{b}\cap U$ and that, arguing by induction on the number $k$ of diagonal blocks, in order to prove that $|M\cap U|=|M^{b}\cap U|$ we can reduce to the case $k=2$.

Now, $M\cap U=(I_{n}+V)\cap U=I_{n}+(V\cap U_{0})$, where $V$ is the $F_{q}$-space spanned by the set of pairs of elementary matrices $\{e_{i,j},e_{j,i}\}$, where $1\leq i<j\leq a_{1}$ or $a_{1}+1\leq i<j\leq n$, and the $F_{q}$-space $U_{0}$ is spanned by the $e_{i,j}$’s with $i<j$. Observing that for every pair $i,j$ we have $e_{i,j}^{b}=e_{\pi(i),\pi(j)}$ and that the pair $\{e_{i,j},e_{j,i}\}^{b}=\{e_{\pi(i_{,}\pi(j)},e_{\pi(j),\pi(i)}\}$ contains exactly one element in $V^{b}\cap U$, we conclude that the $F_{q}$-spaces $V^{b}\cap U_{0}$ and $V\cap U_{0}$ have the same dimension. Therefore, $|C\cap U|=|M\cap U|=|M^{b}\cap U|=|C^{b}\cap U|$

We finish by noticing that $|C_{U(n,q)}(d)|=\prod_{i}q^{\binom{a_{i}}{2}}$ is independent on the choice of the elements $d_{1},\ldots,d_{k}$ and that every element in $D_{\delta}$ is conjugate by a permutation matrix to a ’standard-form’ element $d$ as above. ∎

Before proving Theorem B, we recall some elementary facts.

For any fixed prime number $p$ and an element (of finite order) $g$ of a group $G$, we can write in a unique way $g=xy$, where $x$ and $y$ commute, $x$ is a $p$-element and $y$ has order coprime to $p$. We call $x$ the $p$-part of $g$ and $y$ the $p^{\prime}$-part of $g$.

Also, if $g,h\in G$ are elements of coprime order and they commute, then $C_{G}(gh)=C_{G}(g)\cap C_{G}(h)$.

Theorem 4.3**.**

Let $T=T(n,q)$, $U=U(n,q)$ and $D=D(n,q)$. Then,

[TABLE]

where the sum runs over all partitions $\delta$ of $n$ with length $\leq q-1$, $D_{\delta}=\{d\in D\mid\tau(d)=\delta\}$ and $C_{\delta}=C_{U}(d_{\delta})$ for some $d_{\delta}\in D_{\delta}$.

Proof.

We first prove that

[TABLE]

where $(dv)^{U}=\{(dv)^{y}\mid y\in U\}$ is an orbit under the action by conjugation of $U$; we call it a $U$-class.

To prove (4.1), let us consider an element $x=dv$ on the right hand side where $d\in D$ and $v=u^{p}$ for some $u\in C_{U}(d)$. Since $(p,|D|)=1$ we can write $d=d_{0}^{p}$ for some suitable $d_{0}\in D$; note that $u\in C_{U}(d_{0})=C_{U}(d)$. Hence, $x=d_{0}^{p}u^{p}=(d_{0}u)^{p}\in T^{p}$. Since $T^{p}$ is $U$-invariant, this proves that $T^{p}$ contains the union on the right side of (4.1). Conversely, consider $x=t^{p}$ with $t\in T$. Now write $t=d_{0}u$ where $u$ and $d_{0}$ are the $p$-part and $p^{\prime}$-part of $t$, respectively. So, in particular, $u$ and $d_{0}$ commute. Let $y\in U$ such that $d_{0}^{y}\in D$ (such an element certainly exists, as $D$ is a $p$-complement of $T$ and the $p$-complements of $T$ are a single orbit under the action of $U$). Write $d_{1}=d_{0}^{y}$ and $v=u^{y}$. Then $x^{y}=(t^{p})^{y}=(t^{y})^{p}=((d_{0}u)^{y})^{p}=(d_{1}v)^{p}=d_{1}^{p}v^{p}$. Now $d=d_{1}^{p}\in D$ and $C_{U}(d)=C_{U}(d_{1})$. This proves the other inclusion.

Next, we observe that for elements $u,v\in U$ and $c,d\in D$ which satisfy $[u,c]=1=[v,d]$ and $(cu)^{U}=(dv)^{U}$ then $c=d$. Let $y\in U$ such that $x=dv=c^{y}u^{y}$. Note that $v$ and $d$ are the $p$-part and $p^{\prime}$-part of $x$, respectively, and that the same is true for $u^{y}$ and $c^{y}$. By uniqueness of $p$ and $p^{\prime}$-parts, we hence get $v=u^{y}$ and $d=c^{y}$. In particular, $c^{-1}d=c^{-1}c^{y}=[c,y]\in D\cap U=1$, so $c=d$.

We also have that if $cu\in(dv)^{U}$, for $c,d\in D$, $u\in U$ and $v\in C_{U}(d)^{p}$, then $cu=du=dv^{y}$ for some $y\in C_{G}(d)$. Therefore, the family of $U$-classes in $T^{p}$ is in bijection with the set of pairs $(d,v_{d,i})$, where $d\in D$ and $v_{d,1},\ldots,v_{d,m_{d}}$ is a set of representatives of the $C_{U}(d)$-classes in $C_{U}(d)^{p}$. For a fixed $d\in D$, write $C=C_{U}(d)$ and let $v\in C^{p}$. Observe that $C_{U}(dv)=C_{U}(d)\cap C_{U}(v)=C_{C}(v)$, because $d$ and $v$ are commuting elements of coprime order, so $|(dv)^{U}|=|U:C||v^{C}|$. Hence, we have

[TABLE]

By Lemma 4.2 we conclude that

[TABLE]

where $C_{\delta}=C_{U}(d_{\delta})$ for any fixed $d_{\delta}\in D_{\delta}$. ∎

We will now prove Theorem B.

Proof of the Theorem B.

Proof the the theorem is obtained by simply substituting the values in the formula obtained the Theorem 4.3 above. The value of $|D_{\delta}|$ is computed in the Lemma 4.1. The value of $[U:C_{\delta}]$ is obtained from Lemma 4.2. Now to obtain the last term $|C_{\delta}^{p}|$ we use the fact that $C_{\delta}\cong\prod_{i=1}^{k}U(a_{i},q)$. This completes the proof. ∎

Next, we apply the formula obtained in Theorem B to compute some examples.

Example 4.4**.**

Let $n=3$, $q=5=p$. Let $T=T(3,5)$ and we want to compute $|T^{5}|$. In this case the partitions $\delta$ such that $1\leq l(\delta)\leq\min(n,q-1)=3$) are $(3)$, $(2,1)$ and $(1,1,1)$. Now, $|\Delta_{(3)}|=4$, $|\Delta_{(2,1)}|=3\cdot(4\cdot 3)=2^{2}3^{2}$ and $|\Delta_{(1,1,1)}|=4\cdot 3\cdot 2=2^{3}3$. Further $|d^{T}|=[U:C_{U}(d)]$ is, according to type, as follows: $1$ for $(5)$, $5^{2}$ for $(2,1)$ and $5^{3}$ for $(1,1,1)$. Hence,

[TABLE]

Example 4.5**.**

Let $n=6$, $q=p=3$, $T=T(6,3)$, $D=D(6,3)$ and $U=U(6,3)$.

The partitions $\delta$ of $6$ of length at most two are $(6),(5,1),(4,2),(3,3)$ and for $d_{\delta}\in D_{\delta}$ we have the following

[TABLE]

where we have used the fact that $C_{U}(d_{(6)})\cong U(6,3)$ and that $|U(6,3)^{3}|=585$ (by direct computation).

Hence, we get

[TABLE]

We finish by proving Corollary C.

Proof of Corollary C.

We will consider just the partitions (of length $2$) $(n-i,i)$ for $1\leq i\leq\lfloor{n/2}\rfloor$. Hence, by Theorem A and Theorem B we have

[TABLE]

Hence,

[TABLE]

∎