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If G is a general linear group GL( n, q ) of characteristic p, and U is a Sylow p-subgroup of G, then how many elements of U are transvections?

The size of a centralizer of a transvection is relevant.

If $g = \left[\begin{smallmatrix} 1 & 1 & . & . & . & . \\ . & 1 & . & . & . & . \\ . & . & 1 & . & . & . \\ . & . & . & 1 & . & . \\ . & . & . & . & 1 & . \\ . & . & . & . & . & 1 \\ \end{smallmatrix}\right],$ then $C_{GL}(g) = \left\{\left[ \begin{smallmatrix} a & * & * & * & * & * \\ . & a & . & . & . & . \\ . & * & * & * & * & * \\ . & * & * & * & * & * \\ . & * & * & * & * & * \\ . & * & * & * & * & * \\ \end{smallmatrix}\right]\right\},$ and so $C_U(g) = \left\{\left[ \begin{smallmatrix} 1 & * & * & * & * & * \\ . & 1 & . & . & . & . \\ . & . & 1 & * & * & * \\ . & . & . & 1 & * & * \\ . & . & . & . & 1 & * \\ . & . & . & . & . & 1 \\ \end{smallmatrix}\right]\right\}.$

Then $g^x = \left[ \begin{smallmatrix} 1 & 1 & * & * & * & * \\ . & 1 & . & . & . & . \\ . & . & 1 & . & . & . \\ . & . & . & 1 & . & . \\ . & . & . & . & 1 & . \\ . & . & . & . & . & 1 \\ \end{smallmatrix}\right]$ for all $x = \left[\begin{smallmatrix} 1 & . & . & . & . & . \\ . & 1 & * & * & * & * \\ . & . & 1 & . & . & . \\ . & . & . & 1 & . & . \\ . & . & . & . & 1 & . \\ . & . & . & . & . & 1 \\ \end{smallmatrix}\right].$

The U-conjugacy class of G has $q^{n-2}$ elements. However, there are other transvections, and I'm not so sure how to count them all.

Checking small $(n,q)$ I get a possible formula:

$$ nq^{n-1} - \tfrac{q^n-1}{q-1}$$

In particular, I don't get that each transvection must look like a "row", and I don't get that every element is a transvection. It'd be nice to know what they "look" like as matrices.

Just in case I made a mistake in linear algebra: To check if an element g is a transvection, I check that $\operatorname{Rank}(g-1) = 1$ and $(g-1)^2 = 0$.

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Give that the matrix $g$ is unipotent, you only need ${\rm rank}(g-I) = 1,$ the other condition you are checking will automatically be satisfied. –  Geoff Robinson Jun 19 '12 at 17:57
    
Alperin (2006) ams.org/mathscinet-getitem?mr=2208106 gives a nice formula for conjugacy class sizes in terms of numbers of flags centralized. This is done for the purpose of counting conjugacy classes (which should be a polynomial in q). Robinson (1998) ams.org/mathscinet-getitem?mr=1633196 made more direct headway on this question. Goodwin–Röhrle (2010) ams.org/mathscinet-getitem?mr=2602681 solve this, also for the unipotent radicals of parabolic subgroups. –  Jack Schmidt Jun 20 '12 at 23:35
    
Goodwin-Rohrle solve the question in low dimensions, and for parabolics, not in general –  Geoff Robinson Jun 21 '12 at 0:32

2 Answers 2

up vote 4 down vote accepted

Doesn't it just depend on the lowest non-zero row of $g-I$? Let's say this is row $r$ (starting counting from the bottom). Let's say that the lowest $k$ rows of $g-I$ are zero, but the next one up isn't. If I'm counting right, there are $q^{k}-1$ ways to complete that row with a non-zero vector, then, given that choice, $q^{n-k-1}$ ways to fill out the rows above that so that $g-I$ has rank $1.$ So the answer looks like $\sum_{k=1}^{n-1} (q^{n-1} - q^{n-k-1}).$ I think this agrees with what you wrote.

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Thanks, this is very simple. –  Jack Schmidt Jun 19 '12 at 23:35

This answer is just for reference on transvections in GL( n, K ). One can skip the conjugacy section without affecting the argument. An element $g$ of $U$ has the property that it stabilizes the standard maximal flag:

$$e_k g = e_k + v \qquad \text{for some } v \in \langle e_{k+1}, \dots, e_n \rangle$$

Appearance of transvections

Let $g \in U$ be a transvection. Set $i = \max\{ i : e_i g \neq e_i \}$. Since $g \neq 1$, $i < \infty$, and since $g \in U$, $i < n$. Set $j = \max\{ g e_i - e_i \notin \langle e_j, e_{j+1}, \dots, e_n \rangle \}$. Since $e_ig \neq e_i$, $j \leq n$ and since $g\in U$, $i \leq j$. Write $e_ig = e_i + \alpha e_j + v$ for $v \in \langle e_{j+1}, \dots e_n\rangle$. Then we have the following formula for $e_k g$: $$e_k g = \begin{cases} e_k & \text{ if } k > i \\ e_i + \alpha e_j + v & \text{ if } k = i \\ e_k + \beta_k( \alpha e_j + v) & \text{ if } k < i \end{cases}$$ where the $\beta_k$ follows form the assumption $g-1$ has rank 1.

For example, if $n=8$, $i=3$, and $j=5$, then any choice of $\alpha \in k^\times$ and $\beta \in k^{i-1}$ and $v \in k^{n-j}$ gives a transvection such as the following: $$g(3,5,\alpha;\beta, v) = \left[\begin{smallmatrix} 1 & . & . & . & \beta_1 \alpha & \beta_1 v_1 & \beta_1 v_2 & \beta_1 v_3 \\ . & 1 & . & . & \beta_2 \alpha & \beta_2 v_1 & \beta_2 v_2 & \beta_2 v_3 \\ . & . & 1 & . & \alpha & v_1 & v_2 & v_3 \\ . & . & . & 1 & . & . & . & . \\ . & . & . & . & 1 & . & . & . \\ . & . & . & . & . & 1 & . & . \\ . & . & . & . & . & . & 1 & . \\ . & . & . & . & . & . & . & 1 \\ \end{smallmatrix}\right] $$

In other words, pick a corner $(i,j)$ and a nonzero entry $\alpha$ for it, then fill in the rest of the rectangle (growing up and right) with a rank 1 matrix $[\beta;1] \times [\alpha,v]$. As a block matrix we have:

$$g = \begin{bmatrix} I & u^t v \\ 0 & I \end{bmatrix} \qquad u,v \text{ row vectors }$$

Conjugacy classes of transvections

In this optional section, we show that for each fixed $i,j,\alpha$, all $g(i,j,\alpha;\beta,v)$ form a conjugacy class of size $q^{i+n-j}$ with representative the standard transvection $E_{ij}(\alpha)$ from linear algebra courses.

Set $h$ to be the transvection given by: $$e_k h = \begin{cases} e_j + \alpha^{-1} v & \text{if } k = j \\ e_k & \text{if } k \neq j \\ \end{cases}$$

Calculate $e_k h g h^{-1}$ to be in a nice form: $$e_k h g h^{-1} = \left\{\begin{array}{lllll} (e_j + \alpha^{-1} v) g h^{-1} & = (e_j + \alpha^{-1} v) h^{-1} & = e_k & \text{if } k = j \\ e_k g h^{-1} & = e_k h^{-1} & = e_k & \text{if } k > i, k \neq j \\ e_k g h^{-1} & = (e_i + \alpha e_j + v) h^{-1} & = e_k + \alpha e_j & \text{if } k = i \\ e_k g h^{-1} & = (e_k + \beta_k (\alpha e_j + v) ) h^{-1} & = e_k + \beta_k \alpha e_j & \text{if } k < i \end{array}\right. $$

In particular $g(\alpha,\beta,v)^{h^{-1}} = g(\alpha,\beta,0)$ shows each conjugacy class contains a transvection with $g-1$ having only one nonzero column.

Now assume $v=0$ and redefine $h$ as: $$e_k h = \begin{cases} e_k - \beta_k e_i \ & \text{if } k < i \\ e_k & \text{if } k \geq i \\ \end{cases}$$

Calculate $e_k h g h^{-1}$ to be in a nice form: $$e_k h g h^{-1} = \left\{\begin{array}{lllll} e_k g h^{-1} & = e_k h^{-1} & = e_k & \text{if } k > i \\ e_k g h^{-1} & = (e_k + \alpha e_j) h^{-1} & = e_k + \alpha e_j & \text{if } k = i \\ (e_k - \beta_k e_i ) g h^{-1} & = (( e_k + \beta_k \alpha e_j) - \beta_k( e_i + \alpha e_j ) ) h^{-1} & = ( e_k - \beta_k e_i ) h^{-1} = e_k & \text{if } k < i \end{array}\right. $$

In particular, $g^{h^{-1}} = g(\alpha,0,0)$ so that every transvection is $U$-conjugate to one with $\beta=v=0$, that is, to a standard transvection $E_{ij}(\alpha)$.

Since $i,j$ are defined in terms of the maximal flag, distinct $i,j$ correspond to distinct conjugacy classes, even in $N_G(U)$. [XXX: distinct $\alpha$ are not conjugate in $U$]

[XXX: explicit centralizer of $E_{i,j}(\alpha)$]

Counting the transvections

Hence we get a sum that is basically just nested geometric series:

$$\begin{array}{rl} \#T &= \sum_{i=1}^{n} \sum_{j=i+1}^n \sum_{\alpha \in k^\times} \left|E_{ij}(\alpha)^U\right| \\ &= \sum_{i=1}^{n} \sum_{j=i+1}^n \sum_{\alpha \in k^\times} \sum_{\beta \in k^{i-1}} \sum_{v \in k^{n-j}} 1 \\ &= \sum_{i=1}^{n} \sum_{j=i+1}^n \sum_{\alpha \in k^\times} q^{(i-1)+(n-j)} \\ &= \tfrac{q-1}{q} \sum_{i=1}^{n} q^i \sum_{j=i+1}^n q^{n-j} \\ &= \tfrac{q-1}{q}\sum_{i=1}^{n} q^i \sum_{j=0}^{n-i-1} q^j \\ &= \tfrac{q-1}{q}\sum_{i=1}^{n} q^i \frac{q^{n-i}-1}{q-1} \\ &= \tfrac{q-1}{q}\sum_{i=1}^{n} \frac{q^{n}-q^i}{q-1} \\ &= \tfrac{1}{q}\sum_{i=1}^{n} q^{n}-q^i \\ &= \tfrac{1}{q}\left( n q^{n} - \frac{q^{n+1}-q}{q-1} \right) \\ &= n q^{n-1} - \frac{q^n-1}{q-1} \end{array}$$

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