0
$\begingroup$

Let $p$ be a prime number and $G=GL_n ( \mathbb{Z} / p \mathbb{Z} )$ such that $n\leq p$. Consider the set $U$ of upper-triangular matrices of $G$ having entries of $1$ on the diagonal. The cardinality of $U$ is $p^{\frac {n(n-1)} 2}$ and $U$ is a subgroup of $G$, in particular $U$ is a Sylow $p$-subgroup of $G$. Recall that an element $M$ in $G$ is said to be a cyclic matrix if the characteristic polynomial of $M$ is equal to its minimum polynomial. Let $C$ be the set of cyclic matrices of $U$.

Question: Can $U$ be written as $U=\bigcup_{M\in C}C_{U}(M)$?

Any help would be appreciated so much. Thank you all.

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ In other words, does every element of $U$ commute with a matrix in $M\in U$ such that $M-I_n$ has rank $n-1$? $\endgroup$
    – YCor
    Commented Jun 6, 2020 at 20:17
  • 2
    $\begingroup$ For $n=3$ the centralizer of $1+E_{12}$ in $U$ consists of matrices $1+aE_{12}+bE_{13}$. No such matrix is cyclic. $\endgroup$
    – YCor
    Commented Jun 6, 2020 at 20:21
  • 1
    $\begingroup$ In fact, for $n\leq p$, centralisers of cyclic matrices cover all the elements in $GL(n,p)$. See Proposition 5.1 in [][sci-hub.tw/10.1007/s10801-011-0288-2]. $\endgroup$
    – Nourddine Snanou
    Commented Jun 6, 2020 at 20:28
  • 2
    $\begingroup$ Anyway, my argument is correct. It's true that $1+E_{12}$ belongs to the centralizer to a cyclic matrix $M$, but $M$ cannot be chosen in $U$ (which is your requirement). $\endgroup$
    – YCor
    Commented Jun 6, 2020 at 20:54
  • 2
    $\begingroup$ OK thanks, so I confirm the assertion from this paper doesn't rule out my (trivial) counterexample. That proposition doesn't deal with $U$, but with the whole linear group. Computations of centralizers in $U$ for $n=3$ are quite trivial. $\endgroup$
    – YCor
    Commented Jun 6, 2020 at 20:58

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .