5
$\begingroup$

Let $p$ be a prime number and $G=GL_n ( \mathbb{Z} / p \mathbb{Z} )$. 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 $G$. It is well known that the Sylow $p$-subgroups of a group $G$ are conjugate, and every $p$-subgroup $H$ of $G$ is contained in some Sylow $p$-subgroup of $G$. Then there exists $g\in G$ such that $H\leq gUg^{-1}$, which allows us to compute the number of conjugacy classes of elementary abelian subgroups of rank $2$ ($H=( \mathbb{Z} / p \mathbb{Z} ) ^2$) in the Sylow $p$-subgroup $U$. Any help would be appreciated so much. Thank you all.

$\endgroup$
2
$\begingroup$

I would say the problem is open.

Consider the case where $p>n$, i.e. every element has order $p$ in $U$.

The number of such subgroups is $(|U|·k(U)-1-\frac{|U|-1}{p-1}(p^2-1))\frac{1}{(p^2-1)(p^2-p)}$, where $k(U)$ is the number of conjugacy classes of $U$.

Let me explain the formula. The number of commuting pairs in any finite group G is $|G|⋅k(G)$. Such pairs may

  • Generate a trivial group (only 1 choice)

  • Generate a group of order $p$. There're $\frac{|U|-1}{p-1}$ groups of order $p$ in $U$, and for each group, there're $p^2-1$ choices to generate the group with $2$ elements.

  • Generate a group of order $p^2$. The group is surely rank $2$, and has $(p^2-1)(p^2-p)$ generating pairs. Dividing the number of such pairs in $U$ by $(p^2-1)(p^2-p)$ gives the answer.

So, the question is equivalent with determining the number of conjugacy classes of $U$, which is the Higman Conjecture.

$\endgroup$
  • $\begingroup$ Thank you, sir, but the subgroup U defined in my question is not as in Higman Conjecture. I think there is some confusion. $\endgroup$ – Nourddine Snanou Oct 6 at 18:42
  • 1
    $\begingroup$ What's the difference? Aren't they both Sylow p-subgroups? $\endgroup$ – LeechLattice Oct 7 at 3:41
  • $\begingroup$ So I am confused because I think that the number of conjugacy classes of $U$ is the index of its normalizer in $G=GL_n (\mathbb{Z} / p \mathbb{Z}) $ ($k(U)=[G,N_{G}(U)]$). $\endgroup$ – Nourddine Snanou Oct 7 at 9:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.