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.


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.

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  • $\begingroup$ 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 '19 at 18:42
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    $\begingroup$ What's the difference? Aren't they both Sylow p-subgroups? $\endgroup$ – LeechLattice Oct 7 '19 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 '19 at 9:57

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