# Conjugacy classes in $GL_{n}(Z / pZ)$

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.

• Thank you, sir, but the subgroup U defined in my question is not as in Higman Conjecture. I think there is some confusion. – Nourddine Snanou Oct 6 '19 at 18:42
• What's the difference? Aren't they both Sylow p-subgroups? – LeechLattice Oct 7 '19 at 3:41
• So I am confused because I think that the number of conjugacy classes of $U$ is the index of its normalizer in $G=GL_n (\mathbb{Z} / p \mathbb{Z})$ ($k(U)=[G,N_{G}(U)]$). – Nourddine Snanou Oct 7 '19 at 9:57