# commutators in upper triangular matrices

Consider the group $T_p(n)$ of all non-singular upper triangular matrices with entries in $\mathbb{F}_p.$ Its commutator subgroup is $U_p(n)$ (all elements in $T_p(n)$ with $1$s on the main diagonal). The question is: is every element in $U_p$ a commutator of matrices in $T_p?$

• Need $p>2$, to start. Apr 8, 2016 at 13:03
• I checked for $n=3$ with $p \le 13$, $n=4$ with $p \le 7$ and $n=5$, $p=3$, and the answer was yes, but that was using completely naive checks. Apr 8, 2016 at 14:59
• A theorem of Bier and Holubowski ("A note on commutators in the group of infinite triangular matrices over a ring", Linear and Multilinear Algebra 63 (2015) no. 11, 2301-2310) gives that the answer is "yes" for $n=2$ and $p$ odd; and for $n>2$, every element is a product of at most two commutators of matrices in $T_p(n)$. Apr 8, 2016 at 17:16
• This is easy if $p > n$. Choose $a$ to be a diagonal matrix whose diagonal entries are all distinct. Then, for any elementary matrix $I + e_{i,j}$, we have $[a, e_{i,j}] = I + \lambda_{i,j} e_{i,j}$ for some nonzero $\lambda_{i,j}$. Using this, we can obtain any desired element of $U_p$ as $[a, u]$, for some $u \in U_p$, by working upward from the main diagonal (or, in other words, by working down the central series of $U_p$). Later modifications to $u$ do not change the previously determined entries of $[a, u]$. Apr 8, 2016 at 22:04
• @DaveWitteMorris: This can also be seen as follows: The centralizer of such an $a$ in the matrix ring is all diagonal matrices and thus the centralizer in $U_p(n)$ is $1$, which implies $U_p = [a, U_p]$. Similarly, when $u \in U_p$ has minimal polynomial $(x-1)^n$, then its centralizer in the matrix ring consists of polynomials in $u$, and the centralizer of $u$ in $U_p$ has order $p^{n-1}$. Thus $|[u,U_p]| = |U_p|/ p^{n-1} = |U_p'|$, and thus $[u,U_p] = U_p'$. Apr 9, 2016 at 20:21

Too long for a comment and only a partial answer for $n>2$. A note on commutators in the group of infinite triangular matrices over a ring, by Agnieszka Bier and Waldemar Holubowski, Linear and Multilinear Algebra 63 (2015) no. 11, 2301-2310; seems relevant. It includes results on finite matrices, despite the title.

Let $R$ be an associative ring with $1$, and let ${U}(R)$ be its group of invertible elements. $T(n,R)$ denotes the group of upper triangular $n\times n$ matrices; $UT(n,R)$ the set of upper unitriangular matrices ($1$s in the diagonal), $UT(n,m,R)$ the subgroup containing exactly all those matrices which have zero entries on the first $m$ super diagonals; and the basic commutators $c_k(x_1,\ldots,x_k)$ are defined inductively by $c_1(x_1)=x_1$, $c_{i+1}(x_1,\ldots,x_{i+1}) = [c_i(x_1,\ldots,x_i),x_{i+1}]$; let $\gamma_k(G)=G$ be the $k$th term of the lower central series of $G$. Finally, the Engel words $e_k(x,y)$ are given by $e_2(x,y)=[x,y]$, $e_{m+1}(x,y) = [e_m(x,y),y]$.

The authors prove:

Theorem 1.5 Let $R$ be an associative ring with $1$ such that $U(R)$ is commutative. Then:

1. $\gamma_k(UT(n,R))=UT(n,k,R)$ and every element of $\gamma_k(UT(n,R))$ is a value of the basic commutator $c_k$.
2. Every element of $\gamma_k(UT(n,R))$ is a value of the Engel word $e_k$.

Moreover, if $1$ is a sum of two invertible elements, then

1. $[T(n,R),T(n,R)] = UT(n,R)$ whenever $n\geq 2$.
2. Every element of $\gamma_2(T(2,R)) = UT(2,R)$ is a commutator.
3. Every element of $\gamma_2(T(n,R)) = UT(n,R)$ with $n>2$ is a product of at most two commutators.

In particular, for $R=\mathbb{F}_p$ with $p>2$, then the answer is "yes" for $n=2$ by point 4; but for $n>2$ they only show every element is a product of at most two commutators.

Perhaps this was already known for the special case of matrices over commutative rings/fields?