commutator subgroup of upper triangular matrix I need to compute the commutator subgroup of Unitriangular matrix $UT_{3}(p)$.
My attempt:- 
Since $UT_{3}(p) / Z(UT_{3}(p)) \simeq C_{p^2}$ and  we know that the commutator subgroup is the smallest normal subgroup with abelian quotient hence the commutator subgroup should be $Z(UT_{3}(p))$.
 A: This can be done more generally. For simplicity, we denote by $E_{12}$ the element
$$ 
\begin{bmatrix}
 1 & 1 & & & \\
   & 1 & & & \\
   &   & 1& & \\
   &   &  & \ddots & \\
   &   & & & 1\\
\end{bmatrix}
$$
here the other entries are $0$. In general, then let $E_{ij}$ denote the matrix obtained by inserting $1$ in the $(i,j)$-th place of the identity matrix, and with $i<j$ (so insertion is in upper part). 
These matrices $E_{12}, E_{13}, \cdots, E_{1n}, E_{23},\cdots , E_{2n}, \cdots$ satisfy the following commutator relation
$$[E_{ij}, E_{jk}]= E_{ik} \mbox{ and } [E_{ij}, E_{kl}]=1 \mbox{ for $j\neq k$}.$$
(see Alperin-Bell, Groups and Representations, Chapter on General Linear Groups)
Thus,we have $[E_{12}, E_{23}]=E_{13}$, $[E_{12}, E_{24}]=E_{14}$, and so on. 
You may try to write for size-$4$ matrices. Then we see that the elements 
$$
\begin{bmatrix}
1 &  0& 1 & \\
 & 1 & 0  & \\
 &  & 1 & 0\\
 &  &  & 1
\end{bmatrix}, 
\begin{bmatrix}
1 & 0 &  &1 \\
 & 1 & 0 & \\
 &  & 1 & 0 \\
 &  &  & 1
\end{bmatrix},
\begin{bmatrix}
1 & 0  &  & \\
 & 1 &0  & 1\\
 &  & 1 &0 \\
 &  &  & 1
\end{bmatrix},$$
are inside the commutator subgroup, and they span the subgroup 
$$
N=\begin{Bmatrix}
\begin{bmatrix}
1 & 0  & * & * \\
 & 1 &0  & *\\
 &  & 1 &0 \\
 &  &  & 1
\end{bmatrix}\colon * \in Z_p.\end{Bmatrix}.$$
Moreover $U(4,p)/N$ is abelian, where $N$ is generated by some commutators, by the property of commutator subgroup you mentioned, it follows $N$ is the comutator subgroup. 
