Normal subgroup of upper unitriangular group $U(3,p^2)$

Question

Consider the group $U(3,p^2)$ of $3\times 3$ upper-triangular matrices whose diagonal entries are $1$ and entries are from field of order $p^2$. Then subgroup $$\begin{Bmatrix} \begin{bmatrix} 1 & 0 & *\\ & 1 & 0 \\ & & 1\end{bmatrix}\end{Bmatrix} $$ with $*$ varying over the field, is equal to the center of the group, so it is a normal subgroup of order $p^2$.

Question: Is it the only normal subgroup of order $p^2$?

In general, if we replace field of order $p^2$ by finite field of higher order $p^k$, does similar conclusion holds (that normal subgroup of order equal to order of the center is center itself?)

Answer 1

We can prove that if $N\trianglelefteq G=U(3, K)$, for a field $K$, then either $N\geq Z(G)$ or $N

Since $N$ is normal, $[N,G]\leq N$. Suppose $$ \begin{bmatrix} 1 & a & c \\ & 1 & b \\ & & 1 \end{bmatrix}\in N \text{ and } \begin{bmatrix} 1 & d & f \\ & 1 & e \\ & & 1 \end{bmatrix}\in G.$$ The commutator formula is $$ \left[ \begin{bmatrix} 1 & a & c \\ & 1 & b \\ & & 1 \end{bmatrix}, \begin{bmatrix} 1 & d & f \\ & 1 & e \\ & & 1 \end{bmatrix} \right] = \begin{bmatrix} 1 & 0 & ae-bd \\ & 1 & 0 \\ & & 1 \end{bmatrix}.$$

If both $a=0$ and $b=0$, then $N\leq Z(G)$, so we will suppose there exists an element of $N$ where either $a\ne 0$ or $b\ne 0$. Since $K$ is a field, for all $k\in K$, there exists $e,d\in K$ such that $ae-bd=k$. Therefore, the claim follows.

Therefore, if $K$ is a finite field, there is only one normal subgroup of order $|K|$.