From a Masters Qual. Practice Exam:

Let $G \leq GL(3, \mathbb{F}_3)$, be the group of invertible $3 × 3$ upper triangular matrices over the field with $3$ elements (i.e. entries below the diagonal are zero). Find the order of $G$ and show that $G$ is not a simple group i.e. show that $G$ has a proper non-trivial normal subgroup.

If we didn't have "invertible" condition, this would give us $3^6$ matrices, but I'm not sure how to quickly weed out which are and which aren't invertible.

Intuitively, I think the normal subgroup should be the diagonal matrices (I could be wrong), but I'm not sure how to prove that without some really gross matrix multiplication.

  • $\begingroup$ Since the matrices are upper triangular, the determinant is the product of the $3$ elements in the main diagonal. Thus, if you avoid zeros there, the matrix will be invertible. $\endgroup$
    – math
    Commented Jan 6, 2017 at 2:26
  • $\begingroup$ @MathChat Of course! It hadn't occurred to me to think of it that way. $\endgroup$ Commented Jan 6, 2017 at 2:35
  • $\begingroup$ Also, if you remember the general fact that the center of any general linear group is the subgroup of scalar matrices (so a subset of the diagonal matrices), that would be a good normal subgroup here. $\endgroup$
    – pjs36
    Commented Jan 6, 2017 at 2:43

1 Answer 1


In order that an upper triangular matrix

$$\begin{pmatrix} a & b & c \\ 0 & d& e \\ 0 &0 & f\end{pmatrix}$$

be invertible, it is necessarily and sufficient that the determinant $adf$ be nonzero. This means that $b,c,e$ can be anything you want, while $a, d, f$ cannot be zero. This means there are

$$3 \cdot 3 \cdot 3 \cdot 2 \cdot 2 \cdot 2 = 216 $$

such matrices.

The subgroup $D$ of diagonal invertible matrices is as far from being normal in $G$ as you can get (if $x \in G$, but not in $D$, then $xDx^{-1} \neq D$).

However, if you multiply two upper triangular matrices $x, y \in G$, notice that the entries on the diagonal of $xy$ are obtained by multiplying the corresponding entries on the diagonal of $x$ and $y$. This implies that

$$N = \{ \begin{pmatrix} 1 & b & c \\ 0 & 1& e \\ 0 &0 & 1\end{pmatrix} : b, c, e \in \mathbb{F}_3 \}$$ is a normal subgroup of $G$.

Actually, $G$ is the semidirect product of $N$ and $D$.

  • $\begingroup$ Thanks! Looks great. One question: is there a mathematical definition for "as far from being normal as you can get"? $\endgroup$ Commented Jan 6, 2017 at 2:37
  • $\begingroup$ Yes. If $H$ is a subgroup of $G$, the normalizer $N_G(H) = \{ g \in G : gHg^{-1} = H\}$ contains $H$, and is the unique maximal subgroup of $G$ in which $H$ is normal. So $H$ is normal in $G$ if and only if $N_G(H) = G$, and the other extreme is when $N_G(H) = H$, as we have here. $\endgroup$
    – D_S
    Commented Jan 6, 2017 at 2:44
  • 1
    $\begingroup$ Wow. I was actually joking, but that's really cool! $\endgroup$ Commented Jan 6, 2017 at 2:48

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