I want to determine the group structure of $$G=\left.\left\{ \left(\begin{array}{cc} a & c \\ 0 & b \end{array}\right) \;\right|\; a,b \in \mathbb{F}_3^{×}, c \in \mathbb{F}_3\right\}.$$

My try:

The order of $G$ is $2\cdot 2\cdot 3=12$. And I checked this is not abelian, so this group is isomorphic to $D_{12}$ or $Q_{12}$ or $A_4$. I want to decide which one is. For now, I calculated the order of elements of $G$. There are $2$ elements of order $6$, $2$ elements of order $3$, $7$ elements of order $2$, $1$ elements of order $1$ (if my calculation is right).

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    $\begingroup$ How can there be $3$ elements of order $3$? And you have listed the orders of $3+7+1=11$ elements. What is the order of the last one? $\endgroup$
    – Servaes
    Commented Nov 29, 2021 at 10:24
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    $\begingroup$ (Servaes' point here is that elements of order $3$ always come in in pairs. If $x$ has order $3$, then so does $x^2$.) $\endgroup$
    – MJD
    Commented Nov 29, 2021 at 10:29
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    $\begingroup$ Your counts of elements of given orders is wrong. There are only two of order $3$, for example, and some have order $6$. $\endgroup$
    – Derek Holt
    Commented Nov 29, 2021 at 11:09
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    $\begingroup$ @JyrkiLahtonen That's not a subgroup, it's not closed under multiplication. Did you mean $b=1$? $\endgroup$ Commented Nov 29, 2021 at 12:09
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    $\begingroup$ @kleinfour Unfortunately some people denote the dihedral group of order $2n$ by $D_{2n}$ and others by $D_n$. You seem to have used $D_{2n}$ in your post and $D_n$ in your comment, which is very confusing - please be consistent. I haven't used either notation, and I did not say explicitly that $G$ is isomorphic to the dihedral group of order $12$. I was hoping that you could deduce that yourself from my comments. $\endgroup$
    – Derek Holt
    Commented Nov 29, 2021 at 16:20

3 Answers 3


You have already determined that $G$ is isomorphic to one of $D_{12}$, $Q_{12}$ or $A_4$. You have also determined that $G$ has two elements of order $6$. As $A_4$ has no elements of order $6$, this option is eliminated.

The group $Q_{12}$ has an element of order $4$, but $G$ has no elements of order $4$. This eliminates $Q_{12}$, and so $G\cong D_{12}$.


I think that concentrating on the case $p=3$ and special properties of groups of order $12$ isn't really very illuminating.

So let $p$ be an odd prime. Consider the group $$G= \left\{ \begin{pmatrix} a & b\\ 0 & d \end{pmatrix} \ :\ a,d\in\mathbb{F}_p^{*},\ \ b\in\mathbb{F}_p \right\} $$ which is clearly of order $p(p-1)^2$.

Then the centre of the whole matrix group $Z= \left\{ \begin{pmatrix} a & 0\\ 0 & a \end{pmatrix} \ :\ a\in\mathbb{F}_p^{*} \right\} $ is a cyclic subgroup of $G$ of order $(p-1)$; and so it is a normal subgroup lying in the centre of $G$.

Moreover $G$ has a subgroup $H=\left\{ \begin{pmatrix} a & b\\ 0 & 1 \end{pmatrix} \ :\ a\in\mathbb{F}_p^{*},\ \ b\in\mathbb{F}_p \right\}$ of order $p(p-1)$ which intersects $Z$ trivially, and is normalised by the central $Z$, and so by all of $G$.

Hence $G=Z\times H$.

The group $H$ is known as the affine group: it is clearly isomorphic to the set of mappings $\mathbb{F}_p\to\mathbb{F}_p$ of the form $\{x\mapsto ax+b \mid a,b\in\mathbb{F}_p, a\ne 0\}$.

In the case $p=3$ we clearly have $G=C_2\times S_3$.

  • $\begingroup$ This is essentially the same idea as in my post - locate a normal subgroup and construct the quotient. But how does this classify the group $G$? Which of the 3 groups of this type has this decomposition $Z \times H$? $\endgroup$
    – hm2020
    Commented Dec 1, 2021 at 15:49

Question: I want to determined group structure of $$G=\left.\left\{ \left(\begin{array}{cc} a & c \\ 0 & b \end{array}\right) \;\right|\; a,b \in \mathbb{F}_3^{×}, c \in \mathbb{F}_3\right\},$$

Answer: For finite non-abelian groups of matrices over finite fields one tries to realize the group as a semi direct product of smaller abelian groups:


$$H=\left.\left\{ \left(\begin{array}{cc} 1 & c \\ 0 & 1 \end{array}\right) \;\right|\; c \in \mathbb{F}_3\right\} \cong \mathbb{F}_3,$$


$$N=\left.\left\{ \left(\begin{array}{cc} a & 0 \\ 0 & b \end{array}\right) \;\right|\; a,b \in \mathbb{F}_3^*\right\} \cong (\mathbb{F}_3^*)^2 \cong (\mathbb{Z}/(2))^2.$$

It follows for any $g\in G, h\in H$ that $ghg^{-1}\in H$ hence $H \subseteq G$ is a normal subgroup and the action $\sigma$ of $G$ on $H$ is via the character $\rho:G \rightarrow \mathbb{F}_3^*$ defined by $\rho(g):=a/b$. You get an action

$$\sigma: G \times H \rightarrow H$$

defined by

$$\sigma(g,x):= \rho(g)x:=\frac{a}{b}x \text{ with $x\in \mathbb{F}_3$ and $a,b\in \mathbb{F}_3^*$ }.$$

Since $NH=G$ and $N\cap H=\{e\}$ you may express $G$ as a semi direct product

$$G \cong N \rtimes H \cong (\mathbb{F}_3^*)^2 \rtimes_{\rho(-)} \mathbb{F}_3 \cong (\mathbb{Z}/(2))^2 \rtimes_{\rho(-)} \mathbb{Z}/(3).$$

This express $G$ as a semi direct product of two known abelian groups - maybe this is helpful classifying the group. I'm not familiar with the groups you write down - you should check if this group is one on the list.

Note 1: It seems the alternating group $A_4$ sits in an exact sequence

$$0 \rightarrow (\mathbb{Z}/(2))^2 \rightarrow A_4 \rightarrow \mathbb{Z}/(3) \rightarrow 0,$$

and this may imply there is an isomorphism $G \cong A_4$. It should be possible to write down an explicit isomorphism.

Another approach: You may define the subgroup

$$H_1=\left.\left\{ h:=\left(\begin{array}{cc} u & x \\ 0 & 1 \end{array}\right) \;\right|\; u\in \mathbb{F}_3^*, x \in \mathbb{F}_3\right\} \cong \mathbb{F}_3^* \rtimes \mathbb{F}_3 ,$$

and $H_1 \subseteq G$ is another normal subgroup and you get an exact sequence

$$ 0 \rightarrow H_1 \rightarrow G \rightarrow \mathbb{Z}/(2) \rightarrow 0.$$

As commented: The group $H_1 \cong \mathbb{F}_3^* \rtimes \mathbb{F}_3$ is a non-trivial semi direct product of $\mathbb{F}_3 \cong \mathbb{Z}/(3)$ and $\mathbb{F}_3^* \cong \mathbb{Z}/(2)$ hence is not isomorphic to $\mathbb{Z}/(6)$. The group $\mathbb{F}_3^*$ acts canonically on $\mathbb{F}_3$ via multiplication. Hence $H_1$ is non-abelian.

If you look at the subgroup

$$ N_1=\left.\left\{ g:=\left(\begin{array}{cc} 1 & 0 \\ 0 & b \end{array}\right) \;\right|\; b\in \mathbb{F}_3^* \right\} \cong \mathbb{Z}/(2),$$

it follows $N_1$ acts as follows:

$$ghg^{-1}=\left(\begin{array}{cc} a & x/b \\ 0 & 1 \end{array}\right)$$

hence $g$ acts "through inversion". There is a semi direct product $G \cong H_1 \rtimes N_1$.


  • $\begingroup$ Why order of $G$ is $6$ ? $G$ 's determinant is not supposed to be $1$, so there are two $b$ for given $a$. $b$ is not defined as inverse of $a$, why you assume $b$ is inverse of $a$ ? $\endgroup$ Commented Nov 29, 2021 at 14:59
  • $\begingroup$ Great answer. I'm just a bit confused about the nomenclature. Why is the letter $N$ used for the diagonal matrices while the other one is the one that is normal (and usually called in $N$ in the context of Lie theory)? $\endgroup$
    – Vincent
    Commented Nov 29, 2021 at 15:17
  • $\begingroup$ I ask about about 'note' second one from the bottom. $A_4$ has $4$ conjugacy class, but $G$ has $6$ conjugacy class ( 4 $1$ -dimensional irreducible rep and $2$ 2-dimmensional rep). $\endgroup$ Commented Nov 29, 2021 at 16:00
  • $\begingroup$ I think $G$ and $A_4$ cannot be isom because it's number of conjugacy classes is different. $\endgroup$ Commented Nov 29, 2021 at 16:09
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    $\begingroup$ So for a quick solution, note that $G$ is a nonabelian group of order $12$ with a cyclic subgroup of order $6$ and no element of order $4$. The dihedral group is the only example with those properties. $\endgroup$
    – Derek Holt
    Commented Nov 29, 2021 at 16:27

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