Group structure of given $2\times 2$ matrix group over $\mathbb{F}_3$

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).

• 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? Commented Nov 29, 2021 at 10:24
• (Servaes' point here is that elements of order $3$ always come in in pairs. If $x$ has order $3$, then so does $x^2$.)
– MJD
Commented Nov 29, 2021 at 10:29
• Your counts of elements of given orders is wrong. There are only two of order $3$, for example, and some have order $6$. Commented Nov 29, 2021 at 11:09
• @JyrkiLahtonen That's not a subgroup, it's not closed under multiplication. Did you mean $b=1$? Commented Nov 29, 2021 at 12:09
• @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. Commented Nov 29, 2021 at 16:20

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 $$G=C_2\times S_3$$.

• 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 $Z \times H$? 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:

Let

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

and

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

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 $$G \cong H_1 \rtimes N_1$$.

https://en.wikipedia.org/wiki/Semidirect_product#Inner_semidirect_product_definitions

• 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 $a$ ? Commented Nov 29, 2021 at 14:59
• 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)? Commented Nov 29, 2021 at 15:17
• 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). Commented Nov 29, 2021 at 16:00
• I think $G$ and $A_4$ cannot be isom because it's number of conjugacy classes is different. Commented Nov 29, 2021 at 16:09
• 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. Commented Nov 29, 2021 at 16:27