# Is $\begin{bmatrix} a &b \\0 &1 \end{bmatrix}$ a cyclic group?

I appreciate all the help I can get with this task. $$G=\left \{ \begin{bmatrix} a &b \\0 &1 \end{bmatrix},a,b\in \mathbb{Z}_3, a\neq 0 \right \}$$

1. Is G a cyclic group?
1. Does a subgroup with 3 elements exists?

All elements in G: $$\begin{bmatrix} 1 &0 \\0 &1 \end{bmatrix},\begin{bmatrix} 1 &1 \\0 &1 \end{bmatrix},\begin{bmatrix} 1 &2 \\0 &1 \end{bmatrix},\begin{bmatrix} 2 &0 \\0 &1 \end{bmatrix},\begin{bmatrix} 2 &1 \\0 &1 \end{bmatrix},\begin{bmatrix} 2 &2 \\0 &1 \end{bmatrix}$$

What do I do now?

1. Is this one subgroup of order 3? $$a=1$$ generates 3 matrices. $$\begin{bmatrix} 1 &0 \\0 &1 \end{bmatrix},\begin{bmatrix} 1 &1 \\0 &1 \end{bmatrix},\begin{bmatrix} 1 &2 \\0 &1 \end{bmatrix}$$

From my book

A group is said to be cyclic if it contains an element x such that every member of G is a power of x.

1. Can $$x$$ be one of the six elements above?

EDIT: Thanks for all the help! I appreciate it very much.

• You've already got a subgroup of order $3$ staring you in the face. It's not hard to see that no element has order $6$. This takes 2 minutes to check for yourself. Jan 13 '21 at 13:31
• You could try to figure out the group multiplication law in terms of $a,b$. Jan 13 '21 at 13:32
• Calculate several powers of these. Also note that $2=-1$ in $\Bbb Z_3$. Jan 13 '21 at 13:32

A quicker way to see that $$G$$ is not cyclic: note that $$\pmatrix{2 & 0\\0 & 1} \pmatrix{1 & 1\\0&1} = \pmatrix{2 & 2\\0 & 1}, \\ \pmatrix{1 & 1\\0 & 1} \pmatrix{2 & 0\\0 & 1} = \pmatrix{2 & 1\\0 & 1}.$$ That is, we have found elements $$g,h \in G$$ with $$hg \neq gh$$. Because $$G$$ is not abelian, it cannot be cyclic.

$$\begin{pmatrix} a &b \\0 &1 \end{pmatrix}\begin{pmatrix} c &d \\0 &1 \end{pmatrix}=\begin{pmatrix} ac &ad+b \\0 &1 \end{pmatrix}$$ $$\Rightarrow$$ $$\begin{pmatrix} a &b \\0 &1 \end{pmatrix}^n=\begin{pmatrix} a^n &b(a^n+a^{n-1}+...+1)\\0 &1 \end{pmatrix}$$

which means $$a \neq 1$$ if there exists an element such that $$\langle x\rangle=G$$

$$b=0 \Rightarrow \begin{pmatrix} a &b \\0 &1 \end{pmatrix}^n=\begin{pmatrix} a^n &0 \\0 &1 \end{pmatrix}$$ which implies $$b \neq 0$$ if $$\begin{pmatrix} a &b \\0 &1 \end{pmatrix}$$ is a generator.

$$a=2 \Rightarrow$$ $$\begin{pmatrix}a^n &b(\frac{a^{n+1}-1}{a-1}) \\0 &1 \end{pmatrix}=\begin{pmatrix}2^n &b(2^{n+1}-1) \\0 &1 \end{pmatrix}$$

$$2^{n+1}=1$$ or $$2^{n+1}=2 \Rightarrow b(2^{n+1}-1)=0$$ or $$b(2^{n+1}-1)=b$$

if you choose $$b=1$$ then it will be impossible to find a $$n$$ such that $$\begin{pmatrix} a &b \\0 &1 \end{pmatrix}^n$$=$$\begin{bmatrix} 2 &2 \\0 &1 \end{bmatrix}$$

Also it will not be possible to find a $$n$$ such that $$\begin{pmatrix} a &b \\0 &1 \end{pmatrix}^n$$=$$\begin{bmatrix} 2 &1 \\0 &1 \end{bmatrix}$$ if $$b$$ is chosen as $$2$$

so G is not cyclic