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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?

EDIT: Thanks for your comments.

  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.

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    $\begingroup$ 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. $\endgroup$ Jan 13 '21 at 13:31
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    $\begingroup$ You could try to figure out the group multiplication law in terms of $a,b$. $\endgroup$
    – lisyarus
    Jan 13 '21 at 13:32
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    $\begingroup$ Calculate several powers of these. Also note that $2=-1$ in $\Bbb Z_3$. $\endgroup$
    – Berci
    Jan 13 '21 at 13:32
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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.

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

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

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