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$H$ is a subgroup of $\operatorname{GL}_2(\mathbb{Z}_3)$ and is equal to $\{\pm I_2\} \cup \{A \in \operatorname{GL}_2(\mathbb{Z}_3): \det(A)=1, \operatorname{tr}(A)=0\}$

I found the elements of $H$:

$\begin{bmatrix} 1 & 0 \\0 & 1 \end{bmatrix}$

$\begin{bmatrix} -1 & 0 \\0 & -1 \end{bmatrix}$

$\begin{bmatrix} \bar{0} & \bar{2} \\\bar{1} & \bar{0} \end{bmatrix}$

$\begin{bmatrix} \bar{0} & \bar{1} \\\bar{2} & \bar{0} \end{bmatrix}$

$\begin{bmatrix} \bar{1} & \bar{2} \\\bar{1} & \bar{2} \end{bmatrix}$

$\begin{bmatrix} \bar{1} & \bar{1} \\\bar{2} & \bar{2} \end{bmatrix}$

$\begin{bmatrix} \bar{2} & \bar{2} \\\bar{1} & \bar{1} \end{bmatrix}$

$\begin{bmatrix} \bar{2} & \bar{1} \\\bar{2} & \bar{1} \end{bmatrix}$

But I'm a bit confused, because the question is asking for the order of the elements of $H$, and when I multiply the last 4 matrices by themselves, I get the zero matrix...so there can't be an order, right?

Thanks in advance

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The last 4 matrices are singular and thus not elements of $\,GL_2(\Bbb Z_3)\,$ ... –  DonAntonio Feb 22 '13 at 0:07
    
@DonAntonio Thanks...so there are only 4 elements, right? Since the trace is zero (so the only elements in the diagonal are zero, because otherwise the determinant becomes zero and that can't happen in $Gl_2(Z_3)$, right? –  user58289 Feb 22 '13 at 0:42
    
I wrote an answer below. –  DonAntonio Feb 22 '13 at 2:52

1 Answer 1

up vote 2 down vote accepted

Besides the fact that the last four matrices don't belong to $\,GL_2(\Bbb Z_3)\,$ and thus obviously neither to $\,H\,$ , I think you can find all the relevant matrices by solving some relatively easy equations:

$$\pm I_2\neq\begin{pmatrix}a&b\\c&d\end{pmatrix}\in H\Longleftrightarrow\begin{cases}\operatorname{I}\;\;ad-bc=1\\{}\\\operatorname{II}\;\;\;\;a+d=0\end{cases}\;\;\stackrel{\operatorname{II}}\Longrightarrow \;\;d=-a\stackrel{\operatorname{I}}\Longrightarrow$$

$$\;-a^2-bc=1\Longrightarrow a^2=-1-bc=2(1+bc)$$

but since the only squares in $\,\Bbb Z_3\,$ are $\,0\,,\,1\,$ , the above means that we must have $\,1+bc=0\,,\,2\,$ , so:

$$\begin{align*}(i)\;\;\;bc+1=0\Longrightarrow bc=2\stackrel{\operatorname{I}}\Longrightarrow ad=0\\{}\\(ii)\;\;\;bc+1=2\Longrightarrow bc=1\stackrel{\operatorname{I}}\Longrightarrow ad=2\end{align*}$$

Option (i) , together with II and then I above, gives $\,a=d=0\,\,\,,\,\,bc=2$ , with the only options matrices (3)-(4) in your list.

Option (ii) gives $\,a=1\,,\,d=2\,\,\,\vee\,\,\,a=2\,,\,d=1\,$ , each with the options $\,b=c=1\,,\,b=c= 2\,$ , giving the matrices

$$\begin{pmatrix}1&1\\1&2\end{pmatrix}\;,\;\;\begin{pmatrix}1&2\\2&2\end{pmatrix}\;,\;\;\begin{pmatrix}2&1\\1&1\end{pmatrix}\;,\;\;\begin{pmatrix}2&2\\2&1\end{pmatrix}$$

which you seem to have missed....

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+1 for the answer I removed. Thanks for yours there. –  B. S. Feb 22 '13 at 20:11

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