I'm confusing about this specific problem:

enter image description here

Here, $\Gamma$ is a group of 2x2 - matrices with integer entries, with respect to the usual matrix multiplication, and $det(\Gamma) = 1$.

But will the identity $I = \begin{bmatrix}1&0\\0&1\end{bmatrix}$ fail to prove when $n = 1$ and therefore, $I \not\in \Gamma_n$ ? (Because $1 \equiv 0 \pmod{1}$)

Please, correct me if I'm mistaken.

  • $\begingroup$ The set of all $2 \times 2$-matrices is not a group with respect to matrix multiplication. You probably want $\Gamma$ to be the group of invertible $2 \times 2$-matrices over the integers. $\endgroup$ – Matthias Klupsch Jan 25 at 6:23
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    $\begingroup$ @hardmath : It seems to me the set $\Gamma_1$ is not the trivial subgroup but the whole $\Gamma$, since the conditions are trivially satisfied. $\endgroup$ – Matthias Klupsch Jan 25 at 6:25
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    $\begingroup$ Regarding your problem of proving $I \in \Gamma_1$, ask yourself: is $1 \equiv 1 $ mod $1$ and is $0 \equiv 0 $ mod $1$? This is all you need in order to have $I \in \Gamma_1$. $\endgroup$ – Matthias Klupsch Jan 25 at 6:28
  • $\begingroup$ @MatthiasKlupsch sorry for missing an important info, but it is invertible since $\det(\Gamma) = 1$ $\endgroup$ – Thai Doan Jan 25 at 6:29
  • $\begingroup$ @Matthias: Yes, you are right. It's not clear to me what group $\Gamma$ is, perhaps an additive group or a multiplicative group of matrices. I guess it doesn't matter as far as case $n=1$ goes. $\endgroup$ – hardmath Jan 25 at 6:34

The excerpt explains for you that $\Gamma_2 $ will be the matrices with main diagonal entries odd and minor diagonal entries even. ( This is another way of saying $a\equiv d\equiv1\pmod2$ and $b\equiv c\equiv 0\pmod2$).

Thus the identity, $I=\begin{pmatrix} 1&0\\0&1\end{pmatrix}$ is an element of $\Gamma_2 $.

  • $\begingroup$ Thanks for your help, but it doesn't follow that $I \in \Gamma_n$ for any positive integers $n$ $\endgroup$ – Thai Doan Jan 25 at 7:11
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    $\begingroup$ It actually does. For any $n$ , we have $1\equiv 1\pmod n$ and $0\equiv0\pmod n$. $\endgroup$ – Chris Custer Jan 25 at 7:17
  • $\begingroup$ Nice! I have upvoted your post and thank you for pointing out $1 \equiv 1 \pmod{n}$. $\endgroup$ – Thai Doan Jan 25 at 7:29

My problem is I was mistaken about the "congruence": $1 \equiv 1 \pmod{1}$. They are both in the same equivalence class for 0.


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