3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.