# Abstract algebra: Proving the matrix is a subgroup

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.

• 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. – Matthias Klupsch Jan 25 at 6:23
• @hardmath : It seems to me the set $\Gamma_1$ is not the trivial subgroup but the whole $\Gamma$, since the conditions are trivially satisfied. – Matthias Klupsch Jan 25 at 6:25
• 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 $I \in \Gamma_1$. – Matthias Klupsch Jan 25 at 6:28
• @MatthiasKlupsch sorry for missing an important info, but it is invertible since $\det(\Gamma) = 1$ – Thai Doan Jan 25 at 6:29
• @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. – 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 $$\Gamma_2$$.

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