Abstract algebra: Proving the matrix is a subgroup I'm confusing about this specific problem:



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 $I = \begin{bmatrix}1&amp;0\\0&amp;1\end{bmatrix}$ 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 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,  $I=\begin{pmatrix} 1&amp;0\\0&amp;1\end{pmatrix}$ is an element of $\Gamma_2 $.
My problem is I was mistaken about the "congruence": $1 \equiv 1 \pmod{1}$. They are both in the same equivalence class for 0.