Prove that $Γ(N)$ is a normal subgroup of $\mathrm{SL}_2 ( \mathbb Z )$ 
Consider the group of integral $2 × 2$-matrices of determinant $1$
$$ \mathrm{SL}_2( \mathbb Z) = \left\{M = \left( \begin{array}{cc} a \ \  b \\c \ \ d \end{array} \right)\Biggm| \ a,b,c,d \in \mathbb Z, \ \det(M) = 1 \right\}$$
and, for  $N\in \mathbb Z_{\geq0}$, its subset
$\Gamma(N)=\left\{M = \left( \begin{array}{cc} a \ \  b \\c \ \ d \end{array} \right) \in SL_2(\mathbb Z )\Biggm| \ a-1 \equiv d-1 \equiv c \equiv b \equiv 0\pmod{N} \right\}$
Prove that $\Gamma(N)$ is a normal subgroup of $\mathrm{SL}_2(\mathbb{Z})$.

Maybe using the first isomorphism theorem we could prove that $\Gamma(N)$ is a normal subgroup?
How should I proceed? Thank you. Regards.
 A: Using the isomorphism theorem, you could consider the group of invertible $2\times 2$ matrices with coefficients in $\mathbb{Z}/N\mathbb{Z}$ (they are the ones whose determinant is invertible modulo $N$) and prove that the map
$$\left(\begin{array}{cc}
x&y\\ z&w\end{array}\right)\longmapsto \left(\begin{array}{cc}
x\bmod N & y\bmod N\\
z\bmod N & w\bmod N
\end{array}\right)$$
is a group homomorphism from $\mathrm{SL}_2(\mathbb{Z})$ to $\mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})$ (or even the special linear group modulo $N$); and that the kernel is precisely the group $\Gamma(N)$.
Alternatively, this can be done directly. Note that because the determinant of an element of $\mathrm{SL}_2(\mathbb{Z})$ is $1$, we have that if the matrix is there, then
$$\left(\begin{array}{cc}
a&b\\ c&d\end{array}\right)^{-1} = \left(\begin{array}{rr}
d&-b\\
-c&a\end{array}\right).$$
So we have
$$\begin{align*}
\left(\begin{array}{cc}a&b\\ c&d\end{array}\right)^{-1}& \left(\begin{array}{cc}x&y\\ z&w\end{array}\right) \left(\begin{array}{cc} a&b\\ c&d\end{array}\right) = 
\left(\begin{array}{rr}d&-b\\ -c&a\end{array}\right) \left(\begin{array}{cc}x&y\\ z&w\end{array}\right) \left(\begin{array}{cc} a&b\\ c&d\end{array}\right)\\
&= \left(\begin{array}{rr}
d&-b\\ -c&d\end{array}\right) \left(\begin{array}{cc}
ax+cy & bx+dy\\
az+cw & bz+dw
\end{array}\right)\\
&= \left(\begin{array}{cc}
(ax+cy)d - (az+cw)b & (xb+dy)d - (zb+dw)b\\
(az+cw)a - (ax+cy)c & (zb+dw)a - (xb+dy)c
\end{array}\right).
\end{align*}$$
If $x\equiv w\equiv 1\pmod{N}$ and $y\equiv z\equiv 0\pmod{N}$, then
$$\begin{align*}
(ax+cy)d-(az+cw)b &\equiv ad-cb \equiv 1\pmod{N}\\
(xb+dy)d-(zb+dw)b &\equiv bd-db \equiv 0\pmod{N}\\
(az+cw)a - (ax+cy)c &\equiv ca - ac \equiv 0\pmod{N}\\
(zb+dw)a - (xb+dy)c &\equiv da - bc \equiv 1\pmod{N}
\end{align*}$$
so the result lies in $\Gamma(N)$.
A: Say $R$ is a ring with $1$ and $I$ is a two-sided ideal of $R$. Let $S= 1 + I$. Then

*

*$S$ is closed under multiplication (easy)


*If $s\in S$  is invertible then $s^{-1} \in S$
(indeed: $s^{-1}(1+ i) = 1$ implies $s^{-1} = 1 - i s^{-1} \in S$)


*$u\in R$ invertible then $u S u^{-1} = S$. Indeed, we have $u(1+i) u^{-1} = 1 + u i u^{-1} \in S$.

If $p\colon R\to R/I$ is the natural projection, then $S= p^{-1}(1)$
Now consider $R = M_2(\mathbb{Z})$ and $I = (m)$
