Show that the Quotient Group $Γ_2(p)/Γ_2(p^k)$ is finite. Let $p$ be a prime number. Denote by $Γ_2(p)$ the multiplicative group of all $2×2$ matrices
$$               x =
\begin{pmatrix} 
a & b \\
c & d
\end{pmatrix}
$$
with elements $a, b, c, d ∈ \mathbb{Z}$ such that $ad − bc = 1$ and $x ≡ e_2 (mod \ p)$, where $e_2$ is the
identity $2 ×2$ matrix. 
In other words, the integers $a, d$ are equal to $1$ modulo $p$ while $b$ and
$c$ are multiples of $p$.
For every $k \in \mathbb{N}$, put
$$
Γ_2(p^k) = \{x \in Γ_2(p) : x ≡ e_2 (mod \ p^k)\}
$$
Then $Γ_2(p), Γ_2(p^2), . . .$ is a decreasing sequence of invariant(normal) subgroups of $Γ_2(p)$, and the
intersection of these subgroups contains only the identity $e_2$ of $Γ_2(p)$. So there exists $k \in \mathbb{N}$
such that $g \notin Γ_2(p^k)$.
Show that the quotient group $K = Γ_2(p)/Γ_2(p^k)$ is finite 
$\textbf{HINT:}$   $|K| \leq (p^k)^4 = p^{4k}$.
I have tried almost every possible way but could not identify the elements of this group.
Please help me in proving this.
 A: I will omit the subscript $2$ in $\Gamma_2$, so we have instead $\Gamma$.
Let us consider the short exact sequence
$$
1\to
\Gamma(p^k)
\to
\Gamma(p)
\to
\Gamma(p)/\Gamma(p^k)
\to 
1\ .
$$
$1$ is the trivial group.
The map $\Gamma(p^k)
\to
\Gamma(p)$ is the inclusion of a normal group in a bigger group, one can build the quotient, which is also a group, denoted as usually $\Gamma(p)/\Gamma(p^k)$, and the structural projection from $\Gamma(p)$ to it is
$\Gamma(p)
\to
\Gamma(p)/\Gamma(p^k)
$, a surjective map. It is enough to show that this group is finite.
For this, consider the map of rings from $\Bbb Z\to\Bbb Z/p^k$.
It induces a map at the level of general linear groups $\operatorname{GL}_2$, or even better
$\operatorname{SL}_2$, or even better
$$
\begin{aligned}
G=\operatorname{GL}_2(\Bbb Z)
&\to
\operatorname{GL}_2(\Bbb Z/p^k)
\ ,
\\
\pi:
\Gamma(1)=\operatorname{SL}_2(\Bbb Z)
&\to
\operatorname{SL}_2(\Bbb Z/p^k)
\ .
\end{aligned}
$$
It is easy to show that the subgroup $\Gamma(p)$ is mapped by $\pi$
(surjectively) onto the group of matrices of the shape $1+pX$, $X$ being an arbitrary matrix with coefficients taken modulo $p^k$.
The chasing analysis of
$\require{AMScd}$
\begin{CD}
1 @>>> \Gamma(p^k) @>>> \Gamma(p) @>>> \Gamma(p)/\Gamma(p^k) @>>> 1\\
@. @| @| @VV\pi V @.
\\
1 @>>> 1+p^kM_{2\times2}(\Bbb Z) @>>> 1+pM_{2\times2}(\Bbb Z)  @>>>  
1+pM_{2\times2}(\Bbb Z/p^k\Bbb Z) @>>> 1\\
\end{CD}
shows that the quotient $\Gamma(p)/\Gamma(p^k)$
is isomorphic to the subgroup $1+pM_{2\times2}(\Bbb Z/p^k\Bbb Z)$
inside $\operatorname{SL}_2(\Bbb Z/p^k\Bbb Z)$, which is a finite group of order $(p^{k-1})^4$.
A: Hint: What the hint wants to say (I think) is that the image of $\gamma 
= \begin{pmatrix}a & b \\ c & d \end{pmatrix}$ in the quotient $\Gamma_2(p)/\Gamma_2(p^k)$ is determined by the congruence classes of $a,b,c,d$ mod $p^k$.

 Indeed, take $A \in \Gamma_2(p)$ and $B \in M_2(\mathbb Z)$ such that $A+p^kB \in \Gamma_2(p)$. Then $$A + p^k B = A (e_2 + p^k A^{-1}B)$$ and the factor on the right is in $\Gamma_2(p^k)$.
 Thus the coset $[\gamma]$ depends only on $\gamma \pmod{p^k}$, which means that the projection $\pi : \Gamma_2(p) \to \Gamma_2(p)/\Gamma_2(p^k)$ is well-defined on congruence classes mod $p^k$. There are only finitely many ($p^{4k}$) congruence classes of $M_2(\mathbb Z)$ mod $p^k$, and the finite set $$\left\{(a,b,c,d) \pmod{p^k} : \exists \gamma \equiv \begin{pmatrix}a & b \\ c & d \end{pmatrix} \pmod{p^k} \;\&\; \gamma \in \Gamma_2(p) \right\}$$
 surjects via $\pi$ onto $\Gamma_2(p)/\Gamma_2(p^k)$.

In fact, we see that $\operatorname{SL}_2(\mathbb Z) / \Gamma_2(p^k)$ is finite.
