# 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.

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 $(p^{k-1})^4$.

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.

• Sorry but I have not got the answer.Will you please explain? – Sumit Mittal May 30 '18 at 17:55