For $q \in \mathbb{N}$ let $\Gamma(q)=\operatorname{ker}\left(\mathrm{SL}_2(\mathbb{Z}) \rightarrow \mathrm{SL}_2(\mathbb{Z} / q \mathbb{Z})\right)$
We have to show that, for a prime $p$ and an exponent $t \in \mathbb{N}$ the quotient $\Gamma(p) / \Gamma\left(p^t\right)$ is a $p$-group.
This is my solution, please let me know if this is correct and if not, please give me some suggestions.
Solution:
To show that the quotient $\Gamma(p)/\Gamma(p^t)$ is a $p$-group, we need to show that the order of the quotient group is a power of $p$.
First, let's recall that $\mathrm{SL}_2(\mathbb{Z})$ is the group of $2 \times 2$ matrices with integer entries and determinant 1, and $\mathrm{SL}_2(\mathbb{Z}/q\mathbb{Z})$ is the group of $2 \times 2$ matrices with entries in $\mathbb{Z}/q\mathbb{Z}$ and determinant 1.
Now, consider the homomorphism $\phi: \mathrm{SL}_2(\mathbb{Z}) \rightarrow \mathrm{SL}_2(\mathbb{Z}/p^t\mathbb{Z})$. The kernel of this homomorphism is $\Gamma(p^t)$. By the First Isomorphism Theorem, we have:
$$\mathrm{SL}_2(\mathbb{Z})/\Gamma(p^t) \cong \phi(\mathrm{SL}_2(\mathbb{Z})) \leq \mathrm{SL}_2(\mathbb{Z}/p^t\mathbb{Z})$$
Similarly, we have the homomorphism $\psi: \mathrm{SL}_2(\mathbb{Z}) \rightarrow \mathrm{SL}_2(\mathbb{Z}/p\mathbb{Z})$ with kernel $\Gamma(p)$. Then,
$$\mathrm{SL}_2(\mathbb{Z})/\Gamma(p) \cong \psi(\mathrm{SL}_2(\mathbb{Z})) \leq \mathrm{SL}_2(\mathbb{Z}/p\mathbb{Z})$$
Now, consider the composition of the natural projection $\pi: \mathrm{SL}_2(\mathbb{Z}/p^t\mathbb{Z}) \rightarrow \mathrm{SL}_2(\mathbb{Z}/p\mathbb{Z})$ with the homomorphism $\phi$. We have $\pi \circ \phi = \psi$. By the Second Isomorphism Theorem, we have:
$$\frac{\mathrm{SL}_2(\mathbb{Z})/\Gamma(p^t)}{\mathrm{SL}_2(\mathbb{Z})/\Gamma(p)} \cong \frac{\phi(\mathrm{SL}_2(\mathbb{Z}))}{\psi(\mathrm{SL}_2(\mathbb{Z}))} \leq \frac{\mathrm{SL}_2(\mathbb{Z}/p^t\mathbb{Z})}{\mathrm{SL}_2(\mathbb{Z}/p\mathbb{Z})}$$
Now, we need to find the order of the quotient group $\frac{\mathrm{SL}_2(\mathbb{Z}/p^t\mathbb{Z})}{\mathrm{SL}_2(\mathbb{Z}/p\mathbb{Z})}$. The order of $\mathrm{SL}_2(\mathbb{Z}/p^t\mathbb{Z})$ is given by:
$$|\mathrm{SL}_2(\mathbb{Z}/p^t\mathbb{Z})| = (p^2 - 1)(p^2 - p)$$
And the order of $\mathrm{SL}_2(\mathbb{Z}/p\mathbb{Z})$ is given by:
$$|\mathrm{SL}_2(\mathbb{Z}/p\mathbb{Z})| = (p- 1)^2$$
So, the order of the quotient group $\frac{\mathrm{SL}_2(\mathbb{Z}/p^t\mathbb{Z})}{\mathrm{SL}_2(\mathbb{Z}/p\mathbb{Z})}$ is:
$$\frac{(p^2 - 1)(p^2 - p)}{(p - 1)^2} = p(p + 1)$$
Since $p$ is a prime, $p(p + 1)$ is a power of $p$. Therefore, the quotient group $\frac{\mathrm{SL}_2(\mathbb{Z}/p^t\mathbb{Z})}{\mathrm{SL}_2(\mathbb{Z}/p\mathbb{Z})}$ is a $p$-group.
Now, from the Second Isomorphism Theorem, we have:
$$\frac{\mathrm{SL}_2(\mathbb{Z})/\Gamma(p^t)}{\mathrm{SL}_2(\mathbb{Z})/\Gamma(p)} \cong \frac{\phi(\mathrm{SL}_2(\mathbb{Z}))}{\psi(\mathrm{SL}_2(\mathbb{Z}))} \leq \frac{\mathrm{SL}_2(\mathbb{Z}/p^t\mathbb{Z})}{\mathrm{SL}_2(\mathbb{Z}/p\mathbb{Z})}$$
Since the right-hand side is a $p$-group, it follows that the left-hand side is also a $p$-group. Thus, the quotient group $\Gamma(p)/\Gamma(p^t)$ is a $p$-group.