Index of a subgroup in $SL_2(\mathbb{Z})$ Let $SL_2(\mathbb{Z})$ denote the group (under usual matrix multiplication) of $2\times2$ matrices with integer entries and determinant $1$. Let $H$ be the subgroup of $SL_2(\mathbb{Z})$ consisting of those matrices such that the diagonal entries are all equivalent to $1 \pmod 3$ and the off-diagonal entries are all divisible by $3$. 
What is the index of $H$ in $SL_2(\mathbb{Z})$? There are a total of $3^4=81$ different equivalence classes of matrices in $SL_2(\mathbb{Z})$ modulo $3$ (each of the entries can have $0,1,2$ as remainders). Now, the given condition implies only one of the possible $81$ combinations modulo $3$. How do we proceed? any hints? Thanks beforehand. 
 A: Hint: $H$ is the kernel of the map $\operatorname{SL}_2(\mathbb Z) \to \operatorname{SL}_2(\mathbb Z / 3\mathbb Z )$. What is its image?

 That map is surjective! Hence the kernel has index $|\operatorname{SL}_2(\mathbb Z / 3\mathbb Z )| = 24$. See also Order of general- and special linear groups over finite fields.

The group $H$ is the principal congruence subgroup $\Gamma(3)$. Its index is also stated on Wikipedia: https://en.wikipedia.org/wiki/Congruence_subgroup#Principal_congruence_subgroups
Here's another way to compute the cardinality of $\operatorname{SL}_2(\mathbb Z / 3\mathbb Z )$: it acts on the projective line $\mathbb P^1(\mathbb F_3)$ by homographies. This action is $3$-transitive (this is true for any field). It has kernel $\{\pm I_2\}$. The group $\operatorname{SL}_2(\mathbb Z / 3\mathbb Z )$ is generated by $\begin{pmatrix}1 & 1 \\
0 & 1\end{pmatrix}$ and $\begin{pmatrix}0 & -1 \\
1 & 0\end{pmatrix}$ (this is true for any finite field). Both matrices act by even permutations: the first is a $3$-cycle (that stabilizes $\infty$), the second acts by $(0, \infty)(1, 2)$. So necessarily, the image of the permutation representation is $A_4$. Hence $|\operatorname{SL}_2(\mathbb Z / 3\mathbb Z )| = |\{\pm I_2\}| \cdot |A_4| = 24$.
