Let $SL_2(\Bbb Z)$ denote the group (under usual matrix multiplication) of $2×2$ matrices with integer entries and
determinant $1$. Let $H$ be the subgroup of $SL_2(\Bbb Z)$ consisting of those matrices such that: a. The diagonal entries are all equivalent to 1 mod 3. b. the off-diagonal entries are all divisible by 3.
What is the index of $H$ in $SL_2(\Bbb Z)$?

We will solve this problem for arbitrary $n \in \Bbb N$. Hence we define the subgroup $H_n$ to the subgroup of $SL_2(\Bbb Z)$ consisting of those matrices such that: a. The diagonal entries are all equivalent to 1 mod n. b. the off-diagonal entries are all divisible by n. What is the index of $H_n$ in $SL_2(\Bbb Z)$?

We have a formula for the required index which is given by $$[SL_2(\Bbb Z) : H_n] = n^3 \prod_{p | n}(1-\frac{1}{p^2})$$ where $p$ is prime. For $n=3$, this formula gives the answer $24$.

We will solve this problem for arbitrary $n \in \Bbb N$. Hence we define the subgroup $H_n$ to the subgroup of $SL_2(\Bbb Z)$ consisting of those matrices such that: a. The diagonal entries are all equivalent to 1 mod n. b. the off-diagonal entries are all divisible by n. What is the index of $H_n$ in $SL_2(\Bbb Z)$?

**Solution**:We have a formula for the required index which is given by $$[SL_2(\Bbb Z) : H_n] = n^3 \prod_{p | n}(1-\frac{1}{p^2})$$ where $p$ is prime. For $n=3$, this formula gives the answer $24$.

**Proof of this formula:**Kindly check my next post on principal congruence subgroups of $SL_2(\Bbb Z)$ for a detailed proof.**Share to your groups:**

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