Let \begin{align} \Gamma=\text{SL}_2(\mathbb{Z})=\bigg\{\begin{pmatrix} a & b \\ c & d \end{pmatrix}: a,b,c,d\in \mathbb{Z}, \;ad-bc=1\bigg\}, \end{align} the group of integer matrices with determinant $1$. Let $n\in \mathbb{N}.$ Then through matrix multiplication, $\Gamma$ acts on the left and right of the set \begin{align} M_n=\bigg\{\begin{pmatrix} a & b\\ c&d \end{pmatrix}: a,b,c,d\in \mathbb{Z}, \; ad-bc=n\bigg\}. \end{align} It can also be computed that the number of left orbits is finite (it is equal to $r:=$ the sum of divisors of $n$.) Further, the map \begin{align*} \begin{pmatrix} a &b\\ c &d \end{pmatrix} \mapsto \begin{pmatrix} d &-b\\ -c &a \end{pmatrix} \end{align*} (inverse map, then scaling by $n$) fixes $M_n$ and forms a bijection between the left cosets and right cosets. So there are the same number $r$ of left cosets and right cosets. I'd like to prove the following fact:
There exists a complete set of left coset representatives $\{\alpha_i\}_{i=1}^r$ that is also a complete set of right coset representatives.
Any ideas? I know that for any left and right coset representatives (respectively) $\{\alpha_i\}_{i=1}^R$ and $\{\beta_i\}_{i=1}^r$, we have $M_n=\bigsqcup \alpha_i\Gamma =\bigsqcup \Gamma \beta_i $.
$$M_n = \bigcup_{ad=n, b \bmod d}SL_2(Z) \pmatrix{a & b \\ 0 & d}, \qquad \qquad\scriptstyle\pmatrix{0 & 1\\ -1 & 0}\pmatrix{a & b \\ 0 & d}=\pmatrix{d & 0 \\ -b & a}\pmatrix{0 & 1\\ -1 & 0}$$
Thus
$$M_n =\bigcup_{ad=n, b \bmod d} SL_2(Z)\pmatrix{0 & 1\\ -1 & 0}\pmatrix{a & b \\ 0 & d}=M_n^\top = \bigcup_{ad=n, b \bmod d} \pmatrix{d &0 \\ -b & a} SL_2(Z)\\= \bigcup_{ad=n, b \bmod d} \pmatrix{d &0 \\ -b & a} \pmatrix{0 & 1\\ -1 & 0}SL_2(Z) = \bigcup_{ad=n,b \bmod d}\pmatrix{0 & 1\\ -1 & 0}\pmatrix{a & b \\ 0 & d}SL_2(Z) $$