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Define the groups

$\Gamma_0^0(N,M) = \left.\left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z}) \;\right|\; b = 0 \bmod N, \; c = 0 \bmod M. \right\}$ and $\Gamma_0(NM) = \left.\left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z}) \;\right|\; c = 0 \bmod MN \right\}.$

Are these subgroups conjugate in $SL_2(\mathbb{Z})$ and by which element?

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    As Geoff points out, they are not conjugate in $\mathrm{SL}_2(\mathbb{Z})$ unless $M$ and $N$ are coprime, in which case they are conjugate (because you can find an element of $\mathrm{SL}_2(\mathbb{Z})$ which is the identity mod $N$ and \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} mod $M$). However, they *are* conjugate as subgroups of $\operatorname{SL}_2(\mathbb{R})$, which is good enough for many purposes (e.g. there is a natural bijection between spaces of modular forms for these groups).2012-02-27

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Perhaps you intended that $M$ and $N$ should be relatively prime. Without that assumption, the groups need not be conjugate within ${\rm SL}_{2}(\mathbb{Z}).$ For if $M = N =p$ is a prime, then the images of the two groups after reduction (mod $p$) are not conjugate within ${\rm SL}_{2}(p)$, since one consists exclusively of semisimple matrices, and the other contains non-identity unipotent elements.