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Let $\Gamma(N)$ denotes the principal congruence subgroup of level $N$ and $\beta$ be a $2 \times 2$ matrix with integral entries and deteminant $D$. Prove that $\beta \Gamma(DN) \beta^{-1}$ is contained in $\Gamma(N)$.

Let $\beta=\begin{pmatrix} a &b \\ c &d \end{pmatrix}$ with $ad-bc=D$ and $\beta^{-1}=\dfrac{1}{D}\begin{pmatrix} d &-b \\ -c &a \end{pmatrix}$.

Let $\gamma \in \Gamma(DN)$, $\gamma=\begin{pmatrix} e &f \\ g &h \end{pmatrix} \equiv \begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix} \mod DN$ and $eh-fg=1$.

We have $\beta \gamma \beta^{-1}=\dfrac{1}{D}\begin{pmatrix} acd+bgd-afc-bhc &-acb-b^2g+a^2f+bha \\ ced+d^2g-c^2f-cdh &-ceb-dgb+cfa+dha \end{pmatrix}$

I'm stuck here, for example: As $g,f \equiv 0 \mod DN$ we only need to prove $-acb+bha \equiv 0 \mod DN$ but I cannot figure out.

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If $\beta\in M_2(Z), \det(\beta)=D$ then $\beta = ABA'$ where $A,A'\in SL_2(Z)$ and $B\in M_2(Z)$ is diagonal $\det(B)=D$.

Being the kernel of $SL_2(Z)\to SL_2(Z/nZ)$ then $\Gamma(n)$ is normal in $SL_2(Z)$ thus it stays the same when conjugated by $A,A'$

Thus it suffices to check what happens when conjugating by $\pmatrix{u&0\\ 0 & v},uv=D$.

We find that for $D| n$ and $\pmatrix{na+1&nb\\ nc&nd+1}\in \Gamma(n)$ and $D=uv$ $$\pmatrix{u&0\\ 0 & v}\pmatrix{na+1&nb\\ nc&nd+1}\pmatrix{u&0\\ 0 & v}^{-1}=\pmatrix{na+1&\frac{u}{v}nb\\ \frac{v}{u}nc&nd+1}\in \Gamma(n/D)$$

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