# Principal congruence subgroup $\Gamma(DN)$ and $\Gamma(N)$

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.

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