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The following is from the book Modular Forms by W Stein:

enter image description here

By the very same book "a congruence subgroup is a subgroup of $SL_2(\mathbb{Z})$ that contains $\Gamma(N)$ for some $N$". So $\Gamma(N)$ must be of the form: enter image description here So how $\Gamma(N)$ is contained in $\Gamma_1(N)$ when $a \equiv d \equiv 1$? (it must be $a \equiv d \equiv 0$)

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    $\begingroup$ What you wrote as the zero matrix should be the identity matrix. $\endgroup$ – RghtHndSd Nov 5 '18 at 2:20
  • $\begingroup$ @RghtHndSd, isn't Γ(N) the kernel of $SL_2(\mathbb{Z})$ to $SL_2(\mathbb{Z/NZ})$ map so it id in mod N? $\endgroup$ – user231343 Nov 5 '18 at 2:24
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The identity element of $SL_2(\Bbb Z/n\Bbb Z)$ is the identity matrix $\begin{pmatrix}1&0\\0&1\end{pmatrix}$. Note that the zero matrix is not even an element of $SL_2(\Bbb Z/n\Bbb Z)$, because it has determinant zero.

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