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Is a modular form $f$ of weight $k$ with respect to $\text{SL}_2(\mathbb{Z})$ always a modular form to a congruence subgroup $\Gamma$ (for example $\Gamma_1(N)$)?

If the transformation law $f|_k\gamma=f$ holds for any $\gamma\in \text{SL}_2(\mathbb{Z})$ then it holds also for any $\gamma\in\Gamma\subset \text{SL}_2(\mathbb{Z}).$ But is $f$ holomorphic at all cusps of $\Gamma$?

Thanks in advance.

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Yes, a modular form on $\text{SL}_2(\mathbb{Z})$ is a modular form on any congruence subgroup $\Gamma < \text{SL}_2(\mathbb{Z})$. It's quite clear that the transformation law holds, as you've mentioned.

Suppose that $\alpha \neq \infty$ is a cusp of $\Gamma$. Then since $\alpha$ is equivalent to $\infty$ in $\text{SL}_2(\mathbb{Z})$, i.e. there is some $\gamma \in \text{SL}_2(\mathbb{Z})$ such that $\gamma^{-1} \alpha = \infty$, the behaviour of $f$ at $\alpha$ is the same as the behavior of $f(\gamma z)$ at $\infty$. Since $f(\gamma z)$ is well-behaved at $\infty$, you get that $f(z)$ is well-behaved at $\alpha$, and so $f$ is holomorphic at $\alpha$.

Generally, each modular form on a congruence subgroup $\Gamma' \supset \Gamma$ is also a modular form on $\Gamma$. The forms coming from these larger congruence subgroups are often called "oldforms", while those forms first appearing on $\Gamma$ itself are called "newforms." When studying modular forms of level $N$, it is usually good to look at the newforms of level $N$ as the oldforms have slightly different (and often slightly better) behaviours.

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    $\begingroup$ This is not quite the right definition of "oldforms" and "newforms". E.g. if $\Delta$ is the weight 12 cusp form on $SL_2(\mathbf{Z})$, $\Delta(3z)$ is a cusp form on $\Gamma_0(3)$ which isn't the restriction of any form on any subgroup $\Gamma' \subset SL_2(\mathbf{Z})$ strictly containing $\Gamma_0(3)$, but it still wouldn't normally be considered a newform. $\endgroup$ – David Loeffler May 15 '16 at 9:12
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    $\begingroup$ @DavidLoeffler I didn't intend to give the right definition of an oldform. That's why I kept with the intentionally imprecise heuristic of "coming from larger congruence subgroups." I suppose I was unclear about that. Thank you for helping clear that up. $\endgroup$ – davidlowryduda May 15 '16 at 20:58

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