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I did some calculations on coefficients of $q$-expansion of element of $SL_2(\Gamma_0(19))$ with $a_1=1$, and I think the coefficients satisfies $a_{mn}=a_ma_n$ if $\gcd(m,n)=1$. Does there exist any literature that contains the proof of this (or more general) fact?

Laurence PW
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It is true that the cuspform of weight $2$ on $\Gamma_0(19)$, normalized with $a(1) = 1$, has multiplicative coefficients.

I write the because the vector space of weight $2$ modular forms on $\Gamma_0(19)$ has dimension $2$, and the cuspidal subspace has dimension $1$. Thus there is exactly one, and its $q$-expansion begins $$ q - 2q^3 - 2q^4 + 3q^5 - q^7 + q^9 + 3q^{11} + 4q^{12} - 4q^{13} + O(q^{15}). $$

The general theory is that for any congruence subgroup of $\mathrm{SL}(2, \mathbb{Z})$ (such as $\Gamma_0(19)$), there are particular linear operators called Hecke operators. These operators take cusp forms to cusp forms, hence we can consider their action on the subspace of cusp forms. These operators also commute, and each Hecke operators commutes with its adjoint operator (with respect to the Petersson inner product).

The spectral theorem from linear algebra then guarantees that there is a basis for the space of cusp forms consisting of simultaneous eigenvectors. Computing the action of these Hecke operators shows that a simultaneous eigenform has multiplicative coefficients.

The result is that there is a basis of cuspidal eigenforms for any fixed weight and congruence subgroup. As the space of cuspforms of weight $2$ on $\Gamma_0(19)$ is one-dimensional, it must be the basis and is thus an eigenform --- and hence has multiplicative coefficients.

This result is fundamental in the theory of modular forms, and any introductory book on modular forms should include it. Diamond and Shurman's introduction is good (this result is one of the major pieces of chapter 5 of that book). Bumps Automorphic Forms and Representations includes a concise proof for level $1$ in one section of the first chapter, though a beginner to to the subject might find it too terse. (At least I thought so when I was first learning, though I like it a lot now).