A sufficient condition for $\eta$-quotients to be modular forms Let $f$ be the form :
$$f(\tau)=\prod_{M\mid N}{\eta(M\tau)^{a_M}} \quad (\tau \in \mathcal{H})$$
Generally, we said that $f$ is an $\eta$-quotient when $(a_M)$ is a sequence of integers.  One can find conditions on $(a_M)$ such that $f$ is a modular form of weight $k$ invariant under $\Gamma_0(N)$ (with a charactere $\chi$) :
$$\frac{1}{2}\sum_{M\mid N}{a_M}=k$$
$$\forall c \mid N, \quad\frac{1}{24}\sum_{M\mid N}{\frac{\operatorname{gcd}(c,M)^2}{M}a_M} \in \textbf{Q}_{+}$$
(reference : http://www.beck-shop.de/fachbuch/leseprobe/9783642161513_Excerpt_001.pdf). But with those conditions, for which reason must we impose $\left\{a_M\right\}\in \textbf{Z}$ and not $\left\{a_M\right\}\in \textbf{Q}$ ? 
Thanks !
 A: Sidenote: This answer involves guessing, and I am not particularly fond of it. In fact, I had deleted it, but some users apparently have found it better than nothing and have voted successfully for undeletion. Well then.
Why only integer exponents?


*

*Nothing hinders you from choosing $a\in\mathbb{Q}$ and properly defining a power of the normalized modular discriminant just as the Dedekind eta function has been defined using $a=\frac{1}{24}$. The discriminant is finite and nonzero on $\mathbb{H}$, therefore its fractional powers have no branch points in $\mathbb{H}$, and you get a proper one-valued function as a result.

*Of course, the real period of the resulting function then changes to the (reduced) denominator of $a$, but such considerations have already been mastered for the Dedekind eta, and one could again work out all the necessary details: The new multiplier system for example.

*At least the weight formula you have mentioned would carry over analogously.


But:


*

*Other parts may not carry over. For example, Newman 1958, in working out the $\Gamma_0(N)$ modular forms criteria for eta quotients (presumably not being so lenient as to allow nontrivial multiplier systems), had to distinguish between $N$ coprime to $6$ and the more complicated general case. Expect such things to become more complicated when the denominator of $a$ has factors other than $2$ and $3$.

*If the denominator of $a$ does not divide $24$, the $q$-series for $(q^{-1}\Delta)^a(\tau) = \prod_{n=0}^\infty(1-q^n)^{24a}$, where $q=\exp(2\pi\mathrm{i}\tau)$, no longer has integer-only coefficients.
Now, many concrete investigations work with comparisons of truncated power series, or examine arithmetic properties such as multiplicativity, and having to deal with fractions there would be, well, uhm, not exactly a show stopper (I suppose), but at least nasty. Most researchers would not wrestle with a pig without good reason.


I suppose those are the main reasons.
