# Why does all entire cusp form can be written as $\Delta h$ where $h$ is an entire modular form of weight $k-12$?

I was going over the proof that all entire modular forms are expressible as polynomials of Eisenstein series $$G_4$$ and $$G_6$$. The proof is supported by inducting on the weight of modular forms. And an argument was used to prove that any entire modular form $$f$$ of weight $$k$$ can be written as $$cG_k+\Delta h$$ where $$h$$ is an entire modular function of weight $$k-12$$.

We pick $$c=\frac{f(i\infty)}{G_k(i\infty)}=\frac{c_f(0)}{c_{G_k}(0)}$$ where the $$c(0)$$ are the coefficients in the Fourier expansions of these functions. Thus, $${f-cG_k}$$ is a cusp form. However, I cannot understand the following argument: we can write $$f-cG_k=\Delta h$$ where $$h$$ is an entire modular function of weight $$k-12$$. I tried to think about the ratio $$\frac{f-cG_k}{\Delta h}$$. If it's analytic and it's of weight $$0$$ then it is a constant function. The increase of weights can be explained by the fact that $$\Delta$$ has a zero of multiplicity $$1$$ at $$i\infty$$. However, we must find $$h$$ having zeros of the same multiplicities as $$f$$ at $$i$$, $$\rho$$ and inside the fundamental region $$R_\Gamma$$. How can we be sure that such $$h$$ exists?

The function $$h$$ is necessarily $$h = \frac{f - cG_k}{\Delta}.$$ It's just a matter of checking that $$h$$ really is a modular form of weight $$k - 12$$. That $$h$$ satisfies the weight $$k - 12$$ transformation law is just a consequence of the fact that $$f - cG_k$$ is a weight $$k$$ form and $$\Delta$$ is a weight $$12$$ form. But you still need to check that $$h$$ is holomorphic on the upper half-plane and at $$i\infty$$. There is no possibility of a pole in the upper half-plane because $$\Delta$$ has no zeros in the upper half-plane. Since $$f - cG_k$$ is a cusp form, it has a zero at $$i\infty$$. On the other hand, $$\Delta$$ has an order $$1$$ zero at $$i\infty$$, therefore the quotient $$h$$ is holomorphic at $$i\infty$$.