Show that $$M_{2k}(\Gamma(1))=\left\langle G_{4}^{\alpha}G_{6}^{\beta}: 3\alpha+2\beta=k\right\rangle$$ where $\Gamma(1)= SL_{2}(\mathbb{Z})$ and $M_{2k}$ is the set of modular forms of weight $2k$, $k\in \mathbb{Z}$.
I try to prove it since we have $$\Delta(z)=\frac{1}{1728}(\Delta_{4}(z)^3-\Delta_{6}(z)^2),$$ hence $\Gamma_{1}$ is a polymomial in $E_{4}$ and $E_{6}$ and $f(z)$ be a modular form of weight $\forall 2k$ ($k\geq 2$), with Fourier expansioin $$f(z)=\sum_{n=0}^{\infty}a_{n}e^{2\pi i n z}.$$
But how about the $\Gamma(1)$?
I think it is useful: we may consider the may $$M_{k-12}(\Gamma (1))\rightarrow M_{k}(\Gamma (1)),$$ where we have $M_{12}(\Gamma (1))=\left\langle G_{12}, \Delta\right\rangle$.
We can determine the modular forms of weight $<12$ using above formula. $$\sum v_P + \frac{1}{2} v_i + \frac{1}{3} v_\rho + v_\infty = \frac{2k}{12}$$ They are: constants, $G_4$, $G_6$, $G_4^2$, $G_4G_6$, with weight $0,4,6,8,10$ respectively.
Then use induction: every weight $2k$ modular form $f$ can be made into a cusp form by $f_1 = f - c G_{2k}$ for some constant $c$, so $f_1 / \Delta$ is a modular form of weight $2k-12$.