Let $(a, b; c, d) \in SL_2(\mathbb{Z})$ and $\Delta(z)=q\prod_{n=1}^{\infty} (1-q^n)^{24}$, $q=e^{2\pi i z}$. How to show that $\Delta(\frac{az+b}{cz+d})=(cz+d)^{12}\Delta(z)$? Thank you very much.
I think that $ad - bc =1$ and hence $a=(1+bc)/d$. We can substitute $a$ by $(1+bc)/d$ in $\Delta(\frac{az+b}{cz+d})$. But I am not able to simplify the result.
On
Somehow it turns out to not be reasonable to try to prove this by direct substitution and manipulation.
Proofs were known in the 19th century (see also things about "Dedekind's $\eta$-function", which is the $24$-th root of Ramanujan's $\Delta$).
Succinct, but subtle, proofs were given by Siegel and Weil: these are recalled, with citations to the original sources, in various places, for example my notes http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/08a_product_expansion.pdf
On
First of all, using the fact that $SL(2, \mathbb Z)$ is generated by $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, and the functional equation for the former is obvious, you are reduced to the case of the functional equation for the substitution $\tau \mapsto -1/\tau$.
One approach is to take logarithmic derivatives with respect to $\tau$ (or, up to a $2\pi i$ factor, applying $q \dfrac{d}{dq}$ to $\log \Delta$), so that you are instead studying how the series $$1 - 24 \sum_{n = 1}^{\infty} \bigg(\sum_{d | n} d \bigg) q^n$$ transforms under $\tau \mapsto -1/\tau$.
If you take the logarithmic derivative of the formula $\Delta(-1/\tau) = \tau^{12} \Delta(\tau),$ you will find that it should almost transform like a weight $2$ modular form, but not quite.
The discussion of Eisenstein series in Serre's Course in arithmetic will show that (up to a scalar) this function is an Eisenstein series of weight $2$, namely $$\sum_{(m,n) \neq (0,0)} \frac{1}{(m\tau + n)^2}.$$ If this series converged absolutely, it would actually give a modular form of weight $2$. But it doesn't converge absolutely, so (once you interpret what this series even means), when you make the substitution $\tau \mapsto -1/\tau$ and then rearrange, something slightly non-trivial happens. But you can figure out exactly what does happen, and so derive the correct behaviour under $\tau \mapsto -1/\tau$, and then argue backwards to get the desired modularity property for $\Delta$.
(I think this is all discussed in Course in arithmetic.)
There is a proof (somewhat nontrivial) in Don Zagier's notes.