The following is a question from my assignment in number theory:
$$ \sum_{n=1}^\infty \tau(n)e^{2\pi in\tau} = (2\pi)^{-12}\Delta(\tau) = e^{2\pi i\tau} \prod_{m=1}^\infty (1-e^{2\pi im\tau})^{24}.$$
3. Let $p$ be a prime and let $k$ be an integer, $1\le k\le p-1.$ Show that there exists integer $h$ such that $$ \tau^{12}\Delta\Big(\frac{\tau+h}p\Big) = \Delta\Big(\frac{k\tau-1}{p\tau}\Big) $$ and that $h$ runs through a reduced residue system mod $p$ with $k$.
I am unable to get an idea which result should I use for proving the existence of h. I used 1 st formula in image but unable to prove it.
So, please guide.
Original image: https://i.stack.imgur.com/4nxwY.jpg
There is some $\gamma\in SL_2(Z)$ such that $\gamma(k/p)= i\infty$
$$\gamma(\tau) = \frac{a\tau+b}{p\tau-k},\qquad -ak-bp=1$$
Then $$\Delta(\frac{k\tau-1}{p\tau})=(p\frac{k\tau-1}{p\tau}-k)^{-12}\Delta(\gamma(\frac{k\tau-1}{p\tau}))=\tau^{-12} \Delta(\frac{\tau+r}{p})=\tau^{-12} \Delta(\frac{\tau+r}{p}-\lfloor r/p\rfloor)$$