I am self studying Tom M Apostol Modular functions and Dirichlet series in number theory and I could not think about a step in the proof of the theorem 4.6 in Chapter 4 which is
If $f$ is automorphic under $\Gamma $ and if $p$ is a prime then, $$f_p( \frac {-1 } {\tau } )= f_p(\tau ) + \frac1p f(p \tau ) - \frac1p f( \frac { \tau } {p} ).$$
For this a lemma is proved whose proof I completely understand but I am stating it
Lemma– Let $ T_{\lambda }\tau = \dfrac {\lambda+ \tau } {p} $. Then for each $\lambda $ in interval $[1, p-1]$ , where $\lambda $ is always an integer, there exists an integer $\mu $ in same interval and a transformation $V$ in $\Gamma_0 (p)$ such that $T_\lambda S = V T_\mu $ . Moreover, as $\lambda$ runs through integers $1,2,\ldots,p-1$ so does $\mu$ .
What doubt I have is mentioned at the end of the post. But I am giving whole proof here for the sake of completion.
Proof of theorem – $$ \begin{aligned} pf_p(\frac {-1} {\tau } ) &= \sum_{\lambda =0 }^{p-1} f(\frac {S\tau + \lambda } { p } )\\ & = f(\frac {S\tau } {p}) + \sum_{\lambda=1}^{p-1} f(T_\lambda S\tau )\\ &= f(\frac {-1} {\tau p}) + \ldots\\ &+ \text{( terms involving summation wrt $\lambda $ which are converted into summation}\\ &\text{$\sum_{\mu=0}^{p-1}$ by adding and subtracting $f(\frac {\tau } {p}) $}\\ &= f(\tau p) + pf_p(\tau) - f(\frac { \tau} {p} ) . \end{aligned} $$ My doubt $\to$
how $f(\frac {-1} {\tau p} )$ in step 2 is written as $f(\tau p)$ in step 3.
Can anybody please give some hint.
Let $$A_p = \{ M\in M_2(\Bbb{Z}),\ \det(M)\ = p\}$$ quotient $A_p$ on the left by $\Gamma=SL_2(\Bbb{Z})$ to obtain the disjoint union $$A_p = \bigcup_{j=1}^J \Gamma g_j$$
Let $S=\pmatrix{0&1\\ -1&0}$ or any other element of $\Gamma$ then $$A_p= A_p S= \bigcup_{j=1}^J \Gamma g_j S$$ Since the left cosets are unique it means for some permutation $$\Gamma g_j S=\Gamma g_{\sigma(j)}$$
For $f\in M_k(SL_2(\Bbb{Z}))$ define the $p$-th weight $k$ Hecke operator for $\Gamma$ $$T_p f = \sum_{j=1}^J f|_k \Gamma g_j$$ We find $$T_p f|_k S =\sum_{j=1}^J f|_k \Gamma g_j S=\sum_{j=1}^J f|_k \Gamma g_{\sigma(j)} = T_p f $$
You get your result by setting $f_p = p^{-k}(T_p f -f|_k \pmatrix{p & 0 \\ 0 & 1})$
When $f\in M_k(\Gamma)$ for a congruence subgroup $\Gamma$ we change $A_p$ to $A_p=\Gamma \pmatrix{1 & 0 \\0 & p}\Gamma$ since we don't have $A_p S=A_p$ we don't have $T_p f= T_pf|_k S$ anymore.