I am working through Murty's 'Problems in the Theory of Modular Froms', but I am stuck in the proof of Hecke operators being self adjoint with respect to the Petersson inner product. Before describing my problem, let me introduce some notation.

For $\alpha\in GL_2^+(\mathbb{Q})$ and $f$ a cusp form of weight $2k$ with respect to a congruence subgroup $\Gamma$, that is $f\in S_k(\Gamma)$, define $$ f|\alpha(z) = (\det\alpha)^k(cz+d)^{-2k}f(\alpha.z), \hspace{1cm} \alpha = \begin{pmatrix} a & b \\ c & d\end{pmatrix} $$ and define the Petersson inner product of $f,g\in S_k(\Gamma)$ by $$ (f,g) = \frac{1}{[SL_2(\mathbb{Z}):\Gamma]}\int_\mathcal{F} f(z)\overline{g(z)}(\text{Im } z)^{2k} \frac{dxdy}{y^2} $$ with $\mathcal{F}$ a fundamental domain of $\Gamma$.

Now, my problem involves Exercise 7.1.11, which goes as follows:

Let $\alpha\in GL_2^+(\mathbb{Z})$ have determinant $D$ and define $\alpha' = D\alpha^{-1}$. Show that for $f,g\in S_k(\Gamma)$, with $\Gamma$ a congruence subgroup of $SL_2(\mathbb{Z})$, we have $$ (f|\alpha,g) = (f,g|\alpha') $$ As I understand the exercise, the first inner product is with respect to $\alpha^{-1}\Gamma\alpha\cap\Gamma$, while the second is with respect to $\alpha\Gamma\alpha^{-1}\cap\Gamma$, and this I can prove (it is an application of Theorem 7.1.10, which states $$ (f|\alpha,g|\alpha) = (f,g) $$ where the first inner product is with respect to $\alpha^{-1}\Gamma\alpha\cap\Gamma$, the latter with respect to $\Gamma$, where we note $g|\alpha' = g|\alpha^{-1}$).

In Theorem 7.3.1, where self-adjointness of the Hecke operators is proven, there is a reference to Exercise 7.1.11, where it is used in the sense $$ (f|\alpha,g) = (f,g|\alpha') $$ both inner products with respect to $\Gamma$. This I have not managed to show, so I don't know if this is an error or am I misunderstanding something? Any help appreciated!

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    $\begingroup$ I don't know if this helps, but you can notice that $\Gamma' \subset \Gamma$ implies $(-,-)_{\Gamma'} = (-,-)_{\Gamma'}$ on $S_k(\Gamma)$. $\endgroup$ – Watson May 19 '18 at 10:08
  • $\begingroup$ @Watson Yea, I have noticed, but I can't really seem to put it to use. $\endgroup$ – user114158 May 20 '18 at 6:40

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