[proofplan] We compute the Fourier expansion directly from the defining formula for the normalized prime Hecke operator. The term $p^{k-1}f(pz)$ shifts coefficients from index $m/p$ to index $m$, contributing only when $p \mid m$. The averaging term over translations by $b/p$ kills every Fourier mode whose index is not divisible by $p$, and the surviving modes contribute $a_{pm}(f)$ to the coefficient of $q^m$. Adding the two contributions gives the stated coefficient formula. [/proofplan] [step:Write the normalized prime Hecke operator as two explicit summands] For $z \in \mathbb{H}$, define two functions \begin{align*} A: \mathbb{H} &\to \mathbb{C}, & A(z) &= p^{k-1} f(pz), \\ B: \mathbb{H} &\to \mathbb{C}, & B(z) &= \frac{1}{p}\sum_{b=0}^{p-1} f\left(\frac{z+b}{p}\right). \end{align*} By the normalized definition of the prime Hecke operator on $M_k(SL_2(\mathbb{Z}))$, \begin{align*} T_p f(z) = A(z) + B(z). \end{align*} It remains to compute the Fourier coefficients of $A$ and $B$. [/step] [step:Compute the contribution of $p^{k-1}f(pz)$] Since $q = e^{2\pi i z}$, we have $e^{2\pi i pz} = q^p$. Substituting $pz$ into the Fourier expansion of $f$ gives \begin{align*} A(z) &= p^{k-1} f(pz) \\ &= p^{k-1}\sum_{r=0}^{\infty} a_r(f) e^{2\pi i r p z} \\ &= \sum_{r=0}^{\infty} p^{k-1}a_r(f)q^{pr}. \end{align*} Therefore the coefficient of $q^m$ in $A(z)$ is $p^{k-1}a_{m/p}(f)$ if $p \mid m$, and is $0$ if $p \nmid m$. [/step] [step:Use the translation average to isolate indices divisible by $p$] Let \begin{align*} \zeta_p := e^{2\pi i/p}. \end{align*} For each $b \in \{0,\dots,p-1\}$ and $z \in \mathbb{H}$, \begin{align*} e^{2\pi i (z+b)/p} = q^{1/p}\zeta_p^b. \end{align*} Using the Fourier expansion of $f$ at $(z+b)/p$, we obtain \begin{align*} B(z) &= \frac{1}{p}\sum_{b=0}^{p-1}\sum_{r=0}^{\infty} a_r(f)e^{2\pi i r(z+b)/p} \\ &= \sum_{r=0}^{\infty} a_r(f)q^{r/p}\left(\frac{1}{p}\sum_{b=0}^{p-1}\zeta_p^{br}\right). \end{align*} The finite geometric sum satisfies \begin{align*} \frac{1}{p}\sum_{b=0}^{p-1}\zeta_p^{br} = \begin{cases} 1, & p \mid r,\\ 0, & p \nmid r. \end{cases} \end{align*} Indeed, if $p \mid r$, then every summand is $1$. If $p \nmid r$, then $\zeta_p^r \neq 1$, and the geometric-series identity gives \begin{align*} \sum_{b=0}^{p-1}\zeta_p^{br} = \frac{1-(\zeta_p^r)^p}{1-\zeta_p^r} = \frac{1-1}{1-\zeta_p^r} = 0. \end{align*} Thus only indices $r$ of the form $r = pm$ survive, and \begin{align*} B(z) &= \sum_{m=0}^{\infty} a_{pm}(f)q^m. \end{align*} Hence the coefficient of $q^m$ in $B(z)$ is $a_{pm}(f)$. [/step] [step:Add the two coefficient contributions] Combining the expansions of $A$ and $B$, we have \begin{align*} T_p f(z) &= A(z)+B(z) \\ &= \sum_{m=0}^{\infty} p^{k-1}a_{m/p}(f)q^m + \sum_{m=0}^{\infty} a_{pm}(f)q^m \\ &= \sum_{m=0}^{\infty}\left(a_{pm}(f)+p^{k-1}a_{m/p}(f)\right)q^m, \end{align*} where $a_{m/p}(f)=0$ when $p \nmid m$. This is the desired Fourier coefficient formula for $T_p f$. [/step]