Given a multiplicative subgroup $ \Gamma \subseteq F^*_p $ (multiplicative group of integers modulo prime $ p $), its indicator function $ \Gamma(x) $, and the Fourier transform of a function $ f: F_p \to \mathbb{C} $ defined as
$$ \hat{f}(\xi) = \sum_{x \in F^*_p} f(x) exp(-\xi\cdot2\pi i x / p) $$
we can say say that (by separating the terms for $ \xi =0 $ and $ \xi \neq0 $)
$$ T = \frac{1}{p}\sum_{\xi}\hat{\Gamma}(\xi)^{2k} = \frac{1}{p} \left(\hat{\Gamma}(0)^{2k} + \sum_{\xi\neq0} \hat{\Gamma}(\xi)^{2k}\right) = \frac{|\Gamma|^{2k}}{p} + \frac{1}{p}\sum_{\xi\neq0} \hat{\Gamma}(\xi)^{2k}$$
But I'm not sure how do I estimate $\sum_{\xi\neq0} \frac{1}{p} \hat{\Gamma}(\xi)^{2k}$ in terms of size of $ \Gamma $? I suspect I might need to use properties of $ \Gamma $, but I'm not sure which of them can apply here. Any pointers are greatly appreciated.
First, note that $\hat\Gamma(\xi)$ is real if $\Gamma$ is non-trivial. That follows from Lagrange's theorem, considering that every $\Gamma \subseteq F^\times_p$ is cyclic. From that:
$\hat\Gamma(\xi) = \displaystyle \sum_{x \in \Gamma}{\cos\left(2\pi \xi\dfrac{x}{p}\right)}$
Also, note that if $\xi \in \Gamma$, $\hat\Gamma(\xi) = \hat\Gamma(1)$. If $|\Gamma| = \frac{p - 1}{2}$ and $\xi \not \in \Gamma$, it will be the sum of the other terms, which is $-1 -\hat\Gamma(1)$
$T_k = \dfrac{|\Gamma|^{2k}}{p} + \dfrac{p - 1}{2p}(\hat\Gamma(1)^ {2k} + (-1 - \hat\Gamma(1))^{2k})$
Another particular case, if $\Gamma = F^\times_p$
$T_k = \dfrac{(p - 1)^{2k}}{p} - \dfrac{p - 1}{p}$
In general, we have to break $F^\times_p$ in the cosets, each of size $|\Gamma|$ and get $\hat\Gamma$ for one element $m$ in each coset. Then
$T_k = \dfrac{|\Gamma|^{2k}}{p} + \dfrac{|\Gamma|}{p} \displaystyle \sum_{m_i} \hat\Gamma(m_i)^{2k}$
That's what I was able to get. Hope it helps.