[proofplan] We choose $\Lambda$ to be a maximal dissociated subset of the large spectrum. Maximality forces every large Fourier character to lie in the subgroup generated by $\Lambda$. The size bound comes from comparing two estimates for a trigonometric polynomial supported on $\Lambda$: a lower bound from the large Fourier coefficients of $\mathbb{1}_A$, and an upper bound from Hölder together with the Rudin inequality for dissociated characters. Optimizing the moment exponent gives $|\Lambda| \leq C\rho^{-2}\log(1/\alpha)$. [/proofplan]

[step:Choose a maximal dissociated subset of the large spectrum]A finite set $\Lambda \subseteq \widehat{G}$ is called dissociated if the only choice of coefficients $\varepsilon_\lambda \in \{-1,0,1\}$ satisfying \begin{align*} \prod_{\lambda \in \Lambda} \lambda^{\varepsilon_\lambda} = 1_{\widehat{G}} \end{align*} is the trivial choice $\varepsilon_\lambda = 0$ for every $\lambda \in \Lambda$, where $1_{\widehat{G}}:G\to \mathbb{C}$ denotes the trivial character. Since $\operatorname{Spec}_\rho(A)$ is finite, choose $\Lambda \subseteq \operatorname{Spec}_\rho(A)$ maximal among dissociated subsets of $\operatorname{Spec}_\rho(A)$. We claim that \begin{align*} \operatorname{Spec}_\rho(A) \subseteq \langle \Lambda\rangle. \end{align*} Indeed, let $\gamma \in \operatorname{Spec}_\rho(A)$. If $\gamma \in \Lambda$, then $\gamma \in \langle \Lambda\rangle$. If $\gamma \notin \Lambda$, then maximality implies that $\Lambda \cup \{\gamma\}$ is not dissociated. Therefore there are coefficients $\varepsilon_\gamma \in \{-1,1\}$ and $\varepsilon_\lambda \in \{-1,0,1\}$, not all zero, such that \begin{align*} \gamma^{\varepsilon_\gamma}\prod_{\lambda \in \Lambda}\lambda^{\varepsilon_\lambda} =1_{\widehat{G}}. \end{align*} Solving for $\gamma$ gives \begin{align*} \gamma = \prod_{\lambda \in \Lambda}\lambda^{-\varepsilon_\gamma\varepsilon_\lambda}, \end{align*} so $\gamma \in \langle \Lambda\rangle$.[/step]

[guided]The purpose of dissociativity is to isolate an independent part of the spectrum. A dissociated set has no non-trivial multiplicative relation with coefficients in $\{-1,0,1\}$. Because $\widehat{G}$ is finite, the spectrum $\operatorname{Spec}_\rho(A)$ is finite. Hence we may choose a dissociated subset $\Lambda \subseteq \operatorname{Spec}_\rho(A)$ that is maximal with respect to inclusion. Now take any $\gamma \in \operatorname{Spec}_\rho(A)$. If $\gamma$ already belongs to $\Lambda$, then it belongs to $\langle \Lambda\rangle$. Otherwise, if $\gamma \notin \Lambda$, maximality says that adding $\gamma$ destroys dissociativity. Thus $\Lambda\cup\{\gamma\}$ admits a non-trivial relation \begin{align*} \gamma^{\varepsilon_\gamma}\prod_{\lambda \in \Lambda}\lambda^{\varepsilon_\lambda} = 1_{\widehat{G}}, \end{align*} where $\varepsilon_\gamma,\varepsilon_\lambda \in \{-1,0,1\}$ and not all coefficients are zero. Since $\Lambda$ itself is dissociated, the coefficient of $\gamma$ cannot be $0$; otherwise the displayed relation would already be a non-trivial relation among elements of $\Lambda$. Hence $\varepsilon_\gamma \in \{-1,1\}$. Rearranging gives \begin{align*} \gamma = \prod_{\lambda \in \Lambda}\lambda^{-\varepsilon_\gamma\varepsilon_\lambda}, \end{align*} so $\gamma$ is in the subgroup generated by $\Lambda$. This proves the spanning conclusion.[/guided]

[step:Construct a trigonometric polynomial aligned with the large Fourier coefficients] Let $m := |\Lambda|$. If $m=0$, the desired bound is immediate, so assume $m\geq 1$. For each $\lambda \in \Lambda$, define a phase $c_\lambda \in \mathbb{C}$ by \begin{align*} c_\lambda := \begin{cases} \widehat{\mathbb{1}_A}(\lambda)/|\widehat{\mathbb{1}_A}(\lambda)|, & \widehat{\mathbb{1}_A}(\lambda)\neq 0,\\ 1, & \widehat{\mathbb{1}_A}(\lambda)=0. \end{cases} \end{align*} Since $\lambda \in \operatorname{Spec}_\rho(A)$ and $\rho\alpha>0$, the first case always occurs. Define the trigonometric polynomial \begin{align*} S: G &\to \mathbb{C}\\ x &\mapsto \sum_{\lambda \in \Lambda} c_\lambda \lambda(x). \end{align*} Then \begin{align*} \int_G \mathbb{1}_A(x)\overline{S(x)}\,d\mu_G(x) = \sum_{\lambda \in \Lambda}\overline{c_\lambda}\widehat{\mathbb{1}_A}(\lambda) = \sum_{\lambda \in \Lambda}|\widehat{\mathbb{1}_A}(\lambda)| \geq m\rho\alpha. \end{align*} [/step]

[step:Upper bound the same correlation by Hölder and Rudin's inequality]Let $p\geq 2$ and let $q:=p/(p-1)$ be its Hölder conjugate. Hölder's inequality applied on the probability space $(G,\mu_G)$ gives \begin{align*} m\rho\alpha &\leq \left|\int_G \mathbb{1}_A(x)\overline{S(x)}\,d\mu_G(x)\right|\\ &\leq \|\mathbb{1}_A\|_{L^q(G)}\|S\|_{L^p(G)}. \end{align*} Since $\mathbb{1}_A^q=\mathbb{1}_A$, we have \begin{align*} \|\mathbb{1}_A\|_{L^q(G)} = \left(\int_G \mathbb{1}_A(x)\,d\mu_G(x)\right)^{1/q} = \alpha^{1/q}. \end{align*} We use [Rudin's inequality for dissociated characters](/theorems/???) on the probability space $(G,\mu_G)$: there is an absolute constant $C_R>0$ such that for every finite dissociated $\Lambda\subseteq \widehat{G}$, every coefficient family $(a_\lambda)_{\lambda\in\Lambda}\subseteq \mathbb{C}$, and every $p\geq 2$, \begin{align*} \left\|\sum_{\lambda\in\Lambda}a_\lambda\lambda\right\|_{L^p(G)} \leq C_R\sqrt{p}\left(\sum_{\lambda\in\Lambda}|a_\lambda|^2\right)^{1/2}. \end{align*} Its hypotheses are satisfied here because $\Lambda$ was chosen dissociated, the coefficient family $(c_\lambda)_{\lambda\in\Lambda}$ is finite, and $p\geq 2$. Applying this with $a_\lambda=c_\lambda$ gives \begin{align*} \|S\|_{L^p(G)} \leq C_R\sqrt{p}\left(\sum_{\lambda\in\Lambda}|c_\lambda|^2\right)^{1/2} = C_R\sqrt{pm}. \end{align*} Therefore \begin{align*} m\rho\alpha \leq C_R\alpha^{1/q}\sqrt{pm}. \end{align*} Dividing by $\alpha>0$ and using $1/q=1-1/p$ gives \begin{align*} m\rho \leq C_R\alpha^{-1/p}\sqrt{pm}. \end{align*}[/step]

[guided]We now estimate the same correlation from above. The lower bound used the fact that every $\lambda\in\Lambda$ has large Fourier coefficient. The upper bound uses only the size of $A$ and the independence encoded by dissociativity. Fix $p\geq 2$, and let $q:=p/(p-1)$, so $1/p+1/q=1$. Hölder's inequality on the probability space $(G,\mu_G)$ gives \begin{align*} \left|\int_G \mathbb{1}_A(x)\overline{S(x)}\,d\mu_G(x)\right| \leq \|\mathbb{1}_A\|_{L^q(G)}\|S\|_{L^p(G)}. \end{align*} Because $\mathbb{1}_A$ takes only the values $0$ and $1$, we have $\mathbb{1}_A^q=\mathbb{1}_A$, and hence \begin{align*} \|\mathbb{1}_A\|_{L^q(G)} = \left(\int_G \mathbb{1}_A(x)\,d\mu_G(x)\right)^{1/q} = \alpha^{1/q}. \end{align*} The point of choosing $\Lambda$ dissociated is that trigonometric polynomials supported on $\Lambda$ satisfy a square-root moment bound. Specifically, [Rudin's inequality for dissociated characters](/theorems/???) states that there is an absolute constant $C_R>0$ such that \begin{align*} \left\|\sum_{\lambda\in\Lambda}a_\lambda\lambda\right\|_{L^p(G)} \leq C_R\sqrt{p}\left(\sum_{\lambda\in\Lambda}|a_\lambda|^2\right)^{1/2} \end{align*} for every finite dissociated set $\Lambda\subseteq\widehat{G}$, every coefficient family $(a_\lambda)_{\lambda\in\Lambda}\subseteq\mathbb{C}$, and every $p\geq 2$. Applying it to the coefficients $a_\lambda=c_\lambda$ is valid because $\Lambda$ is finite and dissociated, $(c_\lambda)_{\lambda\in\Lambda}\subseteq\mathbb{C}$, and $p\geq 2$. Since each $c_\lambda$ has modulus $1$, \begin{align*} \|S\|_{L^p(G)} \leq C_R\sqrt{p}\left(\sum_{\lambda\in\Lambda}|c_\lambda|^2\right)^{1/2} = C_R\sqrt{pm}. \end{align*} Combining the lower bound from the previous step with Hölder and Rudin gives \begin{align*} m\rho\alpha \leq C_R\alpha^{1/q}\sqrt{pm}. \end{align*} Since $1/q=1-1/p$, dividing by $\alpha$ gives \begin{align*} m\rho \leq C_R\alpha^{-1/p}\sqrt{pm}. \end{align*} This inequality is the quantitative core of the proof.[/guided]

[step:Optimize the moment parameter to bound the size of the dissociated set]If $\alpha=1$, then $A=G$ and \begin{align*} \widehat{\mathbb{1}_A}(\gamma) = \int_G \overline{\gamma(x)}\,d\mu_G(x) \end{align*} is $1$ for the trivial character and $0$ for every non-trivial character by character orthogonality. Hence $\operatorname{Spec}_\rho(A)=\{1_{\widehat{G}}\}$, so the maximal dissociated subset $\Lambda$ is empty and the claimed estimate holds. Assume now $0<\alpha<1$. Define \begin{align*} L:=\log(1/\alpha)>0 \end{align*} and choose \begin{align*} p:=2+\log(1/\alpha)=2+L. \end{align*} Then $p\geq 2$ and \begin{align*} \alpha^{-1/p} = \exp(L/p) \leq e. \end{align*} From \begin{align*} m\rho \leq C_R\alpha^{-1/p}\sqrt{pm}, \end{align*} we obtain, since $m\geq 1$, \begin{align*} \sqrt{m}\rho \leq eC_R\sqrt{p}. \end{align*} Squaring gives \begin{align*} m \leq e^2C_R^2\rho^{-2}p = e^2C_R^2\rho^{-2}(2+\log(1/\alpha)). \end{align*} This already gives the desired bound when $0<\alpha\leq 1/2$, because then $L\geq \log 2$ and \begin{align*} 2+L \leq \left(1+\frac{2}{\log 2}\right)L. \end{align*} It remains to handle $1/2<\alpha<1$. Since a dissociated set cannot contain the trivial character $1_{\widehat{G}}$, every $\lambda\in\Lambda$ is non-trivial. For non-trivial $\lambda$, character orthogonality gives \begin{align*} \widehat{\mathbb{1}_A}(\lambda) = -\widehat{\mathbb{1}_{G\setminus A}}(\lambda). \end{align*} By [Parseval's identity](/theorems/434) on the finite probability group $(G,\mu_G)$, applied to $\mathbb{1}_{G\setminus A}:G\to\mathbb{C}$, we have \begin{align*} \sum_{\gamma\in\widehat{G}}|\widehat{\mathbb{1}_{G\setminus A}}(\gamma)|^2 = \int_G \mathbb{1}_{G\setminus A}(x)\,d\mu_G(x) = 1-\alpha. \end{align*} Restricting the sum to $\Lambda$ and using $|\widehat{\mathbb{1}_A}(\lambda)|\geq \rho\alpha$ for each $\lambda\in\Lambda$ gives \begin{align*} m\rho^2\alpha^2 \leq \sum_{\lambda\in\Lambda}|\widehat{\mathbb{1}_A}(\lambda)|^2 = \sum_{\lambda\in\Lambda}|\widehat{\mathbb{1}_{G\setminus A}}(\lambda)|^2 \leq 1-\alpha. \end{align*} Since $\alpha>1/2$, this implies \begin{align*} m \leq \rho^{-2}\frac{1-\alpha}{\alpha^2} \leq 4\rho^{-2}(1-\alpha) \leq 4\rho^{-2}\log(1/\alpha), \end{align*} where the last inequality uses $1-\alpha\leq \log(1/\alpha)$ for $0<\alpha<1$. Combining the two density regimes, there is an absolute constant \begin{align*} C:=\max\left\{e^2C_R^2\left(1+\frac{2}{\log 2}\right),4\right\} \end{align*} such that \begin{align*} |\Lambda|=m\leq C\rho^{-2}\log(1/\alpha). \end{align*}[/step]

[guided]The inequality \begin{align*} m\rho \leq C_R\alpha^{-1/p}\sqrt{pm} \end{align*} holds for every $p\geq 2$. We now choose $p$ so that the factor $\alpha^{-1/p}$ is bounded by an absolute constant. First handle the endpoint $\alpha=1$. Then $A=G$, and the [Fourier transform](/page/Fourier%20Transform) of $\mathbb{1}_G$ is supported only at the trivial character. Indeed, \begin{align*} \widehat{\mathbb{1}_G}(\gamma) = \int_G \overline{\gamma(x)}\,d\mu_G(x), \end{align*} which equals $1$ for the trivial character and $0$ for non-trivial characters by orthogonality of characters. Thus the large spectrum is generated by the empty dissociated set, and the estimate is immediate. Now suppose $0<\alpha<1$, and define \begin{align*} L:=\log(1/\alpha)>0. \end{align*} Choose \begin{align*} p:=2+L. \end{align*} This choice is admissible because $p\geq 2$. It also gives \begin{align*} \alpha^{-1/p} = \exp(L/p) \leq e, \end{align*} since $L/p\leq 1$. Therefore \begin{align*} m\rho \leq eC_R\sqrt{pm}. \end{align*} If $m\geq 1$, we divide by $\sqrt{m}$ to get \begin{align*} \sqrt{m}\rho \leq eC_R\sqrt{p}. \end{align*} Squaring yields \begin{align*} m \leq e^2C_R^2\rho^{-2}p = e^2C_R^2\rho^{-2}(2+\log(1/\alpha)). \end{align*} If $0<\alpha\leq 1/2$, then $L\geq\log 2$, so \begin{align*} 2+L \leq \left(1+\frac{2}{\log 2}\right)L. \end{align*} Thus in this density range, \begin{align*} m \leq e^2C_R^2\left(1+\frac{2}{\log 2}\right)\rho^{-2}\log(1/\alpha). \end{align*} This is the only place where the additive $2$ can be absorbed; it works because $\log(1/\alpha)$ is bounded below on $0<\alpha\leq 1/2$. For the high-density range $1/2<\alpha<1$, we use a different estimate. A dissociated set cannot contain the trivial character $1_{\widehat{G}}$, because the one-term relation $1_{\widehat{G}}^1=1_{\widehat{G}}$ would violate dissociativity. Hence every $\lambda\in\Lambda$ is non-trivial. For such $\lambda$, orthogonality of characters gives $\widehat{\mathbb{1}_G}(\lambda)=0$, and since $\mathbb{1}_A=\mathbb{1}_G-\mathbb{1}_{G\setminus A}$, \begin{align*} \widehat{\mathbb{1}_A}(\lambda) = -\widehat{\mathbb{1}_{G\setminus A}}(\lambda). \end{align*} Now apply Parseval's identity on the finite probability group $(G,\mu_G)$ to the function $\mathbb{1}_{G\setminus A}:G\to\mathbb{C}$: \begin{align*} \sum_{\gamma\in\widehat{G}}|\widehat{\mathbb{1}_{G\setminus A}}(\gamma)|^2 = \int_G \mathbb{1}_{G\setminus A}(x)\,d\mu_G(x) = 1-\alpha. \end{align*} Restricting this non-negative sum to $\Lambda$ and using the large-spectrum lower bound gives \begin{align*} m\rho^2\alpha^2 \leq \sum_{\lambda\in\Lambda}|\widehat{\mathbb{1}_A}(\lambda)|^2 = \sum_{\lambda\in\Lambda}|\widehat{\mathbb{1}_{G\setminus A}}(\lambda)|^2 \leq 1-\alpha. \end{align*} Since $\alpha>1/2$, we have $\alpha^{-2}<4$, and since $1-\alpha\leq\log(1/\alpha)$ for $0<\alpha<1$, this yields \begin{align*} m \leq 4\rho^{-2}\log(1/\alpha). \end{align*} Combining the two regimes and taking \begin{align*} C:=\max\left\{e^2C_R^2\left(1+\frac{2}{\log 2}\right),4\right\} \end{align*} gives \begin{align*} m=|\Lambda|\leq C\rho^{-2}\log(1/\alpha). \end{align*} This is the desired size estimate.[/guided]

[step:Combine maximality and the size estimate]The maximal dissociated subset $\Lambda\subseteq \operatorname{Spec}_\rho(A)$ constructed above satisfies \begin{align*} \operatorname{Spec}_\rho(A)\subseteq \langle \Lambda\rangle \end{align*} by maximality, and \begin{align*} |\Lambda|\leq C\rho^{-2}\log(1/\alpha) \end{align*} by the low-density moment estimate together with the high-density Parseval estimate. This proves the theorem.[/step]

[guided]The construction produced a set $\Lambda\subseteq\operatorname{Spec}_\rho(A)$ with two required properties. First, maximality among dissociated subsets forced every character in the large spectrum to lie in the subgroup generated by $\Lambda$: \begin{align*} \operatorname{Spec}_\rho(A)\subseteq \langle \Lambda\rangle. \end{align*} Second, the quantitative argument bounded its size. For $0<\alpha\leq 1/2$, the Hölder-Rudin moment estimate gives the desired bound after absorbing the additive constant into $\log(1/\alpha)$; for $1/2<\alpha<1$, Parseval applied to $\mathbb{1}_{G\setminus A}$ gives the same bound. Thus, with the absolute constant $C$ defined in the preceding step, \begin{align*} |\Lambda|\leq C\rho^{-2}\log(1/\alpha). \end{align*} These are exactly the spanning and size conclusions in the theorem statement.[/guided]