[proofplan] We combine two standard inverse tools in finite-field [additive combinatorics](/page/Additive%20Combinatorics). Plünnecke-Ruzsa bounds the iterated sum-difference sets $2A-2A$ and $3A-2A$ in terms of $K|A|$, while the [finite-field Bogolyubov-Ruzsa lemma](/theorems/4598) places a large subspace inside $2A-2A$. To bound the quotient image of $A$, we count disjoint translates of this subspace inside $3A-2A$ rather than incorrectly treating $2A-2A$ as invariant under the subspace. The affine span of the small quotient image has bounded size, and its pullback is the desired coset of a subspace containing $A$. [/proofplan] [step:Record the two inverse estimates used in the argument] Let $G:=\mathbb F_p^n$ be the additive group of the ambient [vector space](/page/Vector%20Space). For finite subsets $B,C\subset G$, define \begin{align*} B+C&:=\{b+c:b\in B,\ c\in C\},\\ B-C&:=\{b-c:b\in B,\ c\in C\}. \end{align*} Also write $2A-2A:=A+A-A-A$ and $3A-2A:=A+A+A-A-A$. We use the Plünnecke-Ruzsa inequality in the finite abelian group $G$: since $A$ is finite, nonempty, and $|A+A|\le K|A|$, it gives \begin{align*} |2A-2A|\le K^4|A|, \qquad |3A-2A|\le K^5|A|. \end{align*} We also use the finite-field Bogolyubov-Ruzsa lemma in the following quantitative form: for the fixed prime $p$ and the parameter $K$, there is a constant $B_p(K)\ge 1$ such that every finite nonempty $A\subset \mathbb F_p^n$ satisfying $|A+A|\le K|A|$ has a subspace $W\le \mathbb F_p^n$ with \begin{align*} W\subset 2A-2A, \qquad |W|\ge \frac{|A|}{B_p(K)}. \end{align*} Both hypotheses of this lemma are exactly the finiteness and nonemptiness of $A\subset \mathbb F_p^n$ and the doubling bound $|A+A|\le K|A|$. [guided] The proof needs two quantitative facts. First, small doubling for $A$ propagates to controlled bounds for the mixed sum-difference sets $2A-2A$ and $3A-2A$. The Plünnecke-Ruzsa inequality applies in the finite abelian group $G:=\mathbb F_p^n$ because $A$ is finite and nonempty and because the theorem statement gives $|A+A|\le K|A|$. Applied with two positive and two negative summands, and then with three positive and two negative summands, it yields \begin{align*} |2A-2A|=|A+A-A-A|\le K^4|A|, \qquad |3A-2A|=|A+A+A-A-A|\le K^5|A|. \end{align*} The second estimate is the one that will control how many $W$-cosets can meet $A$. Second, small doubling over a fixed finite field forces genuine linear structure inside $2A-2A$. The finite-field Bogolyubov-Ruzsa lemma states, in the quantitative form used here, that for the fixed prime $p$ and the doubling parameter $K$, there is a constant $B_p(K)\ge 1$ such that if a finite nonempty set $A\subset \mathbb F_p^n$ satisfies $|A+A|\le K|A|$, then there is a subspace $W\le \mathbb F_p^n$ satisfying \begin{align*} W\subset 2A-2A, \qquad |W|\ge \frac{|A|}{B_p(K)}. \end{align*} The point of introducing $W$ is that each $W$-coset meeting $A$ contributes a disjoint translate of $W$ inside $3A-2A$, so the number of such cosets is bounded in terms of $p$ and $K$ only. [/guided] [/step] [step:Bound the image of $A$ by counting disjoint $W$-cosets inside $3A-2A$] Let $W\le G$ be the subspace supplied above, and define the quotient map \begin{align*} \pi:G&\to G/W\\ x&\mapsto x+W. \end{align*} Let $R\subset A$ be a set containing exactly one representative from each $W$-coset that meets $A$. Then $|R|=|\pi(A)|$, and the cosets $(r+W)_{r\in R}$ are pairwise disjoint. Since $r\in A$ and $W\subset 2A-2A$ for every $r\in R$, we have \begin{align*} r+W\subset A+(2A-2A)=3A-2A. \end{align*} Therefore the disjoint union $R+W$ is contained in $3A-2A$, so \begin{align*} |\pi(A)|\,|W|=|R|\,|W|=|R+W|\le |3A-2A|\le K^5|A|. \end{align*} Using $|W|\ge |A|/B_p(K)$ gives \begin{align*} |\pi(A)|\le K^5B_p(K). \end{align*} [guided] We now compress the problem by passing to the quotient by the structured subspace $W$. Define the quotient homomorphism \begin{align*} \pi:G&\to G/W\\ x&\mapsto x+W. \end{align*} This is a group homomorphism and also a [linear map](/page/Linear%20Map) of vector spaces over $\mathbb F_p$. The inclusion $W\subset 2A-2A$ does not say that $2A-2A$ is invariant under translation by $W$. Instead, we count cosets meeting $A$ by putting their translates inside the larger set $3A-2A$. Let $R\subset A$ be a set containing exactly one representative from each $W$-coset that meets $A$. Then the quotient map sends distinct elements of $R$ to distinct elements of $G/W$, and every element of $\pi(A)$ is represented by one element of $R$; hence \begin{align*} |R|=|\pi(A)|. \end{align*} The cosets $(r+W)_{r\in R}$ are pairwise disjoint by the choice of one representative per quotient class. For each $r\in R$, we have $r\in A$, and the Bogolyubov-Ruzsa subspace satisfies $W\subset 2A-2A$. Therefore \begin{align*} r+W\subset A+(2A-2A)=3A-2A. \end{align*} Taking the union over all representatives gives a disjoint subset $R+W\subset 3A-2A$. Since each coset $r+W$ has cardinality $|W|$, we obtain \begin{align*} |\pi(A)|\,|W|=|R|\,|W|=|R+W|\le |3A-2A|. \end{align*} The Plünnecke-Ruzsa estimate from the previous step gives $|3A-2A|\le K^5|A|$, and the finite-field Bogolyubov-Ruzsa lemma gives $|W|\ge |A|/B_p(K)$. Combining these two inequalities yields \begin{align*} |\pi(A)|\le \frac{K^5|A|}{|W|}\le K^5B_p(K). \end{align*} This is the key reduction: the image of $A$ in the quotient has cardinality bounded only in terms of $p$ and $K$. [/guided] [/step] [step:Take the affine span in the quotient and pull it back] Choose $a_0\in A$. Define \begin{align*} S:=\pi(A)-\pi(a_0)\subset G/W. \end{align*} Then $0\in S$ and $|S|\le K^5B_p(K)$. Let $H\le G/W$ be the $\mathbb F_p$-linear span of $S$. Since $H$ is generated by at most $|S|$ vectors over $\mathbb F_p$, \begin{align*} |H|\le p^{|S|}\le p^{K^5B_p(K)}. \end{align*} Define the subspace \begin{align*} V:=\pi^{-1}(H)\le G. \end{align*} Since $\pi(a)-\pi(a_0)\in H$ for every $a\in A$, we have $a\in a_0+V$. Thus $A\subset a_0+V$. [guided] The quotient image $\pi(A)$ is small, but the theorem asks for containment in one coset of a subspace of the original group. We therefore take the affine span of $\pi(A)$ in the quotient and pull it back. Choose $a_0\in A$ and define \begin{align*} S:=\pi(A)-\pi(a_0)=\{\pi(a)-\pi(a_0):a\in A\}\subset G/W. \end{align*} The set $S$ contains $0$ because $a_0\in A$, and translation is a bijection on $G/W$, so \begin{align*} |S|=|\pi(A)|\le K^5B_p(K). \end{align*} Let $H\le G/W$ denote the $\mathbb F_p$-linear span of $S$. A vector space over $\mathbb F_p$ generated by $m$ vectors has cardinality at most $p^m$: the linear combination map from $\mathbb F_p^m$ onto the generated subspace has image size at most $p^m$. Applying this with $m=|S|$ gives \begin{align*} |H|\le p^{|S|}\le p^{K^5B_p(K)}. \end{align*} Now define the pullback subspace \begin{align*} V:=\pi^{-1}(H)\le G. \end{align*} It is a subspace because $\pi$ is linear and $H$ is a subspace. For each $a\in A$, the element $\pi(a)-\pi(a_0)$ lies in $S\subset H$, which means $\pi(a-a_0)\in H$. Equivalently, $a-a_0\in \pi^{-1}(H)=V$. Hence $a\in a_0+V$ for every $a\in A$, and therefore \begin{align*} A\subset a_0+V. \end{align*} [/guided] [/step] [step:Compute the size of the pulled-back subspace and define the constants] Since $\ker\pi=W$ and $V=\pi^{-1}(H)$, every element of $H$ has exactly $|W|$ preimages in $V$. Hence \begin{align*} |V|=|W|\,|H|. \end{align*} Because $W\subset 2A-2A$, the Plünnecke-Ruzsa bound gives \begin{align*} |W|\le |2A-2A|\le K^4|A|. \end{align*} Combining this with the bound for $H$ yields \begin{align*} |V|\le K^4p^{K^5B_p(K)}|A|. \end{align*} Define \begin{align*} C_p(K):=K^4p^{K^5B_p(K)}, \qquad r_p(K):=\left\lceil K^5B_p(K)\right\rceil. \end{align*} Then $A\subset a_0+V$ and $|V|\le C_p(K)|A|$, as required. [guided] It remains to check the quantitative bound on the subspace obtained by pullback. Since $\pi:G\to G/W$ has kernel $W$, each point of the quotient has exactly $|W|$ preimages in $G$. Restricting to the preimage of $H$, we obtain \begin{align*} |V|=|\pi^{-1}(H)|=|W|\,|H|. \end{align*} We already bounded $|H|$ by $p^{K^5B_p(K)}$. We also need an upper bound for $|W|$ in terms of $|A|$. The inclusion $W\subset 2A-2A$ and the Plünnecke-Ruzsa estimate give \begin{align*} |W|\le |2A-2A|\le K^4|A|. \end{align*} Therefore \begin{align*} |V|=|W|\,|H|\le K^4|A|\,p^{K^5B_p(K)}. \end{align*} Define the constants promised in the theorem by \begin{align*} C_p(K):=K^4p^{K^5B_p(K)}, \qquad r_p(K):=\left\lceil K^5B_p(K)\right\rceil. \end{align*} They depend only on the fixed field size $p$ and the doubling parameter $K$. The preceding step proved $A\subset a_0+V$, and the displayed estimate proves \begin{align*} |V|\le C_p(K)|A|. \end{align*} This is exactly the asserted coset containment with controlled subspace size. [/guided] [/step]