[proofplan] We encode addition by $B$ as a finite layered directed graph whose bottom layer is $A$ and whose layer-$j$ image of a subset $X \subset A$ is exactly $X+jB$. The hypothesis gives an upper bound $K$ for the first magnification ratio. Plünnecke's magnification inequality then bounds the $k$-th magnification ratio by the $k$-th power of the first one. Choosing a subset of $A$ attaining this minimum gives the required set $A'$. [/proofplan]

[step:Build the layered addition graph generated by $B$]Fix an integer $k \geq 1$. For each integer $j$ with $0 \leq j \leq k$, define the finite set \begin{align*} V_j := A + jB. \end{align*} Here $0B := \{0_G\}$, where $0_G$ is the identity element of the abelian group $G$, so $V_0 = A$. Define a directed layered graph $\Gamma$ whose layer $j$ is a labelled copy of $V_j$. Formally, the vertex set of layer $j$ is $V_j \times \{j\}$, so the layers are disjoint even if the sumsets $V_i$ and $V_j$ overlap inside $G$. For $0 \leq j < k$, there is a directed edge from $(x,j) \in V_j \times \{j\}$ to $(y,j+1) \in V_{j+1} \times \{j+1\}$ exactly when there exists $b \in B$ such that \begin{align*} y = x + b. \end{align*} For each integer $j$ with $0 \leq j \leq k$, define the layer-$j$ image map on subsets of the bottom sumset by \begin{align*} I_j: \mathcal{P}(V_0) &\to \mathcal{P}(V_j) \\ X &\mapsto \{y \in V_j : \text{there is a directed path of length } j \text{ from some } (x,0) \text{ with } x \in X \text{ to } (y,j)\}, \end{align*} where $\mathcal{P}(S)$ denotes the power set of a set $S$. By the definition of the edges, a path of length $j$ from $(x,0)$ with $x \in X$ adds $j$ elements of $B$, and therefore \begin{align*} I_j(X) = X + jB \end{align*} for every $X \subset A$ and every $0 \leq j \leq k$.[/step]

[guided]We want to convert the additive problem into a graph expansion problem. The bottom layer is the original set $A$, and moving upward by one layer means adding one element of $B$. For each integer $j$ with $0 \leq j \leq k$, define \begin{align*} V_j := A + jB. \end{align*} The convention $0B := \{0_G\}$ gives $V_0 = A$. Since $A$ and $B$ are finite, each layer $V_j$ is finite. Now define a directed graph $\Gamma$ whose layer $j$ is the labelled copy $V_j \times \{j\}$. This labelling matters because the same group element can lie in several sumsets $V_j$, and a layered graph must remember which layer the vertex belongs to. For $0 \leq j < k$, we draw an edge from $(x,j) \in V_j \times \{j\}$ to $(y,j+1) \in V_{j+1} \times \{j+1\}$ exactly when there is some $b \in B$ with \begin{align*} y = x + b. \end{align*} Thus one edge records one addition by an element of $B$. For each integer $j$ with $0 \leq j \leq k$, define the layer-$j$ image map on subsets of the bottom sumset by \begin{align*} I_j: \mathcal{P}(V_0) &\to \mathcal{P}(V_j) \\ X &\mapsto \{y \in V_j : \text{there is a directed path of length } j \text{ from some } (x,0) \text{ with } x \in X \text{ to } (y,j)\}, \end{align*} where $\mathcal{P}(S)$ denotes the power set of a set $S$. A directed path of length $j$ starting at $(x,0)$ with $x \in X$ has the form \begin{align*} (x,0),\quad (x+b_1,1),\quad (x+b_1+b_2,2),\quad \dots,\quad (x+b_1+\cdots+b_j,j) \end{align*} for some $b_1,\dots,b_j \in B$. Therefore its endpoint has group coordinate in $X+jB$. Conversely, every element of $X+jB$ has the form $x+b_1+\cdots+b_j$ and is reached by the corresponding path through the labelled vertices. Hence \begin{align*} I_j(X) = X+jB. \end{align*}[/guided]

[step:Relate the first magnification ratio to the doubling hypothesis] For $1 \leq j \leq k$, define the $j$-th magnification ratio of $\Gamma$ by \begin{align*} D_j := \min_{\varnothing \neq X \subset V_0} \frac{|I_j(X)|}{|X|}. \end{align*} The minimum exists because $V_0=A$ is finite and non-empty. Taking $X=A$ in the definition of $D_1$ gives \begin{align*} D_1 \leq \frac{|I_1(A)|}{|A|} = \frac{|A+B|}{|A|} \leq K. \end{align*} [/step]

[step:Apply Plünnecke's magnification inequality to the addition graph]The graph $\Gamma$ is the standard addition graph associated to the abelian group $G$ and the generator set $B$. Its finiteness follows from the finiteness of $A$ and $B$. We verify the two commutative graph axioms required by Plünnecke's magnification inequality. For the upward axiom, suppose $(x,j)$, $(y,j+1)$, and $(z,j+2)$ satisfy $y=x+b_1$ and $z=y+b_2$ for some $b_1,b_2 \in B$. Define $y' := x+b_2 \in V_{j+1}$. Then $(y',j+1)$ is adjacent from $(x,j)$, and the abelian identity \begin{align*} z = x+b_1+b_2 = x+b_2+b_1 = y' + b_1 \end{align*} shows that $(z,j+2)$ is adjacent from $(y',j+1)$. For the downward axiom, suppose $(y_1,j+1)$ and $(y_2,j+1)$ are both adjacent from $(x,j)$, with $y_1=x+b_1$ and $y_2=x+b_2$ for $b_1,b_2 \in B$. Define $z := x+b_1+b_2 \in V_{j+2}$. Then $(z,j+2)$ is adjacent from both $(y_1,j+1)$ and $(y_2,j+1)$, again using commutativity in $G$. We use the following form of Plünnecke's magnification inequality: in a finite commutative layered directed graph with bottom layer $V_0$, if \begin{align*} D_j = \min_{\varnothing \neq X \subset V_0} \frac{|I_j(X)|}{|X|} \end{align*} denotes the $j$-th magnification ratio, where $I_j(X)$ is the set of vertices in layer $j$ reachable from $X$ by directed paths of length $j$, then $D_k \leq D_1^k$ for every $k \geq 1$. Applying this result to $\Gamma$ gives \begin{align*} D_k \leq D_1^k. \end{align*} Using the bound $D_1 \leq K$ from the previous step, we obtain \begin{align*} D_k \leq K^k. \end{align*}[/step]

[guided]Plünnecke's magnification inequality applies to finite commutative layered directed graphs. We verify these hypotheses for $\Gamma$. First, $\Gamma$ is finite because each layer $V_j=A+jB$ is finite: it is the image of the finite Cartesian product $A \times B^j$ under the addition map \begin{align*} A \times B^j &\to G \\ (a,b_1,\dots,b_j) &\mapsto a+b_1+\cdots+b_j. \end{align*} Second, $\Gamma$ is commutative in the sense required by Plünnecke theory. We check the two local axioms. For the upward axiom, assume that $x \in V_j$, $y \in V_{j+1}$, and $z \in V_{j+2}$ satisfy $y=x+b_1$ and $z=y+b_2$ for some $b_1,b_2 \in B$. We need to complete the square by first adding $b_2$ and then adding $b_1$. Define \begin{align*} y' := x+b_2. \end{align*} Then $y' \in V_{j+1}$ and there is an edge from $x$ to $y'$. Since $G$ is abelian, \begin{align*} z = x+b_1+b_2 = x+b_2+b_1 = y' + b_1, \end{align*} so there is an edge from $y'$ to $z$. For the downward axiom, assume $y_1,y_2 \in V_{j+1}$ are both adjacent from $x \in V_j$, so $y_1=x+b_1$ and $y_2=x+b_2$ for some $b_1,b_2 \in B$. Define \begin{align*} z := x+b_1+b_2. \end{align*} Then $z \in V_{j+2}$. The equality $z=y_1+b_2$ gives an edge from $y_1$ to $z$, and the abelian equality \begin{align*} z=x+b_1+b_2=x+b_2+b_1=y_2+b_1 \end{align*} gives an edge from $y_2$ to $z$. Thus both commutative graph axioms hold, and this is precisely where the abelian hypothesis on $G$ is used. We now invoke Plünnecke's magnification inequality in the following precise form. Let a finite layered directed graph have bottom layer $V_0$ and image maps $I_j$ from subsets of $V_0$ to the vertices reachable in layer $j$. If the graph satisfies the upward and downward commutativity axioms checked above, and if \begin{align*} D_j = \min_{\varnothing \neq X \subset V_0} \frac{|I_j(X)|}{|X|}, \end{align*} then $D_k \leq D_1^k$ for every $k \geq 1$. Our graph $\Gamma$ is finite and satisfies these commutativity axioms, so this theorem gives \begin{align*} D_k \leq D_1^k. \end{align*} Since the previous step proved $D_1 \leq K$, monotonicity of the function $t \mapsto t^k$ on $[0,\infty)$ gives \begin{align*} D_k \leq D_1^k \leq K^k. \end{align*}[/guided]

[step:Choose a minimizing subset and translate the graph bound back to sumsets]By the definition of $D_k$ and the finiteness of $A$, there exists a non-empty subset $A' \subset A$ such that \begin{align*} D_k = \frac{|I_k(A')|}{|A'|}. \end{align*} Using $I_k(A') = A' + kB$ from the construction of the graph and the estimate $D_k \leq K^k$, we get \begin{align*} \frac{|A'+kB|}{|A'|} = \frac{|I_k(A')|}{|A'|} = D_k \leq K^k. \end{align*} Multiplying by $|A'|>0$ gives \begin{align*} |A'+kB| \leq K^k |A'|. \end{align*} This is the desired conclusion.[/step]

[guided]The number $D_k$ was defined as a minimum over all non-empty subsets of the finite set $A$. Because there are only finitely many such subsets, the minimum is attained. Choose a non-empty subset $A' \subset A$ such that \begin{align*} D_k = \frac{|I_k(A')|}{|A'|}. \end{align*} The first step identified the graph image with the additive sumset, so for this chosen set $A'$ we have \begin{align*} I_k(A') = A' + kB. \end{align*} Combining this identity with the Plünnecke estimate $D_k \leq K^k$ gives \begin{align*} \frac{|A'+kB|}{|A'|} = \frac{|I_k(A')|}{|A'|} = D_k \leq K^k. \end{align*} Since $A'$ is non-empty, $|A'|>0$, so multiplying both sides by $|A'|$ preserves the inequality and yields \begin{align*} |A'+kB| \leq K^k |A'|. \end{align*} This is exactly the set promised in the theorem statement.[/guided]