Proof. For a list $\mathbf C=(C_1,\ldots ,C_t)$ and $S\in \mathcal S$ , define

$$ \begin{align*} \Phi_S(\mathbf C)=\sum_{i=1}^t\lvert C_iS\rvert. \end{align*} $$

For finite $E,F,S\subseteq G$ and $x\in G$ , the elementary replacement is

$$ \begin{align*} (E,F)\mapsto(E',F')=(E\cup xF,E\cap xF). \end{align*} $$

Since $E'S=ES\cup xFS$ and $F'S\subseteq ES\cap xFS$ , inclusion–exclusion gives

(2.1) $$ \begin{align} \lvert E'S\rvert+\lvert F'S\rvert \le \lvert ES\rvert+\lvert FS\rvert. \end{align} $$

Moreover, $\lvert E'\rvert +\lvert F'\rvert =\lvert E\rvert +\lvert F\rvert $ . Thus, an elementary replacement preserves the total block cardinality and, for the same choice of x, does not increase $\Phi _S$ for any $S\in \mathcal S$ .

Suppose first that G is infinite. Choose $x\notin EF^{-1}$ , so that $E\cap xF=\varnothing $ . Then, the operation replaces the two blocks $E,F$ by one block of size $\lvert E\rvert +\lvert F\rvert $ and an empty block, without increasing any $\Phi _S$ . Repeatedly merging two nonempty blocks proves the first assertion.

Now, suppose that $\lvert G\rvert =N<\infty $ . If $E\ne G$ and $F\ne \varnothing $ , then

$$ \begin{align*} \sum_{x\in G}\lvert E\cap xF\rvert=\lvert E\rvert\lvert F\rvert. \end{align*} $$

Indeed, each pair $(e,f)\in E\times F$ contributes once, for $x=ef^{-1}$ . The average is strictly smaller than $\lvert F\rvert $ , so some x satisfies $\lvert E\cap xF\rvert <\lvert F\rvert $ . Under the replacement above, (2.1) holds simultaneously for all S, the total cardinality is preserved and $\lvert E'\rvert>\lvert E\rvert $ . Repeating elementary replacements on this fixed ordered pair must therefore terminate with $E'=G$ or $F'=\varnothing $ . We call the resulting finite sequence a complete two-block compression. If the original pair has total size m, its final block sizes are

$$ \begin{align*} \min\{m,N\} \quad\text{and}\quad \max\{m-N,0\}. \end{align*} $$

Apply complete two-block compressions repeatedly to pairs of nonempty proper entries in the list. Each complete compression leaves at most one proper entry in place of two, so the process ends with full blocks, empty blocks and at most one residual block R. Here, ‘exactly q full blocks’ means that exactly q entries of the list equal G; it does not assert the existence of q disjoint subsets of G. The total size $qN+r$ forces exactly those q full entries and $\lvert R\rvert =r$ ; when $r=0$ , there is no residual entry and we set $R=\varnothing $ . For nonempty $S\subseteq G$ , one has $GS=G$ , while $\Phi _S$ has been monotone nonincreasing at every elementary replacement. Consequently,

$$ \begin{align*} qN+\lvert RS\rvert\le \sum_{i=1}^t\lvert C_iS\rvert \end{align*} $$

simultaneously for all $S\in \mathcal S$ , proving the second assertion.

Proof of Theorem 1.3

The map $A\mapsto Ab_0$ is a bijection on the n-element subsets of G. For every $S\subseteq B$ , we have $ AS=(Ab_0)(b_0^{-1}S)$ and $S\mapsto b_0^{-1}S$ bijects the nonempty subsets of B of size at most c with those of $D=b_0^{-1}B$ . Hence, $\mu _{G,B}(n,c)=\mu _{G,D}(n,c)$ , so we may replace B by D.

Let $A\subseteq G$ have size n and decompose it over the left cosets of H that it meets:

$$ \begin{align*} A=\bigsqcup_{i=1}^t x_iC_i, \quad C_i\subseteq H, \quad \sum_{i=1}^t\lvert C_i\rvert=n. \end{align*} $$

For every nonempty $S\subseteq D$ , the products $x_iC_iS$ lie in distinct left cosets of H and, hence,

(3.1) $$ \begin{align} \lvert AS\rvert=\sum_{i=1}^t\lvert C_iS\rvert. \end{align} $$

Assume first that H is infinite. Apply Lemma 2.1 in H to the family

$$ \begin{align*} \mathcal S_c(D)=\{S\subseteq D:1\le\lvert S\rvert\le c\}. \end{align*} $$

It gives an n-element set $C\subseteq H$ such that $\lvert CS\rvert \le \lvert AS\rvert $ for every $S\in \mathcal S_c(D)$ . Taking maxima and then minima yields

$$ \begin{align*} \mu_{H,D}(n,c)\le\mu_{G,D}(n,c). \end{align*} $$

The reverse inequality follows by regarding every n-element subset of H as a subset of G.

Now, let $\lvert H\rvert =N<\infty $ and write $n=qN+r$ , $0\le r<N$ . The finite part of Lemma 2.1, together with (3.1), produces $R\subseteq H$ , $\lvert R\rvert =r$ , such that

$$ \begin{align*} qN+\lvert RS\rvert\le\lvert AS\rvert \quad(S\in\mathcal S_c(D)). \end{align*} $$

By the definition of the extremal parameter,

$$ \begin{align*} \mu_{H,D}(r,c) \le \max_{S\in\mathcal S_c(D)}\lvert RS\rvert, \end{align*} $$

because the particular set R supplied by compression need not minimise the maximum. Hence, for the arbitrary A chosen above,

$$ \begin{align*} qN+\mu_{H,D}(r,c) \le qN+\max_{S\in\mathcal S_c(D)}\lvert RS\rvert \le \max_{S\in\mathcal S_c(D)}\lvert AS\rvert. \end{align*} $$

Taking the minimum over all n-element $A\subseteq G$ proves $qN+\mu _{H,D}(r,c)\le \mu _{G,D}(n,c)$ .

For the reverse inequality, if $r=0$ , take $R=\varnothing $ . If $r>0$ , an extremal r-element subset $R\subseteq H$ exists because H is finite. Choose q complete left cosets of H and, when $r>0$ , place a translate of R in one further left coset. To justify that there are enough cosets, put $M=[G:H]$ . If M is infinite, the claim is immediate. If $M<\infty $ , then ${n=qN+r\le MN}$ implies $q\le M$ when $r=0$ and $q<M$ when $r>0$ . For every nonempty $S\subseteq H$ , one has $HS=H$ and the selected cosets are distinct. The resulting maximum is therefore exactly

$$ \begin{align*} qN+\max_{S\in\mathcal S_c(D)}\lvert RS\rvert =qN+\mu_{H,D}(r,c). \end{align*} $$

This proves the reverse inequality and, hence, the formula.

Proof of Corollary 1.4

We use additive notation. Fix $b_0\in B$ and put

$$ \begin{align*} D=B-b_0, \quad H=\langle D\rangle\le\mathbb Z. \end{align*} $$

The based affine-core of $\lambda B$ at $\lambda b_0$ is $(\lambda H,\lambda D)$ . Since $\lambda \ne 0$ , the map

$$ \begin{align*} H\to \lambda H, \quad h\mapsto\lambda h, \end{align*} $$

is an isomorphism carrying D onto $\lambda D$ . For subsets $A,S\subseteq H$ , this map preserves cardinalities and satisfies

$$ \begin{align*} \lambda(A+S)=\lambda A+\lambda S. \end{align*} $$

It therefore bijects the admissible families for $(H,D)$ with those for $(\lambda H,\lambda D)$ and shows that their core parameters agree. The two core groups also have the same order. Applying Theorem 1.3 to B and $\lambda B$ now gives the claimed equality.