Read-back: every elementary amenable group has the packing property

The statement in one sentence

Let $G$ be a group. If $G$ is elementary amenable in the inductive sense spelled out in §1 below, then $G$ has the packing property spelled out in §2 below. The statement is an implication only, not an equivalence.

Setting and binders

• $G$ is an arbitrary type equipped with a group structure (multiplication $\cdot$, identity $1$, inverses). $G$ is implicit; it is not required to be finite, countable, or anything else beyond being a group. Being a type, $G$ is nonempty (it contains $1$). The type $G$ lives in an arbitrary universe; this only matters in one place, noted in §1.
• The single hypothesis is that $G$ satisfies the predicate "elementary amenable" of §1.
• The conclusion is the predicate "has the packing property" of §2.

There are no other hypotheses: no decidability, no finiteness, no countability, no choice of generating set.

1. The hypothesis: $G$ is elementary amenable

"Elementary amenable" is defined as an inductive predicate on groups. That is, it is the smallest property $\mathcal{P}$ of groups that is closed under the seven rules listed next; a group satisfies it exactly when there is a finite derivation tree that establishes it from these rules. Nothing else makes a group elementary amenable.

The rules are:

1. Finite groups. If $G$ is finite (there is a bijection between $G$ and $\{0,\dots,n-1\}$ for some natural number $n$), then $\mathcal{P}(G)$.

2. Commutative groups. If $G$ carries a commutative group structure, then $\mathcal{P}(G)$ holds for $G$ with the group structure underlying that commutative structure.

3. Isomorphism. If $G$ and $H$ are groups, $e : G \to H$ is a group isomorphism (a bijection with $e(ab) = e(a)e(b)$), and $\mathcal{P}(G)$, then $\mathcal{P}(H)$.

4. Subgroups. If $H$ is any subgroup of $G$ and $\mathcal{P}(G)$, then $\mathcal{P}(H)$, where $H$ carries the group structure inherited from $G$.

5. Quotients. If $N$ is a normal subgroup of $G$ (i.e. $g n g^{-1} \in N$ for all $n \in N$, $g \in G$) and $\mathcal{P}(G)$, then $\mathcal{P}(G/N)$. Here $G/N$ is the set of left cosets — two elements $x, y$ give the same coset exactly when $x^{-1} y \in N$ — with the usual quotient group structure.

6. Extensions. If $N$ is a normal subgroup of $G$, $\mathcal{P}(N)$ and $\mathcal{P}(G/N)$, then $\mathcal{P}(G)$.

7. Directed unions. Let $I$ be an index set and $(H_i)_{i \in I}$ a family of subgroups of $G$ such that

   • the family is directed: for all $i, j \in I$ there exists $k \in I$ with $H_i \subseteq H_k$ and $H_j \subseteq H_k$;
   • the subgroup generated by $\bigcup_{i \in I} H_i$ is all of $G$ (literally: the supremum of the family in the lattice of subgroups equals the whole group);
   • $\mathcal{P}(H_i)$ for every $i \in I$.

   Then $\mathcal{P}(G)$.

   Remarks on rule 7. When $I$ is nonempty, directedness makes $\bigcup_i H_i$ already a subgroup, so the second condition then says exactly that every element of $G$ lies in some $H_i$. When $I$ is empty, directedness holds vacuously, the supremum is the trivial subgroup $\{1\}$, and the second condition holds only if $G$ is the trivial group; so the empty case adds nothing beyond rule 1. Nothing requires the $H_i$ to be nested, distinct, proper, or finite.

Universe fine print. In every rule, every group mentioned ($H$ in rules 3 and 4, $N$ and $G/N$ in rules 5–6, the index set $I$ in rule 7) is required to live in the same universe (the same "size class" of sets) as $G$. In particular the index set $I$ of a directed union is a set of the same size class as the group. For a mathematician working in ordinary set theory this is no restriction.

Degenerate instances of the hypothesis. The trivial group satisfies the hypothesis (by rule 1 or rule 2). Every finite group and every abelian group satisfies it outright. The hypothesis is not vacuous; but note that the theorem asserts nothing about groups that fail it.

2. The conclusion: $G$ has the packing property

$G$ has the packing property means:

$$\text{for every finite subset } F \subseteq G \text{ there exist subsets } S, X \subseteq G \text{ with } F \subseteq S,\ S \text{ finite, and } (S, X) \text{ a packing of } G.$$

Here "$F$ finite" and "$S$ finite" mean the subset is in bijection with $\{0,\dots,n-1\}$ for some natural number $n$ (so $n = 0$, the empty set, is included). $F \subseteq S$ means every element of $F$ is an element of $S$.

$(S, X)$ is a packing of $G$ means, literally, that the multiplication map

$$m : G \times G \to G, \qquad (s, x) \mapsto s \cdot x$$

restricted to the Cartesian product $S \times X = \{(s,x) : s \in S,\ x \in X\}$ is a bijection from $S \times X$ onto all of $G$. Unfolded, this is the conjunction of three conditions:

• (maps into) for every $(s,x) \in S \times X$, $s \cdot x \in G$ — automatically true;
• (injective on $S \times X$) for all $(s,x), (s',x') \in S \times X$, if $s \cdot x = s' \cdot x'$ then $(s,x) = (s',x')$, i.e. $s = s'$ and $x = x'$;
• (surjective onto $G$) for every $g \in G$ there exist $s \in S$ and $x \in X$ with $s \cdot x = g$.

Equivalently: every element of $G$ can be written in exactly one way as $s \cdot x$ with $s \in S$ and $x \in X$. Equivalently again: the left translates $\{ s X : s \in S\}$ are pairwise disjoint and their union is $G$ (and, symmetrically, the right translates $\{ S x : x \in X \}$ are pairwise disjoint with union $G$). The order of the factors matters: the element of $S$ is on the left and the element of $X$ on the right.

What the conclusion does and does not require.

• $X$ is not required to be finite; only $S$ is. No bound on the size of $S$ is asserted, and nothing relates $|S|$ to $|F|$ beyond $F \subseteq S$.
• Because $G$ is nonempty and $m$ must be onto $G$, both $S$ and $X$ are forced to be nonempty. Since $F \subseteq S$ with $S$ finite, $F$ itself must be finite, which is the standing hypothesis on $F$.
• The case $F = \varnothing$ is included: then the requirement is only that some finite $S$ and some $X$ form a packing.
• Nothing constrains $X$ in terms of $F$; nothing says $S$ or $X$ is a subgroup, a transversal, or unique; the quantifier is "there exist", not "there exists a unique".
• If $G$ happens to be finite, a packing $(S,X)$ forces $|S| \cdot |X| = |G|$; if $G$ is infinite and $S$ is finite, $X$ must be infinite.
• For the trivial group $G = \{1\}$, the conclusion holds with $S = X = \{1\}$ for every (necessarily empty or equal to $\{1\}$) $F$.

3. Assembled statement

For every group $G$: if $G$ can be built from finite groups and commutative groups by finitely many applications of isomorphism, passing to subgroups, passing to quotients by normal subgroups, extension by a normal subgroup with quotient already built, and directed unions of already-built subgroups (rules 1–7 of §1), then for every finite set $F$ of elements of $G$ there is a finite set $S \supseteq F$ of elements of $G$ and a set $X$ of elements of $G$ such that every element of $G$ is $s \cdot x$ for exactly one pair $(s, x) \in S \times X$.