Read-back: residually elementary amenable groups have the packing property

The statement

Let $G$ be any group (any set, in any size universe, with a group structure: an associative multiplication, an identity $1$, and inverses). The declaration asserts one implication:

$$G \text{ is residually elementary amenable} \;\Longrightarrow\; G \text{ has the packing property},$$

where both notions are the specific ones defined below. There are no further hypotheses on $G$: it need not be countable, finitely generated, infinite, or nontrivial.

Hypothesis: "residually elementary amenable"

$G$ is called residually elementary amenable when the following holds:

For every element $x \in G$ with $x \neq 1$, there exists a normal subgroup $K \trianglelefteq G$ such that $x \notin K$ and the quotient group $G/K$ is elementary amenable (in the sense defined in the next section).

Here "normal" means: for every $n \in K$ and every $g \in G$, $g n g^{-1} \in K$. The quotient $G/K$ is the set of left cosets $gK$, with the group structure induced from $G$ using normality. Nothing is required of $K$ beyond $x \notin K$ and elementary amenability of $G/K$; in particular $K$ need not have finite index, and the subgroup $K$ may depend on $x$.

Edge cases the quantifier includes:

• If $G$ is the trivial group, there is no $x \neq 1$, so the hypothesis holds vacuously.
• The requirement $x \notin K$ forces $K \neq G$, so $K$ is a proper normal subgroup; but $K = \{1\}$ is allowed (then the condition says $G$ itself is elementary amenable).

The class "elementary amenable" used here

"Elementary amenable" is defined as the smallest class of groups $\mathcal{E}$ (a property of a group together with its group structure) closed under the following seven rules. That is, a group is elementary amenable if and only if it can be shown to be so by finitely many applications of these rules, and nothing else is elementary amenable.

1. Finite groups. Every finite group is in $\mathcal{E}$. ("Finite" means: there is a natural number $n$ and a bijection between the group and $\{0, 1, \dots, n-1\}$.)

2. Abelian groups. Every abelian group is in $\mathcal{E}$. (Formally: any group whose structure comes from a commutative-group structure.)

3. Isomorphism. If $G \cong H$ via a bijection $e : G \to H$ with $e(xy) = e(x)e(y)$, and $G \in \mathcal{E}$, then $H \in \mathcal{E}$.

4. Subgroups. If $G \in \mathcal{E}$ and $H \le G$ is any subgroup (with the group structure inherited from $G$), then $H \in \mathcal{E}$.

5. Quotients. If $G \in \mathcal{E}$ and $N \trianglelefteq G$ is any normal subgroup, then $G/N \in \mathcal{E}$.

6. Extensions. If $N \trianglelefteq G$ is a normal subgroup, $N \in \mathcal{E}$ (as a group in its own right), and $G/N \in \mathcal{E}$, then $G \in \mathcal{E}$.

7. Directed unions. Let $I$ be an index set (living in the same size universe as $G$), and let $(H_i)_{i \in I}$ be a family of subgroups of $G$ such that

   • the family is directed: for every $i, j \in I$ there is $k \in I$ with $H_i \subseteq H_k$ and $H_j \subseteq H_k$; and
   • the subgroup generated by $\bigcup_{i \in I} H_i$ is all of $G$; and
   • every $H_i \in \mathcal{E}$ (each with its inherited group structure).

   Then $G \in \mathcal{E}$.

   Edge case: $I$ may be empty. Then directedness holds vacuously, the subgroup generated by the empty union is the trivial subgroup $\{1\}$, so the second condition is satisfiable only if $G$ is trivial. When $I$ is nonempty and the family is directed, the subgroup generated by the union is just the union itself.

In rule 2, "abelian" is the only content: there is no restriction to finitely generated or torsion-free abelian groups. In rules 4, 6 and 7, subgroups are regarded as groups with the multiplication restricted from $G$. In rules 5 and 6, the quotient carries the group structure induced from $G$ via the given normality.

Conclusion: "the packing property"

$G$ is said to have the packing property when:

For every finite subset $F \subseteq G$, there exist subsets $S, X \subseteq G$ such that

$$F \subseteq S, \qquad S \text{ is finite}, \qquad \text{and } (S, X) \text{ is a packing of } G,$$

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

$$S \times X \longrightarrow G, \qquad (s, x) \longmapsto s \cdot x$$

is a bijection from the Cartesian product $S \times X$ onto all of $G$. Spelled out fully, this is the conjunction of three conditions:

• (maps into) for all $s \in S$, $x \in X$: $s x \in G$ — automatically true;
• (injective on $S \times X$) for all $s, s' \in S$ and $x, x' \in X$: if $s x = s' x'$ then $s = s'$ and $x = x'$;
• (surjective onto $G$) every $g \in G$ can be written as $g = s x$ with $s \in S$ and $x \in X$.

Equivalently: every element of $G$ has exactly one factorization $g = s x$ with $s \in S$, $x \in X$; that is, the sets $sX$ for $s \in S$ partition $G$ (equivalently, the sets $Sx$ for $x \in X$ partition $G$).

Points to note in the quantifiers:

• $S$ and $X$ are arbitrary subsets of $G$, not required to be subgroups, and $X$ is not required to be finite. Since $G$ is nonempty (it contains $1$), the bijectivity condition forces both $S$ and $X$ to be nonempty.
• $F$ ranges over all finite subsets including $F = \varnothing$; for that $F$ the condition "$F \subseteq S$" is automatic.
• Only $S$ is required to contain $F$ and to be finite; there is no size bound on $S$ relative to $F$, and no requirement that $S$ or $X$ contain $1$.
• The choice of $S$ and $X$ may depend on $F$.

Summary in one sentence

For every group $G$: if for every non-identity $x \in G$ there is a normal subgroup $K$ with $x \notin K$ and $G/K$ elementary amenable (in the smallest-class sense of the seven rules above), then for every finite $F \subseteq G$ there are a finite set $S \supseteq F$ and a set $X \subseteq G$ such that every element of $G$ is uniquely a product $s x$ with $s \in S$, $x \in X$.