Read-back: the packing property descends from quotients that separate finite sets

Setting

Let $G$ be a group. Nothing else is assumed about $G$: it may be finite or infinite, abelian or not, trivial or not. (The group is allowed to live in any set-theoretic universe; this has no mathematical content.)

All of the notions below are defined in the accompanying definition files rather than being standard library notions, so each is spelled out in full.

The notions used

Packings. For two subsets $S, X \subseteq G$, say that $(S, X)$ is a packing of $G$ when 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 $G$. Unfolded, this is the conjunction of three conditions:

1. (maps into $G$) for every $(s,x) \in S \times X$, the product $s x$ lies in $G$ — automatically true;
2. (injective on $S \times X$) whenever $s, s' \in S$ and $x, x' \in X$ satisfy $s x = s' x'$, one has $(s, x) = (s', x')$, i.e. $s = s'$ and $x = x'$;
3. (onto $G$) every $g \in G$ can be written as $g = s x$ for some $s \in S$ and some $x \in X$.

Equivalently: every element of $G$ has exactly one factorization $g = s x$ with $s \in S$ and $x \in X$. The order matters: the element of $S$ is the left factor and the element of $X$ is the right factor. Neither $S$ nor $X$ is required by the definition to be finite, nonempty, a subgroup, or anything else; the definition is purely the bijectivity of the multiplication map. (Since $G$ contains its identity element, condition 3 forces both $S$ and $X$ to be nonempty whenever a packing exists.)

The packing property. Say that $G$ has the packing property when: for every finite subset $F \subseteq G$ there exist subsets $S, X \subseteq G$ such that

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

Here "finite" means finite in the usual sense (in bijection with $\{0, 1, \dots, n-1\}$ for some natural number $n$, with $n = 0$ allowed). Only $S$ is required to be finite; $X$ may be infinite. The subset $F = \varnothing$ is included in the quantifier, so having the packing property entails in particular that some finite $S$ and some $X$ form a packing of $G$.

Normal subgroups. A subgroup $K \le G$ is normal when for every $n \in K$ and every $g \in G$, the conjugate $g\,n\,g^{-1}$ lies in $K$.

Quotient groups and the canonical map. For a subgroup $K \le G$, the quotient $G/K$ is the set of left cosets $gK$, i.e. the set of equivalence classes of $G$ under the relation $a \sim b \iff a^{-1} b \in K$. When $K$ is normal (as it is throughout the statement below), $G/K$ carries the usual quotient group structure, $(aK)(bK) = (ab)K$, and it is this group structure with respect to which "$G/K$ has the packing property" is asserted. The canonical map is

$$\pi_K : G \longrightarrow G/K, \qquad g \longmapsto gK,$$

and $\pi_K(a) = \pi_K(b)$ holds exactly when $a^{-1} b \in K$.

Injective on a subset. A map $f$ is injective on a subset $F$ of its domain when for all $a, b \in F$, $f(a) = f(b)$ implies $a = b$. Nothing is said about points outside $F$. For the canonical map this reads: for all $a, b \in F$, if $a^{-1} b \in K$ then $a = b$; equivalently, no two distinct elements of $F$ lie in the same coset of $K$.

The statement

Hypothesis. For every finite subset $F \subseteq G$ there exists a normal subgroup $K \trianglelefteq G$ (allowed to depend on $F$) such that both of the following hold:

• the quotient group $G/K$ has the packing property, i.e. for every finite subset $\bar F \subseteq G/K$ there are subsets $\bar S, \bar X \subseteq G/K$ with $\bar F \subseteq \bar S$, $\bar S$ finite, and $(\bar S, \bar X)$ a packing of $G/K$; and
• the canonical map $\pi_K : G \to G/K$ is injective on $F$: for all $a, b \in F$, $a^{-1} b \in K$ implies $a = b$.

Conclusion. $G$ has the packing property: for every finite subset $F \subseteq G$ there exist $S, X \subseteq G$ with $F \subseteq S$, $S$ finite, and every element of $G$ uniquely of the form $s x$ with $s \in S$, $x \in X$.

The statement is an implication (hypothesis $\Rightarrow$ conclusion), not an equivalence.

Edge cases and fine print

• Dependence of $K$ on $F$. The hypothesis is "for each finite $F$ there is some normal $K$"; a single $K$ working for all $F$ at once is not required, and nothing relates the subgroups chosen for different $F$.
• $F = \varnothing$. The empty set is a finite subset, so the hypothesis includes: there is some normal subgroup $K$ with $G/K$ having the packing property (the injectivity condition is vacuous for $F = \varnothing$).
• Singletons. For $F = \{a\}$ the injectivity condition is again automatic, so the hypothesis for singletons also only asks for a normal $K$ with $G/K$ having the packing property.
• Extreme choices of $K$. The trivial subgroup $K = \{1\}$ is normal, $\pi_K$ is injective on every subset, and $G/\{1\}$ is a group isomorphic to $G$. The whole group $K = G$ is normal with $G/G$ the one-element group, and $\pi_G$ is injective on $F$ exactly when $F$ has at most one element. Both choices fall within the quantifier "there exists a normal subgroup $K$".
• The packing property of the quotient is quantified over subsets of $G/K$, not over subsets of $G$: the sets $\bar F, \bar S, \bar X$ in the first bullet of the hypothesis are sets of cosets.
• Finiteness. In both the hypothesis and the conclusion, finiteness is demanded only of $F$ (given) and $S$ (to be found); $X$ is unconstrained.
• No uniqueness or bound. The conclusion asserts existence of some $S$ and $X$; it does not say $S$ or $X$ is unique, does not bound the size of $S$ in terms of $F$, and does not relate the $S$, $X$ found for $G$ to any packing of a quotient.