Read-back: the packing property passes to a directed union of subgroups

Setting and quantified data

The statement is universally quantified over the following data.

• A group $G$. Nothing further is assumed about $G$: it may be finite or infinite, abelian or not, trivial or not. Its group operation is written multiplicatively below.
• An index type $I$. Nothing is assumed about $I$: it may be empty, finite, countable, or of any cardinality, and it carries no order of its own. (Formally, $G$ and $I$ may live in different set-theoretic universes; no relation between their sizes is imposed.)
• A family $(H_i)_{i \in I}$ of subgroups of $G$, one for each index $i \in I$. The subgroups need not be distinct, proper, nontrivial, or normal.

Throughout, "$H_i \le H_k$" between subgroups means inclusion of their underlying sets: every element of $H_i$ is an element of $H_k$.

The packing property (a definition used in the statement)

For any group $K$, say that a pair of subsets $S, X \subseteq K$ is a packing of $K$ when the multiplication map

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

is a bijection from $S \times X$ onto all of $K$. Spelled out, this means all three of:

1. (Maps into $K$.) For every $s \in S$ and $x \in X$, the product $s x$ lies in $K$. This condition is automatic, since $K$ is closed under multiplication.
2. (Injective on $S \times X$.) Whenever $s, s' \in S$ and $x, x' \in X$ satisfy $s x = s' x'$, one has $s = s'$ and $x = x'$ (that is, the pairs $(s, x)$ and $(s', x')$ are equal).
3. (Surjective onto $K$.) Every element $g \in K$ can be written as $g = s x$ for some $s \in S$ and $x \in X$.

Equivalently, $K$ is the disjoint union of the left translates $s X$ over $s \in S$, and every element of $K$ has exactly one representation as a product $s x$ with $s \in S$, $x \in X$. Note the order of the factors: $S$ is on the left and $X$ on the right. Neither $S$ nor $X$ is required to be finite, to be a subgroup, to contain the identity, or to be nonempty by this definition alone (although if $K$ is nonempty, surjectivity forces both to be nonempty).

Say that a group $K$ has the packing property when: for every finite subset $F \subseteq K$, there exist subsets $S, X \subseteq K$ such that

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

Here "finite" means finite as a set (in bijection with $\{0, 1, \dots, n-1\}$ for some natural number $n$, with $n = 0$ allowed). The subset $F$ ranges over all finite subsets of $K$, including the empty set and the singleton $\{1\}$. The set $X$ is not required to be finite. The quantifier is $\exists$, not $\exists!$: no uniqueness of $S$ or $X$ is asserted.

When this definition is applied to a subgroup $H_i$ of $G$ regarded as a group in its own right (with the multiplication inherited from $G$), it is entirely internal to $H_i$: the finite set $F$ is a subset of $H_i$, the sets $S$ and $X$ are subsets of $H_i$, and the bijection is required to be onto $H_i$ (not onto $G$).

Hypotheses

(H1) Directedness. For every pair of indices $i, j \in I$ there exists an index $k \in I$ with

$$H_i \le H_k \quad \text{and} \quad H_j \le H_k .$$

Since $I$ carries no order, nothing is said about how $k$ relates to $i$ and $j$ as indices; only the inclusions of subgroups matter. The index $k$ may equal $i$ or $j$. The family is not assumed to be a chain, and $I$ is not assumed to be nonempty. Degenerate cases: if $I$ is empty, (H1) holds vacuously; if $I$ has exactly one element, (H1) holds automatically (take $k$ to be that element).

(H2) The family generates $G$. The supremum of the family in the lattice of subgroups of $G$, i.e. the smallest subgroup of $G$ containing every $H_i$ (equivalently, the subgroup generated by $\bigcup_{i \in I} H_i$), is the whole group $G$:

$$\bigvee_{i \in I} H_i \;=\; G .$$

Two remarks on what this hypothesis amounts to:

• If $I$ is empty, the supremum of the empty family is the trivial subgroup $\{1\}$, so (H2) asserts that $\{1\} = G$, i.e. that $G$ is the one-element group. In that case hypothesis (H3) below is also vacuous, and the conclusion concerns the one-element group.
• If $I$ is nonempty and (H1) holds, then the union $\bigcup_{i \in I} H_i$ is itself a subgroup, and (H2) is equivalent to the statement that every element of $G$ belongs to $H_i$ for at least one $i \in I$.

(H3) Each member has the packing property. For every index $i \in I$, the group $H_i$ (with the multiplication inherited from $G$) has the packing property, in the sense defined above: for every finite subset $F \subseteq H_i$ there exist $S, X \subseteq H_i$ with $F \subseteq S$, $S$ finite, and $(S, X)$ a packing of $H_i$. If $I$ is empty this hypothesis is vacuous.

Conclusion

Under (H1), (H2), and (H3), the group $G$ has the packing property: for every finite subset $F \subseteq G$ there exist subsets $S, X \subseteq G$ with $F \subseteq S$, $S$ finite, and such that $(s, x) \mapsto s x$ is a bijection from $S \times X$ onto $G$.

What is quantified, in one line

For every group $G$, every index type $I$, and every family $(H_i)_{i \in I}$ of subgroups of $G$: if (H1) the family is directed under inclusion, (H2) the subgroup generated by all the $H_i$ is $G$, and (H3) every $H_i$ has the packing property, then $G$ has the packing property.