Read-back: the packing property for free groups

Setting

Let $A$ be an arbitrary set. No assumption whatever is placed on $A$: it may be empty, finite, or infinite of any cardinality, and it lives in an arbitrary (but fixed) set-theoretic universe. Nothing else is assumed — there is no decidability, nonemptiness, or ordering hypothesis on $A$.

Let $G = F(A)$ denote the free group on $A$, in the following concrete model. A word is a finite sequence of letters, each letter being a pair $(a, \varepsilon)$ with $a \in A$ and $\varepsilon$ one of two signs (the sign records whether the letter stands for $a$ or for $a^{-1}$). Two words are identified when one is obtained from the other by a finite sequence of insertions or deletions of an adjacent cancelling pair $(a,\varepsilon)(a,\bar\varepsilon)$, where $\bar\varepsilon$ is the opposite sign; the elements of $G$ are the resulting equivalence classes. The product $g \cdot h$ of two classes is the class of the concatenation of a word for $g$ followed by a word for $h$ (so the left factor's letters come first). The identity is the class of the empty word, and the inverse of a class is the class of the word read backwards with every sign flipped. This is a group, and it is this group structure that the statement refers to.

Two degenerate values of $A$ are included in the statement: if $A$ is empty then $G$ is the one-element group; if $A$ has exactly one element then $G$ is infinite cyclic.

Subsets below are arbitrary subsets of $G$ (no finiteness or other restriction unless stated). A subset is called finite when there is a natural number $n$ and a bijection between the subset and $\{0, 1, \dots, n-1\}$; in particular the empty set is finite ($n = 0$).

The assertion

For every set $A$ as above, the group $G = F(A)$ has the following property, which is what the statement calls the packing property:

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

1. $F \subseteq S$;
2. $S$ is finite;
3. the map

is a bijection from the Cartesian product $S \times X$ onto all of $G$.

Item 3 is, spelled out literally, the conjunction of three clauses:

• (3a) for every $s \in S$ and $x \in X$, the product $s \cdot x$ lies in $G$ (this clause is automatically true, since $G$ is the whole group; it is present only because the definition of "bijection onto a subset" in use includes a "maps into" clause);
• (3b) injectivity on $S \times X$: whenever $s_1, s_2 \in S$ and $x_1, x_2 \in X$ satisfy $s_1 \cdot x_1 = s_2 \cdot x_2$, one has $(s_1, x_1) = (s_2, x_2)$, i.e. $s_1 = s_2$ and $x_1 = x_2$;
• (3c) surjectivity onto $G$: for every $g \in G$ there exist $s \in S$ and $x \in X$ with $s \cdot x = g$.

Equivalently, (3) says that every element of $G$ can be written in exactly one way as a product $s \cdot x$ with $s \in S$ and $x \in X$, the factor from $S$ on the left and the factor from $X$ on the right. In yet other words, $G$ is the disjoint union of the right translates $S x = \{ s x : s \in S\}$ over $x \in X$, with distinct $x \in X$ giving disjoint translates (each translate $S x$ is in bijection with $S$ via $s \mapsto s x$, which is automatic in any group).

The order of quantifiers is: $A$ is universally quantified first; then $F$ is universally quantified (ranging over finite subsets of $G$); then $S$ and $X$ are existentially quantified and may depend on $F$ (and on $A$). Existence only is asserted: no uniqueness of $S$ or $X$ is claimed.

What the quantifiers include and what is not required

• $F$ may be empty. The conclusion is then still required: some finite $S$ (any finite $S$ satisfies $\varnothing \subseteq S$) and some $X$ must exist with $\mu$ bijective.
• $F$ may contain the identity; it is not assumed to be a subgroup, to be symmetric, or to be closed under anything.
• $X$ is not required to be finite; it is an arbitrary subset of $G$, and no bound on its size is asserted.
• $S$ is not required to be minimal, symmetric, or to contain the identity; it is only required to be a finite superset of $F$ for which (3) holds.
• Clause (3c) applied to $g =$ the identity element (which always exists in $G$) forces both $S$ and $X$ to be nonempty; so although the statement does not say "nonempty" anywhere, the existential quantifier can only be satisfied by nonempty $S$ and $X$.
• Nothing is asserted about the cardinality of $S$ relative to $F$, about $X$ being a subgroup or a transversal of anything, about left translates $x S$, or about products in the other order $x \cdot s$.
• The statement concerns the one concrete group $F(A)$ described above; it says nothing directly about other groups isomorphic to it, nor about subgroups or quotients of $F(A)$.