Read-back: the packing property passes to extensions

This is a rendering of what the declaration literally asserts. It is self-contained: every definition the statement depends on is spelled out below, and every hypothesis, including the ones the formal statement carries silently, is listed.

The setting

Throughout, $G$ is a group. No further structure is assumed on $G$: it may be finite or infinite, countable or uncountable, abelian or not; nothing is assumed about its cardinality or its universe level.

$N$ is a normal subgroup of $G$. Normality is taken in the usual sense: for every $n \in N$ and every $g \in G$, the conjugate $g\,n\,g^{-1}$ lies in $N$. The two extreme cases $N = \{1\}$ and $N = G$ are both allowed by the statement.

Two further groups are built from this data, and the statement treats each as a group in its own right:

• $N$ as a group. Its elements are the elements of $N$, and its multiplication is the multiplication of $G$ restricted to $N$ (so for $x, y \in N$, the product of $x$ and $y$ in the group $N$ is the element $xy$ of $G$, which lies in $N$).

• The quotient group $G/N$. Its elements are the left cosets $gN$ with $g \in G$; two elements $a, b \in G$ represent the same coset exactly when $a^{-1} b \in N$. Every element of $G/N$ is of the form $gN$ for some $g \in G$. The multiplication is $(aN)(bN) = (ab)N$, which is well defined because $N$ is normal.

Definition: packing

Let $H$ be any group and let $S, X \subseteq H$ be two subsets (any subsets: no finiteness, no closure property, no requirement that they contain the identity). The pair $(S, X)$ is called a packing of $H$ when the multiplication map

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

is a bijection from the cartesian product $S \times X$ onto the whole of $H$. Unwinding the bijection, this says exactly two things:

1. (Every element is reached.) For every $h \in H$ there exist $s \in S$ and $x \in X$ with $s\,x = h$.
2. (Reached only once.) If $s, s' \in S$ and $x, x' \in X$ satisfy $s\,x = s'\,x'$, then $s = s'$ and $x = x'$.

Together: every element of $H$ can be written as $s\,x$ with $s \in S$ and $x \in X$ in exactly one way. The order of the factors matters: it is $s$ on the left and $x$ on the right, and nothing is said about products $x\,s$.

An equivalent way to say the same thing: the left translates $sX$, for $s$ ranging over $S$, are pairwise disjoint and their union is all of $H$. (Disjointness of the translates gives $s = s'$ in condition 2, and then $x = x'$ follows by cancellation.)

Edge cases that follow directly from the definition:

• Since $H$ contains its identity element, condition 1 forces both $S$ and $X$ to be nonempty. Neither the empty set nor a pair with an empty member is ever a packing.
• The pair $S = \{1\}$, $X = H$ is a packing of every group $H$.
• $X$ is not required to be finite, to be a subgroup, or to be related to $S$ in any way beyond conditions 1 and 2.

Definition: the packing property

A group $H$ has the packing property when the following holds:

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

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

Points to note in this definition:

• "Finite" means: there is a natural number $n$ and a bijection between the set and $\{0, 1, \dots, n-1\}$. The empty set is finite.
• The finite set $F$ is required to be contained in $S$ (the left factor), not in $X$.
• $S$ must be finite; $X$ is unconstrained apart from $(S, X)$ being a packing. In particular $X$ may be infinite, and when $H$ is infinite and $S$ is finite, $X$ is necessarily infinite.
• The case $F = \varnothing$ is included. For it, $S = \{1\}$ and $X = H$ always work, so that instance of the condition is satisfied in every group.
• The quantifier is $\exists$, not "exists uniquely": there is no claim about how many such pairs $(S, X)$ exist.

When this definition is applied to the group $N$ (as a group in its own right), the sets $F$, $S$, $X$ are subsets of $N$, products are computed in $N$ (equivalently, in $G$), and the packing must cover all of $N$, not all of $G$. When it is applied to $G/N$, the sets are sets of cosets and products are coset products.

The statement

Let $G$ be a group and $N$ a normal subgroup of $G$. Assume:

• (H1) the group $N$ has the packing property, and
• (H2) the quotient group $G/N$ has the packing property.

Then $G$ has the packing property. Written out in full, the conclusion is:

For every finite subset $F \subseteq G$ there exist subsets $S, X \subseteq G$ with $F \subseteq S$, with $S$ finite, and such that every element of $G$ can be written as $s\,x$ with $s \in S$, $x \in X$ in exactly one way.

Complete list of what is quantified and assumed

• $G$: an arbitrary group (universally quantified; the group structure is part of the data).
• $N$: an arbitrary subgroup of $G$ (universally quantified).
• The hypothesis that $N$ is normal in $G$.
• Hypothesis (H1): $N$ has the packing property.
• Hypothesis (H2): $G/N$ has the packing property.

There are no other hypotheses. Nothing is assumed about finiteness, countability, finite generation, commutativity, or torsion of $G$, $N$, or $G/N$.

Degenerate instances covered by the statement

• $N = \{1\}$. Then $N$ is the trivial group, which has the packing property (for any finite $F \subseteq \{1\}$ take $S = X = \{1\}$), and $G/N$ is isomorphic to $G$ via $g \mapsto gN$.
• $N = G$. Then $G/N$ is the trivial group, which has the packing property, and hypothesis (H1) is the packing property for $G$ itself.
• $G$ finite. Then every subset is finite; for any $F$ one may take $S = G$ and $X = \{1\}$, so every finite group has the packing property, and each of (H1), (H2), and the conclusion is satisfied.

The statement is an implication, not an equivalence: it does not assert that the packing property of $G$ implies anything about $N$ or $G/N$.