Read-back

The file contains one declaration, a theorem. Its proof is not part of what is being read back here; only the statement is rendered.

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The statement

Let $B$ be a group. ($B$ ranges over all groups, in any universe; neither the way it is supplied nor the way its group structure is supplied restricts it in any manner beyond its being a group.)

Assume:

1. $B$ is finitely generated as a group. Unfolded: there is a finite subset $S \subseteq B$ such that the smallest subgroup of $B$ containing $S$ is all of $B$. (The subgroup generated by a set is defined as the intersection of all subgroups containing it.)

2. $A$ is a subgroup of $B$. This is the one argument that is explicit; the theorem is quantified universally over it.

3. $A$ is normal in $B$. Unfolded: for every $n \in A$ and every $g \in B$, one has $g n g^{-1} \in A$.

4. $A$ is abelian. Unfolded: the multiplication of $B$, restricted to $A$ (where $A$ is regarded as a group in its own right, with the multiplication inherited from $B$), is commutative — that is, $xy = yx$ for all $x, y \in A$. Nothing is assumed about commutativity anywhere else in $B$.

5. The quotient group $B/A$ is polycyclic, in the sense spelled out in §"Polycyclic" below. Here $B/A$ is the set of left cosets of $A$ in $B$, carried to a group by hypothesis 3.

6. $B$ does not have exponential growth, in the sense spelled out in §"Exponential growth" below.

Then:

$B$ is polycyclic, in the same sense as in hypothesis 5.

Item 2 is a universally quantified variable; items 1, 3, 4, 5 and 6 are hypotheses — 1, 3 and 4 supplied as background structural assumptions and 5 and 6 as ordinary named ones. Logically all five hypotheses stand on equal footing, and the statement carries every one of them.

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Polycyclic

A group $G$ is polycyclic, in the sense used here, when:

There exist a natural number $t$ and a family of subgroups

$$A_0,\; A_1,\; \dots,\; A_t \;\le\; G$$

(that is, a function from the index set $\{0, 1, \dots, t\}$, which has $t+1$ elements, to the subgroups of $G$) such that

• $A_0 = G$ (the whole group),
• $A_t = \{1\}$ (the trivial subgroup),
• and for every index $i$ with $0 \le i \le t-1$, both of the following hold:
  • $A_{i+1} \le A_i$, and
  • the subgroup $A_{i+1} \cap A_i$, viewed as a subgroup of $A_i$, is normal in $A_i$, and the quotient group $A_i / (A_{i+1} \cap A_i)$ formed with respect to that normality is cyclic.

Several points of fine print:

• The subgroup appearing in the quotient is literally the preimage of $A_{i+1}$ under the inclusion $A_i \hookrightarrow G$, i.e. $\{x \in A_i : x \in A_{i+1}\} = A_{i+1} \cap A_i$. Because $A_{i+1} \le A_i$ is asserted alongside it, this is the same as $A_{i+1}$ itself regarded inside $A_i$; but the two conjuncts are asserted independently, and the intersection form is what the quotient is actually taken by.

• The normality assertion is part of the existential claim, not a side condition: the statement asserts that such a chain exists together with proofs that each step is normal in the previous one. Since normality is a proposition (proof-irrelevant), this is equivalent to the conjunction "$A_{i+1} \cap A_i$ is normal in $A_i$, and the resulting quotient is cyclic".

• Only subnormality is claimed. The $A_i$ are not required to be normal in $G$, only each one normal (after intersecting) in its predecessor.

• Cyclic here means: there exists an element $g$ of the quotient group such that the map $\mathbb{Z} \to A_i/(A_{i+1}\cap A_i)$, $k \mapsto g^k$, is surjective. This includes the trivial group (take $g = 1$), and allows either a finite cyclic group or an infinite cyclic group.

• Nothing forces the chain to descend strictly. Repetitions $A_{i+1} = A_i$ are permitted (the quotient is then trivial, hence cyclic), and the length $t$ is not required to be minimal or to bound anything.

• Degenerate case $t = 0$. Then the family has a single member $A_0$, the index set of steps is empty so no step conditions are imposed, and the two remaining conditions read $A_0 = G$ and $A_0 = \{1\}$. So "polycyclic with $t = 0$" says exactly that $G$ is the trivial group. This case is genuinely included, and the trivial group is polycyclic.

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Word balls and exponential growth

For a group $G$, a subset $S \subseteq G$ and a natural number $n$, the word ball $B_S(n)$ is

$$B_S(n) \;=\; \Bigl\{\, g \in G \;:\; \exists\, k \le n \text{ and } x_1,\dots,x_k \in G \text{ with } \bigl(x_j \in S \text{ or } x_j^{-1} \in S\bigr) \text{ for each } j, \text{ and } g = x_1 x_2 \cdots x_k \,\Bigr\}.$$

Fine print on this set:

• The word is a finite list of elements, read in order, and the product is taken in list order $x_1 x_2 \cdots x_k$ (the product is formed by folding from the right against the identity, so a list $[x_1,x_2,x_3]$ gives $x_1(x_2(x_3 \cdot 1))$ — by associativity, the product in the written order).

• The generating set $S$ is not assumed symmetric. Each letter is allowed to lie in $S$ or to have its inverse in $S$, so the ball is the ball with respect to $S \cup S^{-1}$.

• The length condition is $k \le n$, a closed ball. In particular the empty list ($k = 0$, empty product $= 1$) always qualifies, so $1 \in B_S(n)$ for every $n$, and $B_S(0) = \{1\}$ exactly. The balls are nested increasing in $n$.

A group $G$ has exponential growth, in the sense used here, when:

There exists a finite subset $S \subseteq G$ such that

• the subgroup generated by $S$ is all of $G$, and
• there exists a real number $c$ with $c > 1$ (strictly) such that

$$c^{\,n} \;\le\; \bigl|B_S(n)\bigr| \qquad \text{for every natural number } n,$$

the cardinality being taken as a natural number and then compared in $\mathbb{R}$.

Fine print:

• The inequality is demanded for all $n$, not merely for all sufficiently large $n$, and it is non-strict ($\le$). The $n = 0$ instance is $c^0 = 1 \le |B_S(0)| = 1$, so it is automatic and imposes nothing.

• The cardinality is the natural-number cardinality, which is $0$ by convention for an infinite set. Here no such junk value can arise: for a finite $S$ the ball $B_S(n)$ is a finite set, so the cardinality is the honest count.

• The quantifier over $S$ is existential: exponential growth asks only that some finite generating set exhibit the exponential lower bound.

Hypothesis 6 is the negation of this. Unfolded, it says:

For every finite subset $S \subseteq B$ whose generated subgroup is all of $B$, and for every real $c > 1$, there exists a natural number $n$ with

$$\bigl|B_S(n)\bigr| \;<\; c^{\,n}.$$

Note the strength of this by contrast with the existential in the positive form: the negation is a statement about every finite generating set of $B$, and for each it needs only a single witnessing radius $n$ where the bound fails.

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Satisfiability

None of the hypotheses is vacuous or impossible. They are simultaneously satisfiable — for instance by the trivial group $B$ with $A$ the trivial subgroup, or by $B = \mathbb{Z}$ with $A = \mathbb{Z}$ — so the theorem is not vacuously true for want of a model.

Nothing requires $A$ to be proper or nontrivial: $A = \{1\}$ and $A = B$ are both permitted by hypotheses 2–4, and in the latter case hypothesis 5 concerns the trivial quotient and hypothesis 4 says $B$ itself is abelian.

Nothing in the statement asserts a bound on the length $t$ of the resulting chain for $B$, nor any relation between it and the length of the chain furnished for $B/A$ by hypothesis 5.