Read-back: fg_lcsFactor_and_growth_bounds_of_isNilpotent

The objects the statement is built from

Throughout, $\Gamma$ is a group. Three constructions have to be spelled out before the assertion can be stated; each is expanded here exactly as the formal text defines it.

Commutators and the lower central series. For elements, $[p,q] := p\,q\,p^{-1}q^{-1}$. For two subgroups $H,K \le \Gamma$, $[H,K]$ denotes the subgroup generated by the set $\{\,[p,q] : p \in H,\ q \in K\,\}$ (not that set itself). The series used here is indexed from $0$ and starts at the whole group:

$$\Gamma_0 = \Gamma, \qquad \Gamma_{k+1} = [\Gamma_k, \Gamma].$$

So $\Gamma_1$ is the commutator subgroup $[\Gamma,\Gamma]$. These subgroups are nested decreasingly, $\Gamma_{k+1} \le \Gamma_k$.

The $k$-th factor group. For each natural number $k$, write $\Gamma_k^{\mathrm{ab}}$ for the abelianization of $\Gamma_k$, i.e. the quotient of $\Gamma_k$ by the subgroup generated by all commutators $[a,b]$ with $a,b \in \Gamma_k$, and write $\pi_k : \Gamma_k \to \Gamma_k^{\mathrm{ab}}$ for the canonical projection. The object of interest is

$$Q_k := \Gamma_k^{\mathrm{ab}} \big/ \pi_k\!\left(\Gamma_{k+1} \cap \Gamma_k\right).$$

Two points of literal detail. First, the subgroup being divided out is described in the formal text as "$\Gamma_{k+1}$ viewed inside $\Gamma_k$", which by definition is the intersection $\Gamma_{k+1}\cap\Gamma_k$ regarded as a subgroup of $\Gamma_k$; because the series is decreasing this intersection is $\Gamma_{k+1}$ itself, so $Q_k$ is $\Gamma_k^{\mathrm{ab}}$ modulo the image of $\Gamma_{k+1}$. Second, the quotient is formed as a group quotient, which is legitimate for every subgroup here because $\Gamma_k^{\mathrm{ab}}$ is abelian and hence all of its subgroups are normal. $Q_k$ is again an abelian group.

At $k=0$ this reads: the abelianization of $\Gamma$ modulo the image of $[\Gamma,\Gamma]$.

Rank of the $k$-th factor. Let $A_k$ be the abelian group $Q_k$ written additively, viewed as a $\mathbb{Z}$-module in the only way possible (the module structure coming from the additive group structure). Put

$$r_k := \operatorname{rank}_{\mathbb{Z}} A_k,$$

where this rank is defined as follows: take the supremum of the cardinalities of all $\mathbb{Z}$-linearly independent subsets of $A_k$, and convert it to a natural number, with the convention that the value is $0$ if that supremum is infinite. In particular $r_k = 0$ both when $A_k$ has infinite rank and when $A_k$ is a torsion group (over $\mathbb{Z}$, rank $0$ is equivalent to being torsion). For a finitely generated abelian group this $r_k$ is the ordinary torsion-free rank, i.e. the number of infinite cyclic summands. Nothing in the definition of $r_k$ presupposes that $Q_k$ is finitely generated.

Nilpotency class. Let $Z_0 = 1$ and $Z_{n+1} = \{x \in \Gamma : [x,y] \in Z_n \text{ for all } y \in \Gamma\}$ be the upper central series, and let $c$ be the least $n$ with $Z_n = \Gamma$ (equivalently, the least $n$ with $\Gamma_n = 1$); by convention $c = 0$ if no such $n$ exists, though under the hypotheses below one does. Define two natural numbers

$$d_1 := \sum_{k=0}^{c-1} (k+1)\, r_k, \qquad\qquad d_2 := \sum_{k=0}^{c-1} 2^{k}\, r_k,$$

both sums being over the $c$ indices $k = 0, 1, \dots, c-1$ (empty, hence $0$, when $c=0$).

Growth function. For a finite subset $S \subseteq \Gamma$ and a natural number $m$, let

$$\beta_S(m) := \#\left\{\, g \in \Gamma \;\middle|\; \begin{array}{l}\text{there is a finite list } (x_1,\dots,x_\ell) \text{ of elements of } \Gamma \text{ with } \ell \le m,\\ \text{with } x_i \in S \text{ or } x_i^{-1} \in S \text{ for each } i, \text{ and } x_1x_2\cdots x_\ell = g\end{array} \right\}.$$

The product of the empty list is the identity, so the identity always belongs to this set, and $\beta_S(m) \ge 1$ for every $m$ including $m=0$. The condition is $\ell \le m$, not $\ell = m$, so this is a closed ball of radius $m$ and the sets increase with $m$. The generators are used together with their inverses ($x_i \in S \cup S^{-1}$), and repetitions in the list are allowed. The counting is by a cardinality function that returns $0$ for an infinite set; that convention is not triggered here, since the set is finite whenever $S$ is finite.

What the theorem asserts

Let $\Gamma$ be any group (in any universe) which is

• finitely generated — some finite subset of $\Gamma$ generates $\Gamma$; and
• nilpotent — some term of its upper central series is all of $\Gamma$.

Let $S$ be a finite subset of $\Gamma$ and assume

• the subgroup generated by $S$ (the smallest subgroup of $\Gamma$ containing $S$) is all of $\Gamma$.

(The last assumption already exhibits a finite generating set, so the finite-generation assumption is separately listed but not independent of it. Both are present.)

Then both of the following hold.

(A) For every natural number $k$, the group $Q_k$ is finitely generated, i.e. some finite subset of $Q_k$ generates it. This is asserted for all $k \in \mathbb{N}$ without restriction — including $k = 0$ and including every $k$ beyond the nilpotency class, where $\Gamma_k$ is trivial and so is $Q_k$.

(B) There exist real numbers $c_1$ and $c_2$ such that

$$0 < c_1, \qquad c_1 \le c_2,$$

and such that for every natural number $m$ with $m \ge 1$, both inequalities

$$c_1 \, m^{\,d_1} \;\le\; \beta_S(m) \qquad\text{and}\qquad \beta_S(m) \;\le\; c_2 \, m^{\,d_2}$$

hold, where $\beta_S(m)$ is read as a real number, $m$ is the natural number $m$ read as a real number, and the powers are ordinary integer powers of a real number with exponents the natural numbers $d_1$ and $d_2$ defined above (so $m^0 = 1$).

Precision notes on (B)

• The two constants are quantified before $m$: a single pair $(c_1,c_2)$ must work for all $m \ge 1$ simultaneously. They may depend on $\Gamma$ and on $S$.
• $m = 0$ is excluded from the inequalities; nothing is claimed there.
• The inequalities are non-strict ($\le$) in both directions. The strict inequality in the statement is only $0 < c_1$; there is no requirement that $c_2$ be strictly positive beyond what $c_1 \le c_2$ and $0 < c_1$ give, and no upper bound on $c_2$.
• The lower bound uses the exponent $d_1$ and the upper bound uses the exponent $d_2$. The statement asserts no relation between $d_1$ and $d_2$ — not that they are equal, and not that $d_1 \le d_2$.
• The existence claim is plain existence, not uniqueness, and the exponents $d_1, d_2$ are fixed quantities determined by $\Gamma$, not existentially quantified.
• The growth function is taken with respect to the given generating set $S$; no claim is made about other generating sets.
• Degenerate case: if $\Gamma$ is trivial then $c = 0$, so $d_1 = d_2 = 0$ (empty sums) and $\beta_S(m) = 1$ for all $m$; the assertion reduces to the existence of $0 < c_1 \le c_2$ with $c_1 \le 1 \le c_2$.
• Degenerate case: if every $r_k$ with $k < c$ is $0$ (for instance if all the factors $Q_k$ are finite), then again $d_1 = d_2 = 0$ and the assertion reduces to a two-sided bound of $\beta_S(m)$ by constants for all $m \ge 1$.
• $S$ is allowed to be empty, but then the generation hypothesis forces $\Gamma$ to be trivial; the hypotheses are not vacuous — the trivial group satisfies all of them.
• (A) and (B) are joined by "and": the theorem asserts the conjunction, so both parts are claimed under the same hypotheses.