What the statement asserts

Statement. Let $G$ be any group (in any universe; no topology, finiteness or countability is assumed). Suppose $G$ is exponentially bounded in the sense defined below. Then $G$ is amenable in the sense defined below.

The only data are the group $G$ with its group structure, and the single hypothesis that $G$ is exponentially bounded. The conclusion is an implication in one direction only.

The hypothesis: "exponentially bounded"

For a subset $S \subseteq G$ and a natural number $n \ge 0$, let the word ball $B_S(n)$ be the set of all $g \in G$ for which there is a finite sequence $x_1, \dots, x_k$ of elements of $G$ with

• $k \le n$,
• for each $i$, $x_i \in S$ or $x_i^{-1} \in S$ (that is, each factor lies in $S \cup S^{-1}$), and
• $g = x_1 x_2 \cdots x_k$.

The empty sequence ($k = 0$) is allowed and its product is the identity $1$, so $1 \in B_S(n)$ for every $n$, and $B_S(0) = \{1\}$. The set $S$ is not required to contain $1$ or to be closed under inverses; inverses of elements of $S$ are admitted as factors regardless.

$G$ is exponentially bounded if there exists a finite subset $S \subseteq G$ such that

1. $S$ generates $G$ as a group: the smallest subgroup of $G$ containing $S$ is all of $G$; and
2. for every real number $c > 1$ there is a natural number $N$ such that for every natural number $n \ge N$,

$$\#B_S(n) \;\le\; c^{\,n},$$

where $c^n$ is the ordinary real power and $\#B_S(n)$ is the number of elements of $B_S(n)$, read as a real number.

Remarks on this hypothesis:

• The count $\#B_S(n)$ is taken with the convention that an infinite set has count $0$. Since $S$ is finite, $B_S(n)$ is a set of products of at most $n$ factors from the finite set $S \cup S^{-1}$, so it is finite and the convention never actually applies here.
• The existence of a finite generating set is part of the hypothesis, so the hypothesis can only hold for finitely generated groups.
• Only one finite generating set $S$ is required to satisfy the growth bound (the quantifier on $S$ is existential, and it comes before the quantifier on $c$, so the same $S$ serves for every $c > 1$). The threshold $N$ may depend on $c$.
• Degenerate cases: for the trivial group, $S = \varnothing$ works (the subgroup generated by the empty set is $\{1\}$, which is all of $G$), and then $\#B_S(n) = 1 \le c^n$ for all $n$. For any finite group, taking $S = G$ gives $\#B_S(n) \le |G|$, which is eventually below $c^n$ for each $c>1$; so every finite group satisfies the hypothesis.

The conclusion: "amenable"

$G$ is amenable if there exists a function

$$m : \mathcal{P}(G) \to [0, \infty]$$

defined on every subset of $G$ (no measurability restriction), with values in the extended non-negative reals, such that:

1. Finite additivity: $m(\varnothing) = 0$, and for all subsets $A, B \subseteq G$ with $A \cap B = \varnothing$,

$$m(A \cup B) = m(A) + m(B),$$

where addition in $[0,\infty]$ has $\infty + x = x + \infty = \infty$. Only additivity over two disjoint sets is required (hence, by induction, over finitely many); nothing is required of countable unions. 2. Normalization: $m(G) = 1$. 3. Left invariance: for every $g \in G$ and every subset $A \subseteq G$,

$$m(gA) = m(A), \qquad\text{where } gA = \{\, g a : a \in A \,\}.$$

The translation is by left multiplication by $g$ (the group acting on itself on the left), not right multiplication and not conjugation.

Since $m(A) + m(G \setminus A) = m(G) = 1$ by condition 1 and 2, every value of such an $m$ lies in $[0, 1]$; the allowance of the value $\infty$ in the codomain is therefore never used by an $m$ that meets all three conditions.

Summary

For every group $G$: if there is a finite generating set $S$ of $G$ such that, for every real $c > 1$, the number of elements of $G$ expressible as a product of at most $n$ factors from $S \cup S^{-1}$ is at most $c^n$ for all sufficiently large $n$, then there exists a finitely additive, left-translation-invariant function $m$ on all subsets of $G$, with values in $[0,\infty]$, $m(\varnothing)=0$ and $m(G) = 1$.