Why isn't the growth of $\mathrm {Iso}(\mathbb R ^2)$ sub-exponential?

83 Views Asked by At

Given a finite subset $S$ of a group, denote by $\gamma_S(n)$ the number of distinct elements that are words of length $n$ in $S\cup S^{-1}$.

A group is of sub-exponential growth if for every finite subset $S$, the number $\gamma_S(n)$ satisfies $\sqrt[n]{\gamma_S(n)} \to 1$.

I have proved the false claim that the isometries of the plane are a group of sub-exponential growth.

This must be false, because a group of sub-exponential growth does not admit a paradox, while the group of isometries of the plane does admit a paraox. ( The Sierpinski-Mazurkiewicz paradox.)

Which of the following parts is my mistake?

  1. A semidirect product of two groups of sub-exponential growth is again sub-exponential.

(Proof: if $G=H \rtimes Q$, take a word $h_1q_1h_2q_2\ldots $ and make it into a word of the form $q_1\ldots q_n\tilde h_1\ldots \tilde h_n $ using the conjugations $h_iq_j=q_j\tilde h_i$.)

  1. The group of isometries is $\mathrm {Iso}(\mathbb R ^2) = O(2) \ltimes \mathbb R^2$.

  2. $O(2)=SO(2) \rtimes {\pm 1}$.

Alternative formulation It is known that the group $\mathrm {Iso}(\mathbb R ^2) = O(2) \ltimes \mathbb R^2$ contains a free semigroup of order $2$. Why don't claims 1,2,3 disprove this?

1

There are 1 best solutions below

0
On BEST ANSWER

The mistake is that $\tilde h_i$ does not depend only on $h_i$, but also on $q_j$. (This is unlike the case of isometries of $\mathbb R^1$.) Thus the set of possible $\tilde h_i$ is no longer finite.