$S_{\infty}$ and $\mathbb{F}_{n}$ for $n \geq 2$ are ICC groups.

Question

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Recall that the infinite symmetric group can be described in the following way: $S_{\infty}=\bigcup_{n \in \mathbb{N}}{S_n}$. Moreover, $\sigma \in S_{\infty}$ if then $\exists N \in \mathbb{N}$ such that $\forall n > N$ $\sigma(n)=n$. Prove that $S_{\infty}$ has ICC property.

Also, prove that for $n \geq 2$, $\mathbb{F}_{n}$ is an ICC group.

Answer 1

For $S_{\infty}$ we will prove this in the following way: let $e \neq \sigma \in S_n \subset S_{\infty}$ such that $\sigma(i) \neq i$. For $j > n$ let $s_j$ be a permutation swapping i and j. We will prove that $|\{s_j \sigma s_j^{-1}: j > n \}|=\infty$. Since $\{s_j \sigma s_j^{-1}: j > n \} \subset \{g \sigma g^{-1} : g \in S_{\infty}\}$.

Say $\sigma(i)=k$ where $k \leq n$. Then we have that $\sigma s_j^{-1}(i)=k$ and $\sigma s_j^{-1}(j)=i$ and $\forall N > n$ $\sigma s_j^{-1}(N)=N$. Then consider $s_j \sigma s_j^{-1}(i)=i$ and $s_j \sigma s_j^{-1}(j)=k$. In other words as we let j vary for all $j > n$ we end up with distinct permutations in the conjugacy class sending j to k. Hence, we conclude that $|\{s_j \sigma s_j^{-1} : j > n\}|=\infty$ and thus $S_{\infty}$ is ICC.

Recall that $\mathbb{F}_2 = \langle a,b : \text{ a and b have no relations.} \rangle$. We have 4 options on the how reduced words end in the group; namely with $a^{k},b^{k},a^{-k}$, and $b^{-k}$ where $k \geq 1$. For example say that $w(a,b)$ ends with $a^{k}$ then it is not hard to see that conjugating by $g=b^{n}a^{k}$ for $n \geq 1$ gives distinct reduced words ending in $b^{-n}$. If our reduced words are of the form $w(a,b)=a^{k},b^{k},a^{-k},b^{-k}$ the same trick will work. Thus, we conclude $\mathbb{F}_2$ is ICC. This approach can be generalized for $\mathbb{F}_n$ where $n \geq 2$ and noting that we have 2n options for how a word can end in $\mathbb{F}_n$. Thus, we conclude that in general, $\mathbb{F}_n$ is ICC.

Answer 2

For the first question, note that the fixed point set $\mathbb{N}^g = \{n\in \mathbb{N}:\, gn = n\}$ has $\mathbb{N}^{g^h} = h^{-1}(\mathbb{N}^g)$ for $g, h\in S_\infty$.

For the second, fix $g\not =1$ in $F_n$ with $n \geq 2$, and choose $h\in F_n$ with $h\notin\langle{g}\rangle$ and $g\not\in \langle{h}\rangle$. (To prove that such an $h$ exists, use word-length to show that the latter condition holds for only finitely many $h$.) If the set $\{g^{h^n}:\,n \geq 0\}$ is finite, then $g^{h^n} = g$ for some $n$; that is, $[h^n, g] = 1$. Hence $\langle{h^n, g}\rangle$ is abelian, contradicting the fact that any subgroup of a free group is itself free.