Let $F_n = \langle u_1,...,u_n \rangle$ be the free group on $n$ generators. I found this post on showing that the conjugacy class of an non-identity element is infinite.

One suggestion is to show that $C_{F_n}(g)$ is an infinite cyclic group; but that doesn't seem helpful. The other suggestion is to use the fact that "elements of the free group are presented by free words without relations other than $a^{−1}a=1$ and $aa^{−1}=1$."; but that doesn't seem very rigorous. I am reading the wikipedia page on free groups and it doesn't seem to define them as groups without relations other than those relations which follow from group theory axioms. The definition is a little more complicated, albeit more precise, than that.

How does one rigorously prove that $F_n$ has the ICC property? One thing I tried was to exhibit an infinite collection of distinct cosets of $C_{F_n}(g)$, but my methods were not rigorous (e.g., my proofs were relying on the ostensibly imprecise fact that $F_n$ contains no relations other than those that follow from the group axioms).

  • $\begingroup$ Presumably, $n\ge2$. $\endgroup$ – Gerry Myerson May 4 at 0:08
  • $\begingroup$ @GerryMyerson Ah, yes. Thanks for pointing that out. $\endgroup$ – user193319 May 4 at 0:16
  • $\begingroup$ The reason showing that $C_{F_n}(g)$ is infinite cyclic solves the problem is that the number of elements in the conjugacy class is equal to the index of the centralizer, and if the centralizer is infinite cyclic, then it is of infinite index in the free group of rank at least 2. $\endgroup$ – Arturo Magidin May 4 at 5:46
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    $\begingroup$ Free groups as reduced words make this rather easy: pick an $x_i$ different from the first letter in $g$. Then each of $g$, $x_igx_i^{-1}$, $x_i^2gx_i^{-2},\ldots,x_i^mgx_i^{-m},\ldots$ is different, since there is no cancellation at the beginning of the word, and they all have different initial segments. (You just need to worry a bit if $g$ ends in $x_i$, but that case can be handled easily...) $\endgroup$ – Arturo Magidin May 4 at 5:51
  • $\begingroup$ To “rigorously prove” something about free groups, it would be good to know exactly what your working definition of a free group is. There are several (equivalent) ways of thinking about free groups, but the way to phrase a proof about free groups varies depending on which form you take (e..g, the Nielsen-Schreier theorem proof is very different if you think of free groups as fundamental groups of bouquets of circles than if you think of them as reduced words under concatenate-and-simplify). $\endgroup$ – Arturo Magidin May 4 at 17:50

A free group $F_S$ over a set $S$ is ICC unless $S$ has cardinal 1.

Clearly it's enough to do it for $S$ finite.

To prove that a group is ICC it's enough to embed it as a dense subgroup into a connected (or just with no proper open subgroup of finite index) Hausdorff topological group with trivial center. Indeed $F_S$ for $2\le |S|$ can be embedded in this way, for instance into $\mathrm{PSL}_2(\mathbf{R})$.

Of course whether this approach is helpful depends on your background.


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