In a free group $F_n$, $n\geq 2$, all conjugacy classes except that of the identity are infinite

Question

Let $F_n$ be a free group on n generators, $n\geq 2$. Studying a paper, I came across this sentence:

In $F_n$, all conjugacy classes except that of the identity are infinite

while I do not know how he claims it.

Please help me. Thanks in advance.

Answer 1

What you need to know is that elements of the free group are presented by free words without relations other than $a^{-1} a= 1$ and $a a^{-1}=1$.

Say you have two generators $a,b$, than a typical words look like this: $$x=a b a^{-1} b b b a^{-1} b^{-1} a.$$ This word cannot be further simplified (reduced).

Now it is plausible that the words $$A:=a x a^{-1}, B:=ab x b^{-1} a^{-1},....$$ are all different, because otherwise the "difference" $A B^{-1}$ of two such words $A,B$ could be reduced to the empty words = the identity element.

Try it by examples, than you see how it goes.

Answer 2

If $x\in F_n$, then the conjugacy class of $x$ is in bijection with the set of cosets of $C_{F_n}(x)$ in $F_n$ where $C_{F_n}(x)$ denotes the centralizer of $x$.

Exercise: $C_{F_n}(x)$ is infinite cyclic for all nonidentity $x\in F_n$. (Hint: subgroups of free groups are free.)

Hope this helps!