Show that if $y$ has finite order, then $x^ny=yx^n$ for some $n\in\Bbb{N}^{\ast}$ and all $x$.

Question

Let $G$ be a group and $F$ the set of finite order elements in $G$. If $F$ is finite, prove that there exists $n\in \mathbb{N}^{\ast}$ such that $x^ny=yx^n$ for all $x\in G,y\in F$.

My progress so far:

Having $x\in G ,y\in F$, let $k=ord(y)$.

I found that $(xyx^{-1})^k=e$ and the function $f_x:F\to F,f_x(y)=xyx^{-1}$ is bijective. I do not know what can I do next. Can somebody help me, please? I am thinking about something with $|F|!$ since $f_x$ is a permutation.

Answer 1

Lemma: For every $y\in F$ there exists $n(y)\in\Bbb{N}^{\ast}$ such that $x^{n(y)}y=yx^{n(y)}$ for all $x\in G$.

Proof. Because $y\in F$ there exists $k\in\Bbb{N}^{\ast}$ such that $y^k=e$. Then for all $z\in G$ $$(zyz^{-1})^k=zy^kz^{-1}=zez^{-1}=e,$$ and so $zyz^{-1}\in F$. It follows that for all $x\in G$ and $a\in\Bbb{N}$ also $x^ayx^{-a}\in F$, by plugging in $z=x^a$. Because $F$ is finite, there exist $a,b\in\Bbb{N}$ with $b>a$ such that $$x^ayx^{-a}=x^byx^{-b} \qquad\text{ and hence }\qquad x^{b-a}y=yx^{b-a},$$ and so $n(y):=b-a$ has the desired property.$\hspace{10pt}\square$

Corollary: There exists $n\in\Bbb{N}^{\ast}$ such that $x^ny=yx^n$ for all $x\in G$ and all $y\in F$.

Proof. It follows from the lemma that $x^{mn(y)}=y^{mn(y)}$ for all $m\in\Bbb{N}$. Because $F$ is finite the number $$n:=\operatorname{lcm}\{n(y):\ y\in F\},$$ is well-defined, and has the desired property.$\hspace{10pt}\square$

Answer 2

Alternatively, for a more abstract proof, show that the map $$\varphi:\ G\ \longrightarrow\ \operatorname{Aut}(F):\ x\ \longmapsto\ f_x,$$ is a group homomorphism. You should first show that it is well-defined, i.e. that for every $x\in G$ the map $$f_x:\ F\ \longrightarrow\ F:\ y\ \longmapsto\ xyx^{-1},$$ is an automorphism of $F$. You have already shown that it is bijective; it remains to show that it is a group homomorphism.

Because $F$ is finite, also $\operatorname{Aut}(F)$ is finite. Let $n:=|\operatorname{Aut}(F)|$. Then for all $f\in\operatorname{Aut}(F)$ we have $f^n=\operatorname{id}_F$ which means that for all $x\in G$ we have $$f_{x^n}=\varphi(x^n)=\varphi(x)^n=(f_x)^n=\operatorname{id}_F,$$ and so for all $x\in G$ and all $y\in F$ you have $$x^nyx^{-n}=f_{x^n}(y)=\operatorname{id}_F(y)=y.$$