Is the set of periodic elements of a group a subgroup, if the commutator subgroup is periodic? Let $G$ a multiplicative group such that the commutator subgroup of $G$ (the subgroup generated by the set of all commutators of $G$) is periodic.
I’m wondering if under the previous hypothesis, the set of the periodic elements of $G$ is a subgroup of $G$.
I denote the commutator subgroup of $G$ with $G’$, and the set of periodic elements of $G$ with $T$.

I tried to solve my doubt knowing that $G’$ is a normal subgroup of $G$, but I don’t know how I should continue.
Thanks in advance,
Antonino.
 A: First, recall these three important statements.

*

*If $H$ is an abelian group and $T_H$ is the subset of $H$ made of periodic elements, then $T_H$ is a subgroup of $H$.

*For any group $H$, $1\in T_H$ and $x\in T_H\iff x^{-1}\in T_H$. Thus, to be $T_H$  a subgroup of $H$, we have to check only whether $T_H$ is closed under the multiplication of $H$.

*Even if $G$ is not abelian, the quotient group $G/G'$ is abelian. Indeed, for any $x,y\in G$ we have $xy=yx[x,y]$, so $$xG'yG'=xyG'=yx[x,y]G'=yxG'=yG'xG'.$$ This implies that $T_{G/G'}$ is a subgroup of $G/G'$ because of 1).

Let now $x,y\in T_G$. Note that $xG'$ and $yG'$ are periodic elements of $G/G'$. We deduce from facts 1) and 3) that $(x G')(yG')=xyG'$ is also a periodic element. Thus there exists $n\in \mathbb N$ such that $(xyG')^n=G'$, i.e. $(x y)^n\in G'$.
By the hypothesis, $G'$ is periodic, so there exists $m\in \mathbb N$ such that $[(x y)^{n}]^m=1$. We have proved that $xy$ is periodic, so $xy\in T_G$. Since $x,y\in T_G \Rightarrow xy\in T_G$, we conclude that $T_G$ is stable under the multiplication of $G$.
