We prove, that the torsion group $T$, which has elements of finite order, $T\subset G$, is a normal subgroup of $G$.
Definition: A subgroup T, is normal if $gTg^{-1}= T \ \forall g\in G$.\
UPDATE 2: By discussion below, I simplify the original approach to the following.
For a case of finite order, we have:
\begin{equation} (aga^{-1})^{n}=(a^{-n}a^{n}g^{n}a^{n}a^{-n})=((a^{-1}ga))^n \end{equation}
which applies for $g$:
\begin{equation} (gg^{-1})^n=(g^{-n}g^n)=(g^{-1}g)^n \end{equation}
Update 3: Trying with some examples n=(2,3)
\begin{equation} (aga^{-1})^2=aga^{-1}aga^{-1}= agega^{-1}=ag^2a^{-1} \end{equation}
\begin{equation} (aga^{-1})^3=(aga^{-1}aga^{-1})aga^{-1}= agega^{-1}e=ag^3a^{-1} \end{equation}
Is this fine? Thanks