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Suppose group $G$ has only finitely many elements with finite order, call this torsion subset $H$. Then there exists $n$ such that $a^n b=ba^n$ for all $a\in G$ and $b\in H$.

Question is how to show the above statement.

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The conjugacy action $x\mapsto gxg^{-1}$ acts on $H$ defines a group homomorphism from $G$ to the symmetric group of degree $k$, where $k$ is the cardinality of $H$, i.e., $$ \varphi = g\mapsto(x\mapsto gxg^{-1}):G\to S_k. $$ Then let $n=|S_k|=k!$, we have $\varphi(g^n)=(\varphi(g))^n=1$ by Lagrange's theorem, where $1$ denotes the unit element in $S_k$ (The identity permutation). Hence $g^n h g^{-n} = \varphi(g^n)(h) = 1(h) = h$, completes the proof.

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