# Finite order elements commutes with group

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.

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.