Suppose that $G$ is finite and that $g\in G$ . If $h\in G$, show that $o(hgh^{-1})=o(g)$.
Solution:
We have that $(hgh^{-1})^{o(g)}=hgh^{-1}hgh^{-1}...hgh^{-1}=hg^{o(g)}h^{-1}=h1_Gh^{-1}=1_G$, so $hgh^{-1}$ has finite order and $o(hgh^{-1}) \leq o(g)$.
Now, what I don't understand is "A similar argument shows that $$ g^{o(hgh^{−1})} =h^{−1}(hgh^{−1})^{o(hgh^{−1})})h = 1_G ."$$ Which shows that $o(g) \leq o(hgh^{-1})$ and then $o(g) = o(hgh^{-1})$
I don't understand how to get $ g^{o(hgh^{−1})} =h^{−1}(hgh^{−1})^{o(hgh^{−1})})h = 1_G $, what argument do you use to get this?
