Let $G$ be a group and $H$ be a subgroup with $(G:H)=m$. Let $X$ be set of all left cosets of $H$ in $G$.
Define $L_g$ to be a bijection such that $L_g(qH)=gqH$ where $qH$ and $gqH$ are left cosets.
Show the order of $L_g$ as an element of the symetric group on $X$ is a divisor of the order of $g$ in $G$.
So far I just get that the two orders are the same. Suppose the order of $L_g$ is $n$. Then taking $L_g$ on $qH$ $n$ times yields $qH=(g^n)qH$ or $q=(g^n)q$ or $i=g^n$.
Am I doing it right? Or what do you suggest?