A group $G$, acting on a set $A$, is a permutation group such that it has an element of order $30$. Prove that order of $A$ is greater than $10$.
Here, I supposed that, $|A|=6$ and wanted to make a contradiction. The possible order of group $G$ could be $30$, $60$, $120$, $240$, $360$ and $720$. And I supposed $|x|=30$ is that element in $G$. If $|G|=30$ so $G$ be an abelian group, it acts on $A$ transitively so $(G|A)$ is regular and then $|G|=|A|$ which is wrong. The same way could be carried out when $|G|=60$. If $|G|=360$, we would face to contradiction because $A_{6}$ is simple. Honestly, I could not go for the rest. Have I been on a right way? Thanks for any help.