Show that an abelian group $G$ of order $100$ cannot act faithfully on a set with $13$ elements.
I am given the hint: Consider the $5$-Sylow subgroup $H$ of $G$ and determine its centralizer in $S_{13}$.
$|H|=25$ so $|\phi(H)|=25$. Now $\phi(H)$ is abelian and so its centralizer contains at least $25$ elements. I am not sure what to do next. I am not really sure how the centralizer is helpful here.