Every proper subgroup of a simple group containing an order $45$ element has an index of at least $14$.
So far I've supposed that $G$ is a simple group with an element of order $45$. Then $45$ divides the order of $G$. Also, suppose there's a proper subgroup $N$ with index $n$. Then then there's a mapping $\phi:G\rightarrow S_n$. Note that $G$ is simple so the kernel is trivial and this map is injective. Thus $|G|=|\text{Im}(\phi)|$. Since the image is a subgroup of $S_n$, $|\text{Im}(\phi)|$ divides $|S_n|=n!$. Ultimately, $45$ divides $n!$, but I think this shows that $n\geq 6$, not $14$. How can I make the jump to $14$? Any suggestions?