I came across with the following question and I have no idea how to approach it.
Let $G\leq S_{13}$ be a subgroup with an element of order $40$.
Prove that $G$ has a normal proper non-trivial subgroup.
I thought of using the sylow theorem but I don't know order $G$. How should I prove it?
Also, if someone knows from which book the question was taken it will be great (Would like to practice with similar questions).
The only permutations in $S_{13}$ that have order $40$ are the product of a $5$-cycle and a disjoint $8$-cycle. Let's call this element $x$. Then $x^2 \in A_{13}$ but $x \notin A_{13}$, and $G \cap A_{13}$ is a non-trivial proper normal subgroup of $G$.
Edited to add a discussion of motivation: By the way, this isn't something that I had "in the can," so to speak. But since we know that the only non-trivial proper normal subgroup of $S_{13}$ is $A_{13}$, it seemed natural to actually consider what $G$ has to look like when inserted into $S_{13}$, in the hope that we could force a non-trivial intersection with $A_{13}$. Once you ask yourself that question, the answer follows pretty quickly.