# If a group $G\leq S_{13}$ has an element of order $40$, then $G$ has a normal non-trivial subgroup.

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).

• Certainly you mean "nontrivial proper subgroup". Hint: signature.
– YCor
Mar 12, 2019 at 23:14

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.
• Are you familiar with cycle notation for elements of symmetric groups? So for example, we might write $(1~3~5~13~7)(2~4~8~6~12~10~11~9)$ to denote the permutation that sends $1$ to $3$, $3$ to $5$, $5$ to $13$, $13$ to $7$, and $7$ back to $1$ (the $5$-cycle; i.e., a cycle of length $5$) and also cycling the remaining eight numbers as indicated in the $8$-cycle. Mar 12, 2019 at 23:50
• yes I am familiar. so if we have $\sigma=\alpha \cdot \beta \cdot \gamma$ we say that $\sigma$ is a product of a $5-cycle$ if the order of $\sigma$ is $5$? Mar 12, 2019 at 23:52
• Not exactly. It could be that $o(\sigma)=5$ because $\sigma$ is the product of two distinct $5$-cycles. A $5$-cycle is a permutation that cyclically permutes $5$ elements of your set while leaving the remaining elements fixed. Mar 12, 2019 at 23:53
• so if we have a group of order $x=a\cdot b$ and $a$ is prime and $b$ is not, so we should have one cycle with length $b$ and other cycles of length $a$? How many 5-cycles and 8-cycles should the subgroup contain? one 5-cycle and one 8-cycle? Mar 12, 2019 at 23:56