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).

  • 3
    $\begingroup$ Certainly you mean "nontrivial proper subgroup". Hint: signature. $\endgroup$ – YCor Mar 12 '19 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.

  • $\begingroup$ what does it mean "product of a 5-cycle"? $\endgroup$ – vesii Mar 12 '19 at 23:46
  • $\begingroup$ 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. $\endgroup$ – Robert Shore Mar 12 '19 at 23:50
  • $\begingroup$ 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 $5$? $\endgroup$ – vesii Mar 12 '19 at 23:52
  • $\begingroup$ 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. $\endgroup$ – Robert Shore Mar 12 '19 at 23:53
  • $\begingroup$ 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? $\endgroup$ – vesii Mar 12 '19 at 23:56

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