Question: Show that there are cyclic subgroups of order $1,2,3 \ \text{and} \ 4$ in $S_4$ but $S_4$ does not contain any cyclic subgroup of order $ \geq 5$. (Note: I suppose $S_4$ are all permutation groups of length 4.)

My attempt: Obviously $S_4$ contains the cyclic subgroups of order $1,2,3 \ \text{and} \ 4$ since $S_4$ is of order $4$. Notice that a subgroup of $S_4$, take length 3, is closed under the operation and $S_3$ is also closed under taking inverse (I could show this if needed and I could show for the rest too).

For the next part, it is obvious that $S_4$ will not in any way ADD an element to it's permutation no matter how many times you apply the operation...

Am I going in the right direction here?

  • $\begingroup$ There are a couple of things I think you got wrong. First, $S_4$ is a group, and its elements are permutations of the four elements $\{1,2,3,4\}$. Second, $S_4$ is not of order $4$ but of order $4!=24$. I am not sure what you are trying to do with $S_3$. $\endgroup$ – Ludolila Feb 19 '14 at 11:25
  • $\begingroup$ Could I show this using the theorem that states that if G is a finite cyclic group of order m, then for every possible divisor d of m, there exists a subgroup of G order m? I don't know, shit...This course is KILLING me. $\endgroup$ – user3200098 Feb 19 '14 at 11:26
  • $\begingroup$ No, because in your case $G$ is $S_4$, and $S_4$ is not cyclic. Recall that finding a cyclic subgroup of order $t$ is equivalent to finding an element of order $t$. Do you know what the elements of $S_4$ look like? What are their orders? $\endgroup$ – Ludolila Feb 19 '14 at 11:28
  • $\begingroup$ The elements of $S_4$ are $ \{\ 1,2,3,4 \}\ $. The order $ \circ S_4$ is 24. I get that. How do I find the order of the elements? I'm guessing that I need to find $a^n$ such that $a$ is an element and $n$ is the amount of operations I perform to get $e$. But what is $e$? $\endgroup$ – user3200098 Feb 19 '14 at 11:31
  • $\begingroup$ These are not the elements of $S_4$. Try reading the second answer in this post math.stackexchange.com/questions/379841/…, I think it could help. $\endgroup$ – Ludolila Feb 19 '14 at 11:34


1) We can write every permutation in $\;S_n\;$ as a product of disjoint cycles.

2) The order of a product of disjoint cycles is the lowest common multiple of the cycles' orders

3) Thus, in $\;S_4\;$ the maximal order an element can have is four ($\;4\;$ ) .

  • $\begingroup$ Thanks. That cleared it up a bit (I wish I understood the subject better, though). $\endgroup$ – user3200098 Feb 19 '14 at 12:11

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