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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?

user3200098
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  • 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 $S_3$. – Ludolila Feb 19 '14 at 11:25
  • 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.

     – user3200098 Feb 19 '14 at 11:26
  • 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? – Ludolila Feb 19 '14 at 11:28
  • 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 $e$?

     – user3200098 Feb 19 '14 at 11:31
  • These are not the elements of $S_4$. Try reading the second answer in this post http://math.stackexchange.com/questions/379841/how-to-enumerate-subgroups-of-each-order-of-s-4-by-hand?rq=1, I think it could help. – Ludolila Feb 19 '14 at 11:34
  • Okay if I wrote out $S_4$ I'd get 24 different permutations.

    Their orders are $4$...?

     – user3200098 Feb 19 '14 at 11:41
  • "order" means the number of elements in the group. The permutations are elements of $S_4$--it doesn't make sense to talk about their order. – Rustyn Feb 19 '14 at 11:49
  • Not all of the permutations have order $4$, for example a transposition has order $2$. The problem is asking to prove that their orders cannot be greater than $4$, which is not obvious a priori because a generic group of order $24$ might have elements of order up to $24$ (think of $\mathbb Z_{24}$). – dani_s Feb 19 '14 at 11:49
  • Well an order of an element is how many times you have to operate on it to get the identity, isn't it...?

    What theorem would be useful to prove my problem? I need a nudge.

     – user3200098 Feb 19 '14 at 11:58

1 Answers1

2

Hints:

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 ($\;4\;$ ) .

DonAntonio
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