Let $S_4$ be the symmetric group on 4 letters. How many elements of $S_4$ Have order 4?

I am learning group theory all by myself, and couldn't find a way to solve this problem.

I am aware about the fact that any permutation can be written as a product of disjoint cycle, and the order of that permutation is equal to the LCM of the disjoint cycles.

I am basically having the problem to find the permutations

Please help me out

  • 1
    $\begingroup$ $S_4$ is small enough for one to write down all elements, compute all their orders, and count how many have order $4$. $\endgroup$ – Lord Shark the Unknown Oct 3 '18 at 15:27
  • $\begingroup$ Thinking about $\left(\begin{array}{cc} 1 & 2 & 3 & 4 \\ a & b & c & d \end{array}\right)$ where $1 \neq a$, $2 \neq b$, $ 3 \neq c$ and $ 4 \neq d$. After, suppose that at least one element is fixed. $\endgroup$ – Corrêa Oct 3 '18 at 15:30
  • $\begingroup$ Thankyou.. My answer is 6 $\endgroup$ – Akshie Dhiman Oct 3 '18 at 15:48

Think like that-what are the possible cycle structures of the permutations in $S_4$? There are only five possible options:

  1. Four cycles of length $1$

  2. One cycle of length $2$ and two cycles of length $1$

  3. Two cycles of length $2$

  4. One cycle of length $3$ and one cycle of length $1$

  5. One cycle of length $4$

As you know, the order of a permutation equals to the lcm of the lengths of its disjoint cycles. So it is easy to check only the elements of the fifth type (one cycle of length $4$) are elements of order $4$. So all you need to find is how many cycles of length $4$ there are. Well, if you want $\sigma$ to be a $4$-cycle in $S_4$ you have $3$ options to choose the value of $\sigma(1)$, then $2$ options to choose what will be $\sigma(\sigma(1))$, and that's it. Once you know what are $\sigma(1)$ and $\sigma(\sigma(1))$ you know the whole permutation because it is a $4$-cycle and $\sigma(\sigma(\sigma(1)))$ must be the remaining element of $\{1,2,3,4\}$. So the number of such permutations is $3\times 2=6$.

  • $\begingroup$ Thankyou Sir. It really helps me . I caculated the answer = 6 $\endgroup$ – Akshie Dhiman Oct 3 '18 at 16:04

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