# Symmetric group $S_4$ . How many elements are there of order 4

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

• $S_4$ is small enough for one to write down all elements, compute all their orders, and count how many have order $4$. – Lord Shark the Unknown Oct 3 '18 at 15:27
• 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. – Corrêa Oct 3 '18 at 15:30
• Thankyou.. My answer is 6 – 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 $$3\times 2=6$$.

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