Number of different cyclic permutations in a symmetric group

Question

So a permutation cycle is written as $(a,b,c,d)$ where $a\to b\to c\to d\to a$, I would like to know the number of permutation cycles I can express $(a,b,c,d)$ such that the permutation cycles are different.

By different I mean $(a,b,c,d)=(d,a,b,c)$.

So I am looking for the number of different permutation cycles on the permutation $(a,b,c,d)$. The only way I am able to do this is to list them out one by one, and so I got $6$, but I'm not sure how to do that in another way.

Answer 1

There are $4!$ permutations of the four elements, but given a permutation there are $4$ that give you the same cycle, so it's $4!/4=6$. If you want to get all $4$-cycles in $S_n$, this is easily adapted to $$\frac{4!}4\binom n4$$ to account for the number of ways to pick four elements.

Answer 2

Let's say that a cycle, by convention, always begins with $a$. Then $4$-cycles correspond bijectively to the permutations of $b,c,d$, i.e. to the group $S_3$. Now, we have the basic fact that $\;|S_3|=6$.