Elements of order 5 in $A_6$

Question

I am trying to find the elements of order 5 in $A_6$ and I understand that they are of the form $(abcde)$, correct? So the number of elements is $(6*5*4*3*2)/5$=144. I looked somewhere else and it said it was $6!/5$, which is the same thing because the only thing that is left off is multiplying by 1. But why would I have to do $6!$ ?

Answer 1

The sign of a permutation is odd if and only if it has an odd number of even cycles in its decomposition. How many even cycles does a $5-$cycle have?

Since the order of a permutation is the least common multiple of the lengths of its cycles the permutations of order $5$ are exactly the $5$ cycles. to count them you can first select the elements in the cycle in $6$ ways ( this is the same as selecting the element not in the cycle.

Once these have been selected thee are $5!$ ways to order them inside the brackets. However out of these ways to write them there are groups of $5$ that represent the same permutation but are just rotated. So we get $4!$ cycles once the elements in the cycle have been selected.

Thus there are $4!6=\frac{6!}{5}$ elements of order $5$ in $A_6$ (and also in $S_6$)