Number of elements of order $5$ in $S_7$: clarification

Question

In finding the number of elements of order $5$ in $S_7$, the symmetric group on $7$ objects, we want to find products of disjoint cycles that partition $\{1,2,3,4,5,6,7\}$ such that the least common multiple of their lengths is $5$. Since $5$ is prime, this allows only products of the form $(a)(b)(cdefg)$, which is just equal to $(cdefg)$. Therefore, we want to find all distinct $5$-cycles in $S_7$.

Choosing our first number we have seven options, followed by six, and so on, and we obtain $(7\cdot 6 \cdot 5 \cdot 4 \cdot 3)=\dfrac{7!}{2!}$. Then, we note that for any $5$-cycles there are $5$ $5$-cycles equivalent to it (including itself), since $(abcde)$ is equivalent to $(bcdea)$, etc. Thus, we divide by $5$, yielding $\dfrac{7!}{2!\cdot 5}=504$ elements of order $5$.

I understand this...it makes perfect sense. What's bothering me is:

Why isn't the number of elements of order five $\binom{7}{5}/5$? It's probably simple, I'm just not seeing why it doesn't yield what we want.

Answer 1

$\binom{7}{5} = \frac{7!}{5!2!}$ is the number of ways to choose $5$ things from $7$ when order doesn't matter. In this case, the order does matter, since, for example $(12345) \not= (12435)$.

This is an example of the difference between the number of combinations of objects and the number of permutations.

Answer 2

$\binom{7}{5}$ is the number of ways of choosing $5$ numbers, but you still need to take into account the different possible orders. Given $5$ numbers you can start your cycle with any of the $5$ you choose, then there are $4$ choices for the second number, $3$ for the third. Hence the result is $$\binom{7}{5}\cdot 4! = \frac{7!}{5\cdot 2}$$ as you have already computed.

Answer 3

This is your code:

g:=SymmetricGroup(7);

h:=Filtered(Elements(g),x->Order(x)=5);

Size(h);

The size of $h$ is the same as other theoretical approaches suggested. It is $504$.