TITLE: Group Theory and Divisors
QUESTION [6 upvotes]: Here is the problem I'm working at:

Find a divisor $d$ of $6! = 720$ such that $S_6$ does not have a subgroup of order $d$.

Initial Approach: 
So I factored the number $720$ and here are the divisors that I came up with:
$1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 36, 40, 45, 48, 60, 72, 80, 90, 120, 144, 180, 240, 360, 720$
So there are a total of 30 divisors for the number $720$. I believe it would be helpful to know how many total subgroups are there in $S_6$. I did find this link, not sure if it would help out or not. 
Update
So after doing some further research, I found out that $S_6$ has a total of $1455$ subgroups, but a proof is not given as to why this is true. But why is this true? Well the divisors of $1455$ are given as follows:
$1, 3, 5, 15, 97, 291, 485, 1455$

REPLY [12 votes]: I don't see how knowing the number of subgroups in $S_{6}$ really helps, but here is an argument which may help, provided you know the classification of groups of order $pq$ for distinct primes $p, q$. Note that any group of order $15$ must be cyclic, since $15$ is the product of the distinct primes $3$ and $5$, and $5$ is not congruent to $1$ mod $3$. Hence, if $S_{6}$ had a subgroup of order $15$, it must have an element of order $15$. But this cannot be, since the order of any element of $S_{6}$ is the least common multiple of the lengths of the cycles in its disjoint cycle decomposition.