Finding Subgroups Of $S_{15}$ 
Find subgroups of order 45 and 39 in $S_{15}$

For order of $49$ I know that a permutation in $S_n$ can be written as two disjoint cycles and the order of two disjoint cycles $a,b$ is $lcm(a,b)$ 
so I am looking for $a,b<15$ such that $lcm(a,b)=45$ now $lcm(a,b)=\frac{a\cdot b}{gcd(a,b)}$ so I should look for $a,b$ such that $gcd(a,b)=1$ and $a\cdot b=45$ which are $5$ and $9$
In this case we can say that $|\langle a \rangle|=o(a)$?
For subgroup of order $39$ the answer is that it can not be as 1. we can not find a cycle of $39$ with $15$ elements 2. there are no $2$ disjoint cycles with $lcm=39$ but how can we conclude this?
 A: For order $45$ we can indeed find an element of that order. We take, as you say, disjoint cycles of length $5$ and $9$. (Note that we need $5+9\leqslant 15$ to fit them in. So $$a=(1 2 3 4 5)(6 7 8 9 A B C D E)$$ generate the subgroup we need.
For a subgroup of order $39$ things are less straightforward. The only possibilities for a cyclic subgroup would be to find a $39$ cycle, or to find disjoint $3$ and $13$ cycles: neither will fit into $S_{15}$. 
But there is also the possibility of a non-abelian group of order $39$. This has a normal cyclic subgroup of order $13$, generated by $x$, say; extended by a cyclic subgroup of order $3$, generated by $y$ such that $y^{-1}xy=x^3$. [This group exists because $3|(13-1)$.]
So we want to find such a subgroup of $S_{15}$. For a start we can take
$$ x= (1 2 3 4 5 6 7 8 9 A B C D)$$
as all the elements of order $13$ look the same.
Then we have
$$x^{3}= (1 4 7 A D 3 6 9 C 2 5 8 B)$$
and can now read off an appropriate $y$, $$y=(1)(2 4 A)(3 7 6)(5DB)(89C)$$
to give our required subgroup $\langle x, y\rangle$.
