If $S_n$ has cyclic subgroups of these orders or not? This question was asked in my quiz and I am having difficulties in solving it .

Question : For a positive integer $n$ , let $S_n$ denote permutation group on $n$ symbols. Choose correct statements from below.
(a) For a positive integer $n$ and for every $m$ with $1\leq m \leq n , S_n $ has a cyclic subgroup of order $m$;
(b) For every positive integer $n$ and for every $m$ with $n<m <n!$, $S_n$ has a cyclic subgroup of order $m$;
(c) There exists positive integers $n$ and $m$ with $n<m<n!$ such that $S_n$ has a cyclic subgroup of order $m$;
(d) For every positive integer $n$ and for every group $G$ of order $n$, $G$ is isomorphic to a subgroup of order $S_n$ .

Attempt:
(a) is true as I can choose an element of length $m$ which will generate cyclic subgroup of order $m$.
(b) Choose $n=4$ and $m=19$ to contradict (b).
(c) I dont know if there can be found any such case and if yes how to do so .
(d) I dont know about it . kindly tell.
Answer:

 A,C,D are true

 A: (a): You did it yourself correctly.
(b): Every, $m\not|\ \ n!$, works as your example. $(n! - 1)$ always works.
(c): Every permutation can split to disjoint cycles and its order is their orders $lcm$. So it suffices to find such. For $n \leqslant 4$ and $n=6$, there is no such $m$. But for $n=5$ and $n \geqslant 7$ you have always some. You can split all $n$ elements into two subset of order $[n/2], [(n+1)/2]$ (for odd $n$) and $[n/2], [(n-1)/2]$ (for even $n$) and  then product their cycles that have order $n(n \pm 1) / 4$ which always satisfies the criteria.
(d): Since in group, the equation $gx = h$, for every $g,h, \ $ has always a unique solution ($x$), so every element permutes all group elements (including itself), so they behave like permutations on $n$ things (actually themselves). Therefore we can see them as permutation functions, $g(x) = gx$. Now since $h(g(x)) = (hg)(x)$, they respect to group multiplication. Hence group ($G$), is isomorphic to a subgroup of $S_n$.
A: Hints:
For $c)$, let $n$ be an integer with a partition $n=m_1+m_2$ such that $1<m_1,m_2<n-1$ and $m_1,m_2$ are coprime, then the product of two disjoint cycles of lengths $m_1$ and $m_2$ is a permutation of order $m=m_1m_2$. So if you can find an integer $n$ satisfying the above conditions and such that $m_1m_2>n$, you've answered the question positively.
For $d)$, you can embed $G$ in $S_n$ associating to each $g\in G$ left (or right) multiplication by $g$.
