Conditions on cycle types for permutations to generate $S_n$ Consider the following result of Dedekind:

For any polynomial $p \in \mathbb{Z}[x]$ and any prime $q$ not dividing the discriminant of $p$, if $p$ factors modulo $q$ into a product of irreducible polynomials with degrees $d_1, \ldots, d_s$, then the Galois group $\text{Gal}(p)$ contains a permutation with cycle structure $(d_1, \ldots, d_s)$.

In particular, it is known that in $S_r$, $r$ prime, is generated by a $2$-cycle and an $r$-cycle, so if one can find appropriate primes, one can use Dedekind's result to show quickly that a given polynomial of prime degree has full Galois group $S_5$. This condition is rather limited, however, and this motivates the following question:

Suppose we know the cycle types of some elements of a group $G \leq S_n$. Under what conditions on those cycle types can we conclude that $G$ is actually the full group $S_n$?

This line of thinking was motivated, by the way, by this question, which asks about the solvability of the polynomials
$$p_n(x) := x^n - \sum_{k = 0}^{n - 1} x^k$$
that arise when studying the so-called $n$-nacci sequences. Two of the answers there report CAS computations giving that the Galois group of $p_n$ is $S_n$ for $n \leq 12$, and one speculates that that holds for all $n$. For $n = 5$, one can use the primes $q = 3, 5$ to show that the Galois group contains $2$- and $5$-cycles, and hence by the above fact $\text{Gal}(p_5) \cong S_5$. One can do the same, for example, with $n = 13$ and $q = 5, 17$ to conclude that $\text{Gal}(p_{13}) \cong S_{13}$, extending the other answers there.
 A: Well, I do not expect you to find my answer complete but I just let you know another result quite similar to the one you have  given. 
If $G$ is a transitive subgroup of $\mathfrak{S}_n$ and $G$ contains a transposition and a $n-1$-cycle then $G=\mathfrak{S}_n$. the proof goes as follow :
let $\tau$ and $c$ be respectively a transposition and a $n-1$-cycle in $G$. Up to global conjugation (i.e. up to the study of a conjugate of $G$) one can assume that :
$$c=(2,...,n) $$
Let us define $\tau=(i,j)$ then, because $G$ is transitive, you can find $g\in G$ such that $g(i)=1$ so that :
$$g\tau g^{-1}=(1,k)\in G $$
With $k>1$. But now you have that $c$ fixes $1$ and act transitively on $\{2,...,n\}$ so that :
$$\{(1,2),...,(1,n)\}=\{c^l\tau c^{-l}|l\in\mathbb{N}\}\subseteq G $$
That is $G$ contains $(1,2),...,(1,n)$ which is a generating set for $\mathfrak{S}_n$. 
Well, obviously this is not a perfect result but if you assume that your polynomial must be irreducible then you always have $G$ transitive so it is adapted to the Galois theory.
