Subgroup of $S_n$ with elements having cycles of any possible length Let $G$ be a subgroup of the symmetric group $S_n$. 
Suppose that $G$ has the following property:
given any positive integer $k\leq n$, there exists an element of $G$ having a cycle of length exactly $k$.
Is it any true that under this assumption $G=S_n$?
If not, could one make a counterexample with $n$ prime?
Please notice that I am not saying that $G$ has elements which ARE cycles of order $k$.
Could you please remove the duplicate, as if you read it carefully you see that it is a different question? (please see also the answer that I accepted)
 A: The answer is yes. There is a result (I think of Jordan) that a transitive subgroup of $S_n$ that contains an element of prime order $p$ with $n/2 < p < n-2$ contains $A_n$. (This result is used in a probabilistic algorithm for testing whether a given group contains $A_n$.)
For $n \ge 8$ there exists such a prime $p$. By assumption, your group contains an $n$-cycle, so it is transitive, and a $p$-cycle, so it contains $A_n$. (The assumptions say only that $G$ contains an element containing a $p$-cycle, but since $p>n/2$, some power of that element must be a $p$-cycle.) Since it contains an $n$-cycle and an $(n-1)$-cycle, it must contain an odd permutation, and hence it must be $S_n$.
You can check the cases $n\le 7$ individually.
References for the result of Jordan are:
C. Jordan.
Sur la limite de transitivit\'e des groupes non altern\'es.
Bull. Soc. Math. France, 1:40-71, 1873.
Theorem 13.9 of Wielandt's book|"Finite Permutation groups".
(The result actually applies to primitive groups with any prime $p<n-2$. Ths condition $p>n/2$ ensures that the group is primitive.)
The existence of the prime follows from a strong version of  Bertrand's postulate - see the Wikipedia page for example.
