We have all cycles of length $n$. For example when $n=4$ our set is following form: $$G = \{(1234);(1243);(1342);(1324);(1432);(1423)\} $$ Our task is to find order of the group generated by the set $G$.
We know, that $|G|\ge (n-1)!$, but I have no clue how to begin.
Try to help me, please.
On
Note that any subgroup of $S_n$ generated by a complete set of cycles of length $n$ will be a normal subgroup. For $n\gt 4$ there are just two possibilities - $A_n$ or $S_n$ (the trivial subgroup is impossible).
If $n$ is odd, the cycles which generate the group are all even permutations, and can't generate an odd permutation, so the group is $A_n$. If $n$ is even, the cycles are odd permutations, so the group they generate contains an odd permutation, and must be $S_n$.
For $n\leq 4$ it is necessary to use a different approach (e.g. direct computation in a small group).
Note that essentially the same proof applies when the generators are a complete set of elements of any conjugacy class in $S_n$. However, when $n=4$ the elements conjugate to $(12)(34)$ generate the normal subgroup of order $4$ - so this approach cannot work in general for order $4$.
$$(1234)(1243)=(132)(4)$$ and so on, hence all the $3$-cycles belong to the generated group, giving $A_n\subseteq \langle G\rangle$. If $n$ is odd, then $$\langle G\rangle = A_n$$ since no element of $S_n$ with sign $-1$ can belong to $\langle G \rangle$. On the other hand, if $n$ is even there is $\sigma\in\langle G\rangle$ such that $\operatorname{sign}(\sigma)=-1$, so $|\langle G\rangle|>\frac{n!}{2}$, so the generated group is the whole $S_n$:
$$\langle G \rangle = \left\{\begin{array}{rcl} S_n&\text{if}&n\equiv 0\pmod{2},\\ A_n&\text{if}&n\equiv 1\pmod{2}.\end{array}\right.$$