TITLE: Permutation Group as quotient of Free Group
QUESTION [5 upvotes]: I could use some help with the following question:
Let $S_{n}$
  be the permutation group of $\left\{ 1,...,n\right\}$
 , what is the minimal $k\in\mathbb{N}$
  such that $S_{n}$
  is a quotient of the free group $F_{k}$
  (free group with k
  generators).
Thanks a lot!

REPLY [4 votes]: Clearly, $k>1$ (except for which small values of $n$?).
Can you see how $S_n=\langle(1\, 2),(1\,2\,\ldots\,n)\rangle$ and hence $k\le 2$?