# Permutation Group as quotient of Free Group

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!

-
Alternatively, what is the size of the smallest set of generators of $S_n$ – Thomas Andrews Dec 25 '12 at 20:50

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 $k\le 2$?
I wasn't familiar with that set of generators for $S_{n}$ but it certainly clarifies it :) Obviously for n=1,2 it would be k=1 and otherwise 2. Thanks a lot :) – Serpahimz Dec 25 '12 at 21:31
Actually, when $n=1$, you can use the free group on $0$ generators... – Thomas Andrews Dec 25 '12 at 22:15