Permutation Group as quotient of Free Group

Question

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!

Answer 1

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 $k\le 2$?