About the generators of a free group

Question

Suppose we know that $G$ is a free group of rank $n$ and that $\{g_1,...,g_n\}$, with all the $g_i$ distinct, is a system of generators for $G$. Are we sure that it is also a $\textbf{free}$ system of gereators for $G$ ? Why?

Answer 1

Yes! Let $\{h_1, \dotsc, h_n\}$ be a free generating set and consider the homomorphism $G \to G$ that sends $h_i$ to $g_i$ for each $i$. That $\{g_1, \dotsc, g_n\}$ is a generating set guarantees that this map is surjective; it is injective if and only if $\{g_1, \dotsc, g_n\}$ is a free generating set (see below). But finitely generated free groups are Hopfian: any surjection from such a group to itself is injective. See: http://groupprops.subwiki.org/wiki/Finitely_generated_and_free_implies_Hopfian

Edit: $\{g_1, \dotsc, g_n\}$ freely generates if and only if no non-trivial reduced word in those generators and their inverses represents the identity element. Suppose that the map is injective. Let $w$ be a reduced word in the $g_i$ and let $w'$ be the word in the $h_i$ obtained by replacing $g$'s with $h$'s. Then $w'$ is a reduced word, so is not the identity, so nor is its image under the map. But its image is $w$.