Sorry for this possibly trivial or stupid question, but I'm very far from being an expert in Algebra. 


Let G be the fundamental group of the nonorientable surface of even rank n=2k (n generators, 2n if counting their inverses). Does G contain free subgroups? In the affirmative, how many of them have a maximum number of generators, and which is this number? I suspect that the answers are 'yes', k, and k respectively. 


Of course the same question makes sense when n is odd. 


Thanks in advance.



free subgroups of the fundamental group of nonorientable surfaces
A double cover (thus a subgroup of index $2$) of your fundamental group is the fundamental group of an orientable surface, which is either closed or not. In the second case, the fundamental group is free. In the first case, it is not free, but any subgroup of infinite index is free. You can have free subgroups of ANY rank.


