# If every finitely generated subgroup is free, then is the group free too? [duplicate]

If a group $$G$$ has the property that every finitely generated subgroup $$H \leq G$$ is free, then must $$G$$ be a free group?

My intuition for thinking that this is true is that a relation on some generators of $$G$$ always involves only a finite number of generators and therefore would also hold in some subgroup, but I don't see how to make this precise.

• $\mathbb{Q}$ is a counterexample. – Andy Putman Aug 16 at 14:17
• Such groups even have a name, they are called locally free. – Moishe Kohan Aug 16 at 14:31
• One reason your intuition is off is because even free groups have presentations with nontrivial relators. For instance the (additive) subgroup of $\mathbb Q$ generated by $a=\frac{1}{3}$ and $b=\frac{1}{2}$ has presentation $\langle a,b \mid a b a^{-1} b^{-1} = \text{Id}, a^3 = b^2 \rangle$; the actual subgroup is the rank 1 free group generated by $b a^{-1} = \frac{1}{6}$. – Lee Mosher Aug 16 at 14:50
• Good points everyone, thanks. Pretty new to MO, should this be a response so I can accept it? – user32157 Aug 16 at 15:04