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.

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    $\begingroup$ $\mathbb{Q}$ is a counterexample. $\endgroup$ – Andy Putman Aug 16 at 14:17
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    $\begingroup$ Such groups even have a name, they are called locally free. $\endgroup$ – Moishe Kohan Aug 16 at 14:31
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    $\begingroup$ 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 $b a^{-1} = \frac{1}{6}$. $\endgroup$ – Lee Mosher Aug 16 at 14:50
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    $\begingroup$ Good points everyone, thanks. Pretty new to MO, should this be a response so I can accept it? $\endgroup$ – user32157 Aug 16 at 15:04
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    $\begingroup$ By the way, don’t feel bad about this. It was one of my answers to the ancient question about common false beliefs about mathematics: mathoverflow.net/a/23522/317 $\endgroup$ – Andy Putman Aug 16 at 15:28