This is Exercise 6.1.9 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to this search for "locally free finitely generated" in the group theory tag, it is new to MSE.
This is a solution-verification question.
The Details:
I am comfortable with free groups. For a list of different definitions, see here.
A group is locally free if every finite subset is contained in a free subgroup.
Theorem 6.1.1: (The Nielsen-Schreier Theorem) If $W$ is a subgroup of a free group $F$, then $W$ is a free group.
There is more to that theorem. However, I only need the bit I quoted.
The Question:
Prove that a group is locally free exactly when its finitely generated subgroups are free.
Thoughts:
Suppose all finitely generated subgroups of a group $G$ are free. Let $X\subset G$ be a finite set. Then $X\subseteq \langle X\rangle$, the subgroup of $G$ generated by $X$, which is free by hypothesis. Hence $G$ is locally free.
Conversely, suppose $G$ is locally free. Then each finite subset $X\subset G$ is in a free subgroup $H\le G$. Consider $\langle X\rangle$. It is finitely-generated. It is contained in $H$ because one way of defining $\langle X\rangle$ is as the intersection of all subgroups of $G$ containing $X$. Every subgroup of a free subgroup is itself free (by Theorem 6.1.1). Hence $\langle X\rangle$ is free. But $X$ was an arbitrary finite subset of $G$. Hence each finitely generated subgroup of $G$ is free.
Is it really so simple? Have I missed something?
It might seem easy to me due to my experience with combinatorial group theory.
Please help 🙂
Best Answer
This is a CW post in order to close the question.
My proof is correct.
However, as mentioned in the comments, the terminology locally free is idiosyncratic to Robinson's book: it is usually defined to be what the equivalent property here is.