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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 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 :)

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  • $\begingroup$ Why the downvote? $\endgroup$
    – Shaun
    Commented Jan 31, 2022 at 13:41
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    $\begingroup$ Where did this -1 come from?!! and why. Please wait friends $\endgroup$
    – Mikasa
    Commented Jan 31, 2022 at 13:41
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    $\begingroup$ Some problems are simply designed to test if you understood basic definitions and result. You did. $\endgroup$ Commented Jan 31, 2022 at 14:11
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    $\begingroup$ Yes, the proof is as simple as you wrote. (Also, typically a "local" property of a group is one enjoyed by finitely generated subgroups, e.g. "locally finite" means finitely generated subgroups are finite, etc. So this question is phrased weirdly (of course that is the books fault... :-) ) $\endgroup$
    – user1729
    Commented Jan 31, 2022 at 14:11
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    $\begingroup$ Yes the proof is correct. I promise you that I was not the downvoter, but there really is no point at all in asking "why the downvote?". People who downvote generally prefer to remain anonymous. $\endgroup$
    – Derek Holt
    Commented Jan 31, 2022 at 14:21

1 Answer 1

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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.

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