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> This is a crossport of [this question](https://math.stackexchange.com/questions/1099950/categorical-proof-that-subgroups-of-free-groups-are-free) from MSE.
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Is there a categorical proof that subgroups of free groups are free?
How about the result that subgroups of free *abelian* groups are free abelian?


What is it about $\mathsf{Grp}$ and $\mathsf{Ab}$ that makes subobjects of free objects free? Can one characterize such categories? In particular, are there any other ones apart from $\mathsf{Grp}$ and $\mathsf{Ab}$?




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I have received some comments on MSE, one claiming such a proof exists, one doubting it, and another claiming it cannot be proved in $ZF$.



Categorical proof subgroups of free groups are free?
"Subobjects of free algebras are free" is satisfied comparatively rarely for algebraic theories. I'm going to start a list and people should feel free to add. I'm making it CW. 


Before starting, let me say that IMHO a more interesting general question to consider is: when are retracts of free objects free? That can be a very tough question. We had some discussion here: [Is a retract of a free object free?](https://mathoverflow.net/questions/90734/is-a-retract-of-a-free-object-free) 


* Ben Webster got the ball rolling with various categories of modules (although he concentrated more on whether subobjects of projectives are projective). By the way, vector bundles can be seen as projective objects over the ring of smooth functions, but need not be free: some vector bundles are not trivial bundles.
* Submonoids of free monoids need not be free. This fails for even the simplest cases, e.g., the monoid generated by $2, 3$ in $(\mathbb{N}, +)$ isn't free.
* Subalgebras of free commutative $k$-algebras (aka polynomial algebras) need not be free. The subalgebra generated by $t^2, t^3$ in $k[t]$ ($k$ a field) can't be a polynomial algebra as it isn't even a UFD.
* Subalgebras of free Boolean algebras need not be free. For example, a finitely generated free Boolean algebra on $n$ elements has cardinality $2^{2^n}$, isomorphic to a power set of $P([2^n])$. A quotient of the set $2^n$, say one with 3 elements, induces an inclusion $P([3]) \to P([2^n])$ of Boolean algebras.
* Pablo mentioned that subalgabras of free Lie algebras *are* free. (What's a good reference for that?)


