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Suppose that $G$ is a group such that every subgroup $H \subseteq G$ (including $G$ itself) is either free or a non-trivial free product, i.e. $H = H_1 * H_2$ with $H_1, H_2$ both non-trivial. Is there an example of such a $G$ which is not free?

If $G$ is finitely generated then Grushko's theorem implies $G$ must be free, but the infinitely-generated case seems likely to have a non-free example.

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Yes, there's an example.

Kurosh proved that the group $G$ with presentation $$\langle (a_n)_{n\ge 0},(b_n)_{n\ge 1}\mid a_nb_na_n^{-1}b_n^{-1}=a_{n-1},\;\forall n\ge 1\rangle$$

has the following properties: $G$ is torsion-free, isomorphic to $G\ast\mathbf{Z}$, all freely indecomposable subgroups of $G$ are cyclic, but $G$ is not free (it's easily checked not to be residually nilpotent: the intersection of the lower central series contains all $a_n$).

Reference: A. Kurosch. Zum Zerlegungsproblem der Theorie der freien Produkte. Mat. Sbornik 2 (44): 5, 995–1001, 1937.


[Added] Context:

Kurosh asked in 1934 (Math Ann.) the (free) Zerlegungsproblem: does every group have a free product decomposition into freely indecomposable factor?

In the same 1934 paper he proved that if the answer is positive for a given group, then all such decompositions are isomorphic. This was improved by Baer and Levi (Comp. Math. 1936): on an arbitrary group, any two free product decompositions (with possibly decomposable factors) have a common refinement (in a suitable sense). Then in 1937 (Mat. Sbornik) Kurosh found the above example providing in general a negative solution to the Zerlegungsproblem. Later, Grushko (Mat. Sbornik 1940) and then independently B.H. Neumann (J. LMS 1943) proved that the Zerlegungsproblem has a positive answer for finitely generated groups, showing that the generating rank is additive (and not only sub-additive) under free products.


Added remark: there's an embedding of the countable group $G$ into the 3-generator group with 1-relator presentation $$\Gamma=\langle a,x,y\mid a={^x}\!a\;^y\!a\rangle\qquad (\text{where } {^x}\!a\;^y\!a=xax^{-1}yay^{-1}),$$ given by $a_n\mapsto x^{-n}ax^{n}$, $b_n\mapsto x^{-n}yx^{n}$. I don't know if this 1-relator group has ever been studied specifically.

More precisely, $\Gamma$ is the ascending HNN-extension of $G$ associated with the injective endomorphism $i$ given by $a_n\mapsto a_{n+1}$, $b_n\mapsto b_{n+1}$ (note that $G=i(G)\ast\langle b_0\rangle$).

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  • $\begingroup$ I don't know whether your one-relator group has been "explicitly studied", but it's a very easy group to describe: it is the fundamental group of the space obtained from the 3-holed sphere with boundary components $a,b,c$ by gluing $b$ and $c$ to $a$. In particular, it's certainly hyperbolic, and has many nice properties. $\endgroup$
    – HJRW
    Commented Dec 19, 2020 at 18:06
  • $\begingroup$ How do you see that $\langle (a_i)_{n\geq 0},(b_i)_{n\geq 1},(c_i)_{n\geq 1},(d_i)_{n\geq 1}\mid \left[a_n,b_n\right]\left[c_n,d_n\right]=a_{n-1},\forall n\geq 1\rangle$ is not isomorphic to Kurosh's group? $\endgroup$ Commented Dec 22, 2020 at 11:50
  • $\begingroup$ @seldomseen I made no such claim (I didn't mention this group). Feel free to post a question if the answer is not immediate. $\endgroup$
    – YCor
    Commented Dec 22, 2020 at 12:06
  • $\begingroup$ That was a question. And nothing in the question implies that you made such a claim. $\endgroup$ Commented Dec 22, 2020 at 12:07
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    $\begingroup$ You can also see that Kurosh's group is not free by observing that $G$ is freely indecomposable relative to $a_0$, whereas any infinitely generated free group is freely decomposable relative to any element. $\endgroup$ Commented Dec 22, 2020 at 18:46

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