Is there a non-free group $G$ whose subgroups are all freely decomposable? 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.
 A: 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 $G=i(G)\ast\langle b_0\rangle$).
