An interesting fact was relayed to me in [another question of mine](https://mathoverflow.net/questions/57902/fundamental-groups-of-closed-hyperbolic-3-manifolds-are-freely-indecomposableBlockquote) that 



>  If $M$ is any closed manifold with universal cover homeomorphic to $R^n$ for $n>1$ then $\pi\_1(M)$ is freely indecomposable. 




What are some other sufficient conditions for the free-indecomposability of a group? Are there any interesting necessary conditions?



Sufficient Conditions for Free Indecomposability
There are some more-or-less equivalent conditions:


* Bass--Serre theory says that a group is freely indecomposable if and only if it acts on a tree with trivial edge stabilizers and no global fixed point.
* More deeply, Stallings' Ends Theorem asserts that a finitely generated group splits over a *finite* subgroup if and only if it has more than one end.


As these are equivalent (modulo finite=trivial), you might say that they're not very interesting. Truly interesting necessary conditions are rather hard to write down, since being freely indecomposable is, in a sense, generic.


Here's one, much stronger, sufficient condition:



> 
> If a group $\Gamma$ has a finite generating set $S$ such that every element of $S$ is torsion and every element of $S^2$ is torsion then $\Gamma$ doesn't split at all, let alone freely.
> 
> 
> 


This is essentially a consequence of Helly's Theorem for trees.




---


Here's an outline of a proof. Suppose that $\Gamma$ acts on a tree $T$. We want to prove that $\Gamma$ fixes a point.


1. Prove that if $\gamma\in\Gamma$ has finite order then $\gamma$ fixes a point.
2. Prove that if the sets of fixed points of $\gamma,\delta\in\Gamma$ are disjoint then $\gamma\delta$ has infinite order. You can do this by constructing an isometric copy of $\mathbb{R}$ in $T$ that $\gamma\delta$ translates.
3. Prove Helly's Theorem for trees: if $\{T\_i}$ is a finite set of pairwise-intersecting subtrees of $T$ then $\bigcap\_i T\_i\neq \varnothing$.
4. Conclude as follows. Because every $s\in S$ is torsion, each such $s$ has a fixed point. Because $st$ is torsion for $s,t\in S$, the sets of fixed points pairwise intersect. By Helly's Theorem, there is a point fixed by every $s\in S$, and hence by $\Gamma$.


