Sufficient Conditions for Free Indecomposability An interesting fact was relayed to me in another question of mine 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?
 A: 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.


*

*Prove that if $\gamma\in\Gamma$ has finite order then $\gamma$ fixes a point.

*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.

*Prove Helly's Theorem for trees: if $\{T_i}$ is a finite set of pairwise-intersecting subtrees of $T$ then $\bigcap_i T_i\neq \varnothing$.

*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 $\Gamma$.

