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?

  • $\begingroup$ see Henry Wilton's answer to your other question about groups with infinitely many ends. $\endgroup$ – Ian Agol Mar 10 '11 at 4:43
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    $\begingroup$ As this question doesn't have one correct answer, it should probably be community wiki. $\endgroup$ – HJRW Mar 10 '11 at 5:00

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 $\Gamma$.
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  • $\begingroup$ I apologize for the ignorance, but is that a variation/application of Helly's theorem on convex sets? Could you point me to a proof of the sufficient condition above? $\endgroup$ – JeremyKun Mar 11 '11 at 3:30
  • $\begingroup$ One proof is basically Example 6.3.1 in Serre's book Trees. Serre's proof is rather algebraic, but the proof I had in mind is more geometric. See the edit above. $\endgroup$ – HJRW Mar 11 '11 at 22:04

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