Almost free actions on simply-connected spaces Let $G$ be a simply-connected compact topological group (you can think of $SU(n)$ if you like it more concrete), and let $X$ be a finite-dimensional simply-connected $G$-CW-complex. If we know that all the isotropy groups are finite, does this imply they are trivial? And if not, are they at least bounded in size (in some sense)?
 A: EDIT: This is wrong,  as Jens explains below. 
It is enough to show that the isotropy groups of the 0-cells are trivial: higher dimensional cells must have smaller isotropy.
We may suppose, by restricting to $G$-skeleta, that $X$ only has $G$-CW-cells of dimension $\leq 1$. Then $X$ is path-connected if and only if $X/G$ is, in which case its fundamental group is given by the fundamental group of the graph of groups with underlying graph $X/G$ and edges and vertices decorated by their (finite) isotropy groups. The fundamental group of $X$ is therefore obtained as an amalgamated free product followed by HNN extensions (where all homomorphisms are given by conjugation in $G$, so in particular are injective). An HNN extension is never trivial (as it has $\mathbb{Z}$ as a quotient) and an amalgamated free product of non-trivial groups (along injective homomorphisms) is never trivial. Thus if $X$ is simply-connected then the isotropy of the 0-cells must be trivial, as required.
A: Since it was requested, here is the answer (from the comment) again.
The statement is wrong. For each $k$, one can start with a unique $0$-cell $G/(\mathbb Z/k)$ and attach a $2$-cell $G \times D^2$ by a map $G \times S^1 \to G/(\mathbb Z/k)$ that equivariantly extends a generator $S^1 \to G/(\mathbb Z/k)$ of the fundamental group $\pi_1(G/(\mathbb Z/k)) = \mathbb Z/k$. The resulting $G$-CW complex will be simply-connected, with finite, but non-trivial isotropy subgroups.
