Let $G$ be a simplyconnected compact topological group (you can think of $SU(n)$ if you like it more concrete), and let $X$ be a finitedimensional simplyconnected $G$CWcomplex. 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)?
EDIT: This is wrong, as Jens explains below.
It is enough to show that the isotropy groups of the 0cells are trivial: higher dimensional cells must have smaller isotropy.
We may suppose, by restricting to $G$skeleta, that $X$ only has $G$CWcells of dimension $\leq 1$. Then $X$ is pathconnected 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 nontrivial groups (along injective homomorphisms) is never trivial. Thus if $X$ is simplyconnected then the isotropy of the 0cells must be trivial, as required.

3$\begingroup$ I am skeptical about this argument. Why can we assume that there are only Gcells of dimension $\leq 1$? Actually, why can't we start with $G/(\mathbb Z/k)$ for $k$ arbitrarily large (that is the unique 0cell) and attach a 2cell $G \times D^2$ via a map that equivariantly extends a generator $S^1 \to G/(\mathbb Z/k)$ of the fundamental group $\pi_1(G/(\mathbb Z/k)) = \mathbb Z/k$? This example should answer my question in the negative. $\endgroup$ – Jens Reinhold Feb 21 '17 at 20:19

1$\begingroup$ I suppose that is right. I had mistakenly thought that because G is simplyconnected that 2cells and higher did not matter for the fundamental group , but of course the trivial group shows this is not the case... $\endgroup$ – Oscar RandalWilliams Feb 21 '17 at 20:59

$\begingroup$ @JensReinhold, since your comment seems to answer the question, why don't you post it as an answer? $\endgroup$ – HJRW Feb 22 '17 at 21:33
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 simplyconnected, with finite, but nontrivial isotropy subgroups.