TITLE: Almost free actions on simply-connected spaces
QUESTION [6 upvotes]: 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)?

REPLY [2 votes]: 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.