Homotopy groups of non-contractible manifolds Motivated by a proof in a differential geometry book and so far my lack of knowledge in algebraic topology I would like to know the following :

Is it possible to have a compact non-contractible manifold $M$ with finite fundamental group and trivial homotopy groups $\pi_k(M)$ for $k\geq 2$?

Any help or reference is appreciated, thanks in advance.
 A: This is impossible even without assuming compactness (assuming you want $M$ to be connected).  Suppose you had such a manifold $M$.  Then $M$ would be a $K(G,1)$ space for the finite nontrivial group $G=\pi_1(M)$.  However, any nontrivial finite group has infinite cohomological dimension, so $M$ would have nontrivial cohomology (with appropriate coefficients) in infinitely many degrees.  This is a contradiction, since if $M$ is a manifold of dimension $n$ it can't have nontrivial cohomology above degree $n$.
A: This can also be achieved through Miller's theorem which states that the based mapping space $Map_*(BG,X)$ is weakly contractible if $G$ is a Lie group with a finite number of path components and $X$ is a nilpotent $CW$ complex with a finite number of cells.
In particular all maps $BG \rightarrow X$ are nullhomotopic.
Your $M$ is a $K(\pi_1(M),1)$ so also a $B\pi_1(M)$. Obviously $G = \pi_1(M)$ viewed as a discrete Lie group has finite number of path components and $M$ also has to be a finite $CW$ complex since it's compact. Thus the identity map $M =BG \rightarrow M$ is nullhomotopic and $M$ is contractible.
