TITLE: Are all unstable homotopy groups of $U(n)$ torsion?
QUESTION [7 upvotes]: The first few unstable homotopy groups of the unitary groups $U(n)$ were calculated by Borel-Hirzebruch, Toda, and Kervaire, and they are all torsion.  There is a paper by Matsunaga (details below) in which the p-primary parts of the next few homotopy groups are calculated.  This paper claims to be giving a complete description of these homotopy groups, but I don't see any explanation of why there are no free abelian factors.
My question is: Why $\pi_{2n+i} (U(n))$ is a torsion group for $i=3,4,5$?
More generally, is it know whether or not $\pi_* (U(n))$ is torsion for all $*>2n-1$? 
Matsunaga's paper is: 
MR182010 Matsunaga, Hiromichi Unstable homotopy groups of unitary groups (odd primary components). Osaka J. Math. 1 1964 no. 1, 15–24. 
It's freely available here: https://www.jstage.jst.go.jp/article/kyushumfs/15/1/15_1_72/_article
There's actually an errata list for the article that's not contained in the freely available .pdf.  It just corrects various notational errors (though not the obvious notational error in the main theorem, in which part b should refer to the 3-primary summand).

REPLY [16 votes]: Being an $H$-space, $U(n)$ has the rational homotopy type of a product of odd-dimensional spheres. As we know its cohomology, these are $S^1 \times S^3 \times S^5 \times \cdots \times S^{2n-1}$. In particular, its rational homotopy groups vanish above degree $2n-1$.

REPLY [12 votes]: A Sullivan minimal model for $U(n)$ is given by $\Lambda(x_1,x_3,\ldots,x_{2n-1})$ with zero differential. In fact for a Lie group, a minimal model is always a free algebra on odd generators and in this case, you can compute the degree of those generators by computing the cohomology of $U(n)$. 
The graded group $\pi_*(U(n))\otimes\mathbb{Q}$ is the vector space of indecomposable elements in the minimal model. Thus you get one non-torsion factors in the $i$-th homotopy of $U(n)$ for $i=1,3,\ldots,2n-1$ and all the other homotopy groups must be torsion.