How to show that the homotopy group $\pi_4(U(3))$ of a unitary group is finite I have dealt with the stable unitary groups and calculated the groups for $\pi_i(U(n))$ when $i=1,2,3$ and I know that $\pi_4(U(3))\cong \pi_4(U(4))$ and the fibration $U(n)\to U(n+1) \to S^{2n+1}$ which I've tried to use the LES but haven't gotten far. I'm aware of the isomorphism to $SU$, but I don't think that should be necessary here.
 A: From
$$
\begin{aligned}
&
\color{gray}{
\pi_5(U(2))\to
\pi_5(U(3))\to}
\pi_5(S^5)
&\searrow
\\
&&\swarrow
\\
&
\longleftarrow\longleftarrow\longleftarrow\longleftarrow\longleftarrow\longleftarrow\longleftarrow\longleftarrow\longleftarrow
&&
\\
\swarrow\quad
&
\\
\searrow\quad
&
\\
&\pi_4(U(2))\to
\pi_4(U(3))\to
\underbrace{\pi_4(S^5)}_{=0}
\end{aligned}
$$
we get a surjection $\color{blue}{\pi_4(U(2)) \twoheadrightarrow\pi_4(U(3))}$.
Using
$$
\underbrace{U(1)}_{S^1}\to U(2)\to \underbrace{U(2)/U(1)}_{S^3}$$
we get the isomorphisms in higher degrees of the homotopy groups for $U(2)$ and $S^3$.
In particular,
$$
\pi_4(U(2))\cong \pi_4(S^3)\cong \Bbb Z/2\ ,
$$
the last isomorphism taken from the Wiki Table.

To see that in the end $\pi_4(U(3))=0$, one has maybe to investigate the map
$$
\Bbb Z\cong
\pi_5(S^5)\to\pi_4(U(2))\to \pi_4(U(2)/U(1))=\pi_4(S^3)\overset{\text{Hopf}}\cong \pi_4(S^2)
\cong \Bbb Z/2
$$
and show, it is not the zero map.

The catlab-links are providing some more information.
A good reference is the Mimura-Toda paper from 1963.

*

*ncatlab.org/nlab/show/unitary+group

*ncatlab.org/nlab/show/unitary+group#MimuraToda63
A: One can completely compute the rational homotopy groups of an H-space simply by looking at the cohomology. It is always a symmetric algebra (i.e. a tensor product of an exterior algebra on the odd degree generators and a polynomial algebra on the even degree generators), and we have the property that the rank of the nth rational homotopy group is equal to the number of generators of degree $n$.
In the case of $U(3)$ we have that its cohomology is $\Lambda[x_1,x_2,x_3]$ where $|x_i|=2i-1$. Since there is no generator in degree $4$, we see that $\pi_4(U(3)) \otimes \mathbb{Q}=0$ hence $\pi_4(U(3))$ is torsion.
