A proof that every compact Lie Group has torsion second homotopy group I am trying to prove that every compact lie group has torsion second homotopy group but I get stuck. My argument is the following:
Since $\Pi_1(G)$ is finite the universal cover G* of G is also a compact lie group so it suffices to find the second homotopy group of G*. Now since $\Pi_1(G^*)=0$ $\Rightarrow$$H_1(G^*;Z)$=O $\Rightarrow$$H^1(G^*;Z)$=O $\Rightarrow$ $H^1(G^*;Q)$=O 
(By universal coefficient theorem of cohomology)
Now by a theorem of hopf we know the cohomology ring of $G^*$ with Q  coeffiients is isomorphic to an exterior algebra over Q on odd generators $\Rightarrow$ $H^2(G^*;Q)$=O.
 Now I need to somehow deduce that $H_2(G^*;Q)$=O because then using universal coefficient theorem for homology I will get $H_2(G^*;Z)$ $\otimes$Q=$0$ $\Rightarrow$ $H_2(G^*;Z)$ is torsion $\Rightarrow$ $\Pi_2(G^*)$ is torsion by Hurewich theorem.
So is there any way to deduce $H_2(G^*;Q)$=O?
Thanks in advance

In fact over a field $\Bbb F$ with $H_*$ finite dimensional, $H^k(X, \Bbb F) \cong H_k(X,\Bbb F)$ for any CW-complex $X$. This is because $\operatorname{Ext}_{\Bbb F}(A,B)=0$ as $A$ and $B$ are free (all modules over a field are free). So the exact sequence from universal coefficients  $$ 0 \to \operatorname{Ext}_{\Bbb F}(H_{k-1}(X,\Bbb F))\to H^k(X,\Bbb F)\to \operatorname{Hom}(H_k(X,\Bbb F),\Bbb F)\to 0$$
becomes 
$$ 0 \to 0\to H^k(X,\Bbb F)\to \operatorname{Hom}(H_k(X,\Bbb F),\Bbb F)\to 0$$
As $V\cong V^*$ for finite dimensional vector spaces, we get the desired isomorphism.
This done near the beginning of Ch. 3 of Hatcher. 
