# Rational homotopy type of a Lie group

I have to show that Lie groups have the rational homotopy type of a wedge of spheres. Unfortunately, my knowledge of Sullivan models is a bit shaky. The proof I come up with shows that they have the RHT of a product of spheres, which should be something else (I guess):

Let $$G$$ by a Lie group. Therefore, $$G$$ is an H-space and it's cohomology algebra $$H^*(G,\mathbb Q)$$ is a Hopf algebra (see Hatcher p283). By a theorem of Hopf, $$H^*(G,\mathbb Q)\cong \Lambda V$$ is isomorphic to a free graded-commutative algebra (see Hatcher 3C.4). Since $$G$$ is a manifold, it's higher cohomology-groups are trivial, so $$V$$ cannot have generators in even degree.

Now it is easy to see, that $$\Lambda V$$ with the zero-differential is a minimal Sullivan model of $$G$$.

Next, I want to factorize the sullivan model $$\Lambda V \cong \bigotimes\limits_i \Lambda(e_i)$$ , where all $$e_i$$ have odd degree. If I look at the product of spheres $$P=\prod\limits_i S^{|e_i|}$$, then I get a Sullivan model for $$P$$ by taking the tensor product of the Sullivan models for each single sphere (by Example 2 of chapter 12 in Félix/Halperin/Thomas' Rational Homotopy Theory). By Example 1 on the same page, the Sullivan model of $$S^{|e_i|}$$ is exactly $$\Lambda(e_i)$$. So I conclude $$\Lambda V$$ is a Sullivan model for $$P$$ and $$G$$, so $$P$$ and $$G$$ have the same rational homotopy type. But my aim was to show that $$G$$ has the RHT of a wedge of spheres.

Where am I mistaken? Where should I do something else to obtain the desired consequence?