# Is the special orthogonal group really rationally homotopy commutative?

It is a classical result that the rational cohomology of $SO(n)$ is given by:

$$H^*(SO(2m); \mathbb{Q}) = \begin{cases} S(\beta_1, \dots, \beta_{m-1}, \alpha_{2m-1}) & n = 2m \\ S(\beta_1, \dots, \beta_m) & n = 2m+1 \end{cases}$$ where the $\beta_i$ have degree $4i-1$, $\alpha_{2m-1}$ has degree $2m-1$, and $S(V)$ is the free graded commutative algebra on the generators given by the graded vector space $V$. Since all the generators have odd degree, this happens to be an exterior algebra here.

Since $SO(n)$ is a (Lie) group, there is an induced Hopf algebra structure on $H^*(SO(n);\mathbb{Q})$, and the generators $\beta_i$ (and $\alpha_{2m-1}$ if applicable) happen to be primitive. But then, a free (graded) symmetric algebra on primitive generators is (graded) cocommutative, so in this case $H^*(SO(n))$ is cocommutative.

Now, like all Lie groups, $SO(n)$ is formal, and all its rational homotopy theory is contained in its cohomology ring. This is where things start to get muddy for me: if I understand correctly, the coproduct on $H^*(SO(n))$ should be a model for the product on $SO(n)$. Since $H^*(SO(n))$ is cocommutative, then the product on $SO(n)$ is rationally homotopy cocommutative.

Is this correct? If not, where's my mistake? If yes, how to interpret that fact? The group $SO(n)$ doesn't really "look" commutative at all. Maybe I'm assigning a stronger meaning than I should to the commutativity up to rational homotopy?