I should probably add that I hope you are right, since I have a student who hopes to prove that you are (homotopically of course).

Interesting. But I think to give it substance you would have to display a manifold of dimension 62 and Kervaire invariant one.

Surely an alternative proof is not irrelevant! In the Rothenberg-Steenrod spectral sequence, when H_*(G;k) is an exterior algebra, E_2 is a polynomial algebra, there are no differentials for dimensional reasons (if char k is not 2) and it follows immediately that H^*(BG;k) is a polynomial algebra. In char 0, being an exterior algebra is equivalent to being commutative. McCleary's 6.38 is stated for general (perfect) fields, with the evident hypotheses. The universally transgressive hypothesis ensures that H_*(G;k) is an exterior algebra, which is what connects the proofs. Peace.