How do one prove (or disprove) that $\Omega S^{2}$ and $\Omega(S^{2} \vee S^{2})$ are non commutative Hopf spaces?


I thought this is a question for math.stackexchange, but not many people even viewed that question. Seems like it should be easy but right now I have no idea how to do it. Any hint? 


Definition: If $X$ is commutative Hopf space if, for the multiplication $
\mu $ there is a homotopy between $ H: \mu \simeq \mu \circ T $, where $T : X \times X \to X \times X$ is the switch map. 


Clarification: When I say multiplication, I mean multiplication induced by the loop structure. On $\Omega S^{2}$ there is another multiplication which is induced by multiplication on $S^{1}$ as $\Omega S^{2} = \Omega \Sigma S^{1} $ ( This is of course commutative).



Non-commutativity of certain Hopf spaces
Let us run the Serre SS for homology with $\mathbb{Z}/2$ coefficients for the fibration 


$\Omega(S^{2} \vee S^{2}) \to \* \to S^{2} \vee S^{2}$


Let $x,y \in H\_{2}(S^{2} \vee S^{2})$ be the generators. Since these guys has to die, there exists $x\_{1}, y\_{1} \in H\_{1}(\Omega(S^{2} \vee S^{2}))$ which are hit by $x$ and $y$ respectively. 


Notice that there are $4$ terms at $E^{2}\_{2,1}$ namely $x\_{1}x, y\_{1}x, x\_{1}y, y\_{1}y$.


$d\_{2}(x\_{1}y) = x\_{1}y\_{1}$ and $ d\_{2}(y\_{1}x) =y\_{1}x\_{1} $.


So $x\_{1}y + y\_{1}x$ survives if $x\_{1}y\_{1} = y\_{1}x\_{1}$ and never dies. Which is a contradiction. 


this shows that in homology the elements $x\_{1},y\_{1}$ do not commute. So $\Omega(S^{2} \vee S^{2})$ is not a commutative Hopf space. 


I think this is correct. But the case of $\Omega S^{2}$ still remains unanswered. 


Edit: The complete answer is given by Justin in the comments below.


