Does a graded vector space isomorphism between the homology of two loop spaces imply the existence of an algebra isomorphism? Consider ordinary homology with coefficients in a field. For $X$ a path-connected pointed space, the graded vector space $\bigoplus_{q\ge 0} H_q(\Omega X)$ has the structure of an algebra with the multiplication induced as follows: $$ H_p(\Omega X) \otimes H_q(\Omega X) \rightarrowtail H_{p+q}(\Omega X \times \Omega X) \to H_{p+q}(\Omega X)$$
Here, the first map is a monomorphism by Kunneth's theorem. The second map is induced by the concatanetion of loops $\Omega X \times \Omega X \to \Omega X$.
Let two path-connected pointed spaces $X$ and $Y$ be given. Suppose that, for each $q\ge 0$, the two vector spaces $H_q(\Omega X)$ and $H_q(\Omega Y)$ have the same dimension. Hence $\bigoplus_{q\ge 0} H_q(\Omega X)$ and $\bigoplus_{q\ge 0} H_q(\Omega Y)$ are isomorphic as graded vector spaces. Is it necessary that $\bigoplus_{q\ge 0} H_q(\Omega X)$ and $\bigoplus_{q\ge 0} H_q(\Omega Y)$ are isomorphic as algebras?
 A: Here's a partial positive answer, in response to your comment.  Suppose $X=\Sigma X'$ and $Y=\Sigma Y'$ are suspensions of connected spaces whose homology is finitely generated in each degree.  Then if $H_*(\Omega X)$ and $H_*(\Omega Y)$ have the same dimension in each degree, they are isomorphic as algebras.  Indeed, a theorem of Bott and Samelson says that $H_*(\Omega\Sigma X')$ is naturally isomorphic to the tensor algebra on the reduced homology $\tilde{H}_*(X')$.  Under the assumed finiteness conditions, the dimensions of $\tilde{H}_*(X')$ can be recovered from the dimensions of $H_*(\Omega \Sigma X')$, and so the algebra structure on the latter is uniquely determined (up to isomorphism) by its dimensions.
A: No.
Let $X = B\Bbb Z/4$ and $Y = B(\Bbb Z/2 \times \Bbb Z/2)$ be classifying spaces for the two groups of order four. Then, as loop spaces, $\Omega X$ and $\Omega Y$ are homotopy equivalent to the discrete spaces $\Bbb Z/4$ and $\Bbb Z/2 \times \Bbb Z/2$ respectively.  
Their rational homology groups are the same, because the underlying spaces $\Omega X$ and $\Omega Y$ are homotopy equivalent: both of them are homotopy equivalent to a discrete set with four points. However, $H_*(\Omega X)$ is the group algebra $\Bbb Q[\Bbb Z/4]$ and $H_* (\Omega Y)$ is the group algebra $\Bbb Q[\Bbb Z/2 \times \Bbb Z/2]$. Those are nonisomorphic rings: one has three prime ideals, and the other has four.
