Suppose I have a grouplike $A\_{\infty}$-space $G$ that carries an *additional structure* as a topological group (which does not coincide with the $H$-space structure of the $A\_{\infty}$-part). Denote the $H$-space multiplication by $\*$ and the group multiplication by $\cdot$ and suppose that 
$$
(a\*b) \cdot (c \*d) = (a \cdot c) \* (b \cdot d) 
$$
also known as the Eckmann-Hilton condition holds (equality, no "up to homotopy" here). Moreover, the identity of the group is a homotopy unit for the $H$-space structure. 



> 
> Is this enough to deduce that $BG$
>  coincides with the space $B(G\_{\*})$,
>  i.e. that the classifying space of the
>  topological group is (up to homotopy
>  of course) the delooping of the
>  $A\_{\infty}$-space? Do I get homotopy
>  commutativity as well?
> 
> 
> 



Eckmann-Hilton for $A_{\infty}$-spaces?
EDIT: Here is a *counterexample* to the stated question.


I'm going to start with a topological group $G$ which is a product of Eilenberg-Mac Lane spaces. Specifically, I'm going to choose $G \simeq K(\mathbb{Z},2) \times K(\mathbb{Z},7)$. The Dold-Thom theorem allows me to actually construct this as an abelian topological group: I take a based CW-complex with $H\_2 = \mathbb{Z}$, $H\_7 = \mathbb{Z}$, and other (reduced) homology groups zero (e.g. $S^2 \vee S^7$). Then I take the free topological abelian group on it, and this has the homotopy groups I want.


* The classifying space of $G$ using this multiplication is a $K(\mathbb{Z},3) \times K(\mathbb{Z},8)$.
* $G$ has a unique H-space structure. Because Eilenberg-Mac Lane spaces represent cohomology, $[G \times G, G]$ is $H^2(G \times G) \times H^7(G \times G)$. A calculation with the Kunneth formula tells you that the restrictions to each factor makes the map
$$[G \times G, G] \to [G,G] \times [G,G]$$
into an isomorphism. Thus, up to homotopy there is a unique binary multiplication that's homotopy unital; the group multiplication on $G$ is a representative.
* As a result, if I construct any other $A\_\infty$-structure on $G$, the binary multiplication is homotopic to the multiplication of $G$. Since $A\_\infty$-structures are insensitive to such a change by homotopy (yes, this is handwaving) I could equally well assume that the $A\_\infty$-structure starts with the multiplication of $G$. Because $G$ is abelian, this satisfies the interchange law.
* Since $A\_\infty$-structures pass along homotopy equivalences, it suffices for me to find a space $Y$ and a homotopy equivalence $G \to \Omega Y$ where $Y$ is not $K(\mathbb{Z},3) \times K(\mathbb{Z},8)$.
* There is a map $f: K(\mathbb{Z},3) \to K(\mathbb{Z},9)$ classifying the cohomology operation $x \mapsto x^3$ (which is nontrivial, and 2-torsion). I'm going to let $Y$ be the homotopy fiber of $f$. Then $\Omega Y$ is the homotopy fiber of a map $\Omega f: K(\mathbb{Z},2) \to K(\mathbb{Z},8)$ which is 2-torsion, and hence $\Omega f$ is trivial. Therefore $\Omega Y$ is homotopy equivalent to the homotopy fiber of the trivial map, and is a product homotopy equivalent to $G$.
* The map $f$ is the first k-invariant in the Postnikov tower of $Y$, and is nontrivial, so $Y$ is not a product of Eilenberg-Mac Lane spaces.


An exercise for the reader who is not in bed right now, for whatever reason, is to trace through why several simpler choices of cohomology operation (e.g. $Sq^2: H^2 \to H^4$) don't make this argument work.




---


Below follows my original post. I assumed that the interchange law held for the entire $A\_\infty$-structure, meaning that for any of the $n$-ary operations $m\_n$ we would have
$$
m\_n(x\_1 \ast y\_1, \ldots, x\_n \ast y\_n) = m\_n(x\_1, \ldots, x\_n) \ast m\_n(y\_1, \ldots, y\_n).
$$
However, the question only says that the H-space structure itself satisfies the interchange law, and not the higher coherences.


The Eckmann-Hilton condition implies that you can take classifying spaces twice, and the order of taking classifying spaces doesn't matter. You can write $BBG$ for this iterated classifying space. Both $B(G)$ and $B(G\_\*)$ are grouplike, and so the group completion theorem implies that the natural maps $B(G) \to \Omega BBG$ and $B(G\_\*) \to \Omega BBG$ are weak equivalences. Thus, under the standard types of point-set assumptions on $G$, both classifying spaces are weakly equivalent.


In addition, this implies that, because $G$ is grouplike, it is weakly equivalent to the 2-fold loop space $\Omega^2 BBG$ with either H-space structure, and hence both multiplications are homotopy commutative.


This is related to work of Brun, Fiedorowicz, and Vogt in "On the multiplicative structure of topological Hochschild homology." ([preprint version here](http://arxiv.org/abs/math/0410367)) They show that the "tensor product" of the associative operad with the little $n$-cubes operad is weakly equivalent to the little $(n+1)$-cubes operad. This means that having an associative multiplication and an $A\_\infty$-multiplication is, up to homotopy equivalence, basically the same as having an $E\_2$-multiplication.


