When can you desuspend a homotopy cogroup? Any topological group $G$ has a classifying space, whose loopspace is a (homotopy) group which is homotopy equivalent to $G$ in a way that preserves the group structure.  More generally, if $G$ is an $A_\infty$-group (a space with a binary operation which satisfies the group axioms up to coherent homotopy), it similarly can be delooped to a classifying space.
Now suppose you have a cogroup in the category of pointed spaces.  If it is actually literally a cogroup, it's not hard to show it must be a point.  However, up to (coherent) homotopy, the suspension of any space is a cogroup.  Is the converse true?  Can you desuspend any $A_\infty$-cogroup?  Are there any examples of homotopy cogroups (possible not $A_\infty$) which are not suspensions?  More generally, are there any criteria that you can use to prove that a space does not have the homotopy type of a suspension?  The only one I know is that all cup products must vanish, but this also holds automatically for a homotopy cogroup (indeed, for any "co-H-space").
 A: Whilst the n-lab page on co-H-spaces could be described as a little meagre, it does nonetheless contain a reference to a paper in the Handbook of Algebraic Topology.  Various parts of this book are in the "preview" in google books, in particular page 1153 which contains the magical sentence:

We can now construct cell complexes $S^n \cup_\alpha e^m$ which are co-H-spaces but not suspensions.

This follows a theorem which classifies when such spaces are co-H-spaces and suspensions.
The specific article is the aptly named "Co-H-spaces" by Martin Arkowitz, and is chapter 23 of Handbook of Algebraic Topology, edited by Ioan James.  I would recommend this as a first place to look to find out more about these spaces.
Incidentally, I don't believe your remark about delooping.  I'm almost certain that there are H-spaces that can't be delooped so either there are H-spaces that aren't $A_\infty$-groups or your claim is incorrect.  Unfortunately, I'm not currently surrounded by my usual stock of algebraic topology textbooks so can't look this up (and you can do an internet search as well as I can).  Can you supply a reference for the delooping claim?
A: Hopkins' result alluded to above gives a coordinate-free approach a lá Segal. The paper I wrote with Schwänzl and Vogt:
Comultiplication and suspension. Topology Appl. 77 (1997), no. 1, 1–18,
gives an actual co-operadic approach: we show that a 2-connected space which has a homotopy everything co-action by the Stasheff operad (i.e., a co-$A_\infty$-structure) is always a suspension.
This paper answers the first of the questions posed above.
There is an open question about this matter which I'd like to bring up. Hopkins claims his result holds for 1-connected spaces, but I and my co-authors could only get our result to work for 2-connected ones. Hopkins' proof, which was based on the Bousfield spectral sequence, never appeared (our proof involved the higher Blakers-Massey theorems). 
Hopkin's result is stated, but not proved, in: Formulations of cocategory and the iterated suspension. Algebraic homotopy and local algebra (Luminy, 1982), 212-226, Astérisque, 113-114, Soc. Math. France, Paris, 1984. Thus we don't know if the 1-connected case has really been settled or not.
Additional Note: Schwänzl, Vogt and I also show that if the space is highly connected with respect to its CW dimension (i.e., $(n-1)$-metastable), then the existence of a co-$A_n$-structure on it is tantamount to the existence a co-$A_\infty$-structure. It follows that it too desuspends.
A: This is a question I've asked myself, but never found an answer.  By analogy with the story for A-infinity spaces and topological groups, you would like to be able to start with a co-A-infinity space Y, build a "cobar complex" on it, then take the homotopy limit of that complex to obtain a space X, and then hope that ΣX=Y.
If this has any chance, it had better work when Y is already known to be a suspension.  In the paper by Goerss,  "Barratt's desuspension spectral sequence and the Lie ring analyzer", Quart. J. Math Oxford, 44 (1993), something like this is proved.  He builds (actually, he describes how Barratt builds) a cosimplicial space whose n-th space is homotopy equivalent to an n-fold wedge of ΣX, and shows for simply connected X that the Tot of this complex recovers X.  
Given this, I suspect it should be possible to prove that "co-A-infinity"="suspension", at least for 2-connected spaces.
Edit: some of these ideas get discussed in the commentary under the question; since they were hidden, I didn't notice them till now.
A: Mike Hopkins tells me it is indeed true that for any A∞ cogroup space Y, the homotopy limit X of the associated cobar complex is a desuspension of Y (Y = ΣX as A∞ cogroups)--we don't even need any connectivity assumptions on Y.
Edit: I think the way to prove this is to look at the spectral sequence for the homology of X.  It's easy to see that algebraically it degenerates on the E2 page to something which could be the homology of a desuspension of Y.  I don't know how to check that this algebraic convergence has anything to do with X, or how to handle π0 and π1 issues.  However, in the case that Y is the suspension of a discrete pointed set, one can check by hand that X works out to be the original pointed set, which makes it plausible that these low-dimensional issues don't cause problems.
