There are many models for spectra, by which I mean a model category whose homotopy category is triangulated-equivalent to the stable homotopy category. In each model, there are ways to construct Eilenberg-MacLane spectra $HA$, where $A$ is an abelian group. In $S$-modules, this is described in Section IV.2 of this version of EKMM. In symmetric spectra, this is described in Example 1.14 of Stefan Schwede's book project (version 3.0).

Question:Are there models of Eilenberg-MacLane spectra that are fibrant, cofibrant, and (strict) abelian group objects with respect to the addition map $+ \colon HA \times HA \to HA$?

My first candidate was symmetric spectra because there, the construction of $HA$ follows directly from a standard construction of Eilenberg-MacLane spaces $K(A,n)$ as topological abelian groups (or simplicial abelian groups, if working in simplicial sets). In particular, $HA$ is an abelian group object. Moreover, $HA$ is an $\Omega$-spectrum, and those are the fibrant objects in, for instance, the absolute projective stable model structure (Theorem III.4.11 in Schwede's book, or Theorem 3.4.4 in Hovey—Shipley—Smith). However, I'm still missing cofibrancy, and I suspect that a cofibrant replacement would mess up the abelian group object structure.

Another idea would be to use the various stable model structures on symmetric spectra. One could also try in $S$-modules, where every object is fibrant.

For the record, an associative smash product is not crucial to my purposes, though of course it would be nice.

Algebras and Modulesgive conditions so that bifibrant monoids forget to bifibrant objects, and the condition holds for symmetric spectra (as is shown in Mandell-May-Schwede-Shipley). Alternately, you could try to put a model structure on the category of group objects and do cofibrant replacement there, but I've never done so. $\endgroup$ – David White Sep 11 '15 at 12:50Stable homotopical algebra and $\Gamma$-spaces, which might do the trick. I'm dealing with connective spectra, so I can work in $\Gamma$-spaces. Moreover, $HA$ is fibrant in the stable $Q$-model structure. I need to think some more about cofibrancy though. $\endgroup$ – Martin Frankland Sep 11 '15 at 15:56