Let $X$ be a CW complex and let $\Sigma^\infty X$ denote its suspension spectrum. By definition, the $n$th singular homology group of $\Sigma^\infty X$ with coefficients in $\mathbb{Z}$ is $\pi\_n(\Sigma^\infty X \wedge H\mathbb{Z})$.


Now, connective spectra are in bijection with infinite loop spaces. The infinite loop space corresponding to $\Sigma^\infty X$ is $QX := \Omega^\infty \Sigma^ \infty X = \lim\_{n \to \infty} \Omega^n \Sigma^n X$. Is the homology of $\Sigma^\infty X$ (as a spectrum) isomorphic to the homology of $QX$ (as a CW complex)? I would expect this.


But what confuses me is a theorem is Rudyak's *On Thom spectra, orientability, and cobordism* (Theorem 7.11.(i)). Rudyak states that for every rational spectrum $E\_\mathbb{Q}$ the *Hurewicz map* $\pi\_\ast(E\_\mathbb{Q}) \rightarrow H\_\ast(E\_\mathbb{Q})$ is an isomorphism. This seems to prove that the homotopy/homology groups of spectra do not agree with homotopy/homology groups of their corresponding infinite loop spaces.


First, $\pi\_\ast(Q X) \otimes \mathbb{Q}$ is isomorphic to the stable rational homotopy $\pi\_\ast^S(X) \otimes \mathbb{Q}$ of $X$. Then, by the Milnor-Moore theorem, $H\_\ast(QX,\mathbb{Q}) \cong \mathbb{Q}[\pi\_\ast^S(X) \otimes \mathbb{Q}]$. So we'd have that $H\_\ast(X,\mathbb{Q})$ is the free commutative-graded algebra on the rational stable homotopy groups of $X$ for any CW complex $X$. But it is easy to write down examples of CW complexes whose rational homology is not free e.g. projective spaces. So what, then, is the relationship between $\pi\_\ast(\Sigma^\infty X \wedge H\mathbb{Z})$ and $H\_\ast(QX, \mathbb{Z})$?



Homology of spectra vs homology of infinite loop spaces
$\newcommand{\H}{\mathrm{H}} \newcommand{\Z}{\mathbf{Z}}$Let $X$ be a space. Then the $E$-(co)homology of $X$ is the same as the $E$-(co)homology of its suspension spectrum, i.e., $E\_\ast(X) \cong E\_\ast(\Sigma^\infty\_+ X)$ (and you remove the basepoint in $\Sigma^\infty\_+ X$ to get reduced $E$-homology). In the case when $E = \H\Z$, this says that $\H\_\ast(X;\Z) \cong \H\_\ast(\Sigma^\infty\_+ X;\Z)$. In the case when $E$ is the sphere spectrum, this says that the *stable* homotopy groups of $X$, i.e., the *unstable* homotopy groups of $QX$, are the same as the homotopy groups of the *spectrum* $\Sigma^\infty X$.


Let's now turn to the homology of $QX$. We know that $\H\_\ast(QX;\Z) \cong \H\_\ast(\Sigma^\infty\_+ QX;\Z)$, so you need to understand $\Sigma^\infty\_+ QX$. A theorem of Snaith's says that there is an equivalence
$$\Sigma^\infty\_+ QX \simeq \bigvee\_{n\geq 0} (\Sigma^\infty X)^{\wedge n}\_{h\Sigma\_n},$$
so we find that
$$\H\_\ast(QX;\Z)\cong \bigvee\_{n\geq 0} \H\_\ast((\Sigma^\infty X)^{\wedge n}\_{h\Sigma\_n};\Z).$$
The groups $\H\_\ast((\Sigma^\infty X)^{\wedge n}\_{h\Sigma\_n};\Z)$ can be computed by a homotopy orbits spectral sequence:
$$E\_2^{s,t} = \H\_\ast(\Sigma\_n; \H\_\ast((\Sigma^\infty X)^{\wedge n};\Z)) \Rightarrow \H\_\ast((\Sigma^\infty X)^{\wedge n}\_{h\Sigma\_n};\Z).$$
If $\Z$ is replaced by $\mathbf{Q}$, then this spectral sequence degenerates (the $E\_2$-page vanishes for $s>0$), and you find that $\H\_\ast((\Sigma^\infty X)^{\wedge n}\_{h\Sigma\_n};\mathbf{Q}) \cong \H\_\ast(\Sigma^\infty X;\mathbf{Q})^{\otimes n}\_{\Sigma\_n}$, i.e., $\H\_\ast(QX;\mathbf{Q}) \cong \mathrm{Sym}^\ast \H\_\ast(\Sigma^\infty X;\mathbf{Q})$. By the way, something special happens with $\mathbf{Q}$-cohomology of spectra: since the $\mathbf{Q}$-localization of the sphere spectrum is $\H\mathbf{Q}$, you find that $\pi\_\ast(F\_\mathbf{Q}) \cong \H\_\ast(F\_\mathbf{Q};\mathbf{Q}) \cong \H\_\ast(F;\mathbf{Q})$ for any spectrum. (This is more generally true for any *smashing localization*.) (Edit: Whoops, I didn't actually finish my answer.) This means that $\H\_\ast(\Sigma^\infty X;\mathbf{Q}) \cong \pi\_\ast(\Sigma^\infty X\_\mathbf{Q}) \cong \pi\_\ast(\Sigma^\infty X) \otimes \mathbf{Q} = \pi\_\ast^s(X) \otimes \mathbf{Q}$, so you find that
$$\H\_\ast(QX;\mathbf{Q}) \cong \mathrm{Sym}^\ast(\pi\_\ast^s(X) \otimes \mathbf{Q}),$$
as you said in your post. The homology of $QX$ with coefficients in a general homology theory $E$ is a lot more complicated: if $E$ is a structured ring spectrum, it describes what are known as *Dyer-Lashof operations* for $E$.


