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 $H_\ast(QX, \mathbb{Z})$?