The Eilenberg-MacLane spectrum and retractions I want to show that $H\mathbb{Z}$ is not a retract of $ku$, where $ku$ is the connective cover of the complex $K$-theory and $H\mathbb{Z}$ is the Eilenberg-MacLane spectrum.
 A: If you had a map $H\mathbb Z\to ku$ inducing an isomorphism on $\pi_0$ then you could compose with the usual map $ku\to KU$ to get a map $H\mathbb Z\to KU$ inducing an isomorphism on $\pi_0$. This can be ruled out in a number of ways, I imagine, but here's one: Using the periodicity equivalence $ \Sigma^{2j}KU\sim KU$ you would then get maps $\Sigma^{2j}H\mathbb Z\to KU$ for all integers $j$ such that all together they give an equivalence $\coprod_j \Sigma^{2j}H\mathbb Z\sim KU$.
ADDED in response to the reasonable complaint that I merely reduced the question to a similar question:
To me, and perhaps to many of us, $KU$ is a more familiar object than $ku$. I thought it was worth pointing out that if one could split off that one $\mathbb Z$ from $ku$ then one could split off all the $\mathbb Z$'s from $KU$. And "everybody knows that" $KU$ is not a product of Eilenberg-Maclane spectra, i.e. that periodic $K$-theory is truly an extraordinary cohomology theory.
One quick way of verifying this is to observe that while the reduced integral cohomology groups of $\mathbb RP^{2n}$ are killed by $2$ the group $\tilde KU(\mathbb RP^{2n})$ is not, if $n\ge 2$. In fact, let $L$ be the complexification of the nontrivial line bundle; $c_1(L)$ is the nontrivial element of $H^2$, so that $c_2(L\oplus L)=c_1(L)\cup c_1(L)$ is the nontrivial element of $H^4$ and $L\oplus L$ is not stably trivial.
A: If it were the case, then $K(Z,n)$ would be a retract of $\underline{ku}_n$ where $\underline{ku}_n$ is the $n$-th infinite loop space associated to the spectrum $ku$.  However, $ku$ is what is called $BP\langle 1 \rangle$ (at the prime 2), and we know that the (2-local) homology of $BP\langle 1 \rangle _j$ is torsion free by work of Wilson
{\it The Ω-spectrum for Brown-Peterson cohomology. Part II}, American Journal of Mathematics, {\bf 97} 1975, 101-123  whereas the homology of $K(Z,3)$ is not.
Edited after comments by Dylan Wilson and Achim Krause
Here is a simple argument following the original idea of the OP (deleted from the post later).
Consider the cofibration $\Sigma^2ku \to ku \to HZ$.  Take the mod 2 homology to
get a long exact sequence $$HZ/2 _{*-2}(ku) \to HZ/2 _*(ku) \to HZ/2 _*(HZ)\to HZ/2 _{*-3}(ku).$$  By induction on *, we see that the connecting homomorphism
is epi, so that as graded vector spaces $$HZ/2 _*(HZ) \cong  HZ/2 _*(ku)
\oplus HZ/2 _{*-3}(ku).$$
This implies that $HZ/2 _*(HZ)$ can't be a summand of $HZ/2 _*(ku)$.  It also
shows, even without any knowledge of $HZ/2 _*(HZ)$, $HZ/2_3(ku)$ is larger than 
 $HZ/2 _3(HZ)$.  Thus contradicting retraction.
To get the claim in Dylan Wilson's comment for $HZ/2_3HZ$ it suffices to play the same game with the cofibration $HZ \to HZ \to HZ/2$.
Incidentally, by analyzing the long exact sequence which splits as short exact sequences, we end up with the description of $HZ/2_*HZ$ in my comment and that of $HZ/2_*(ku)$ in Achim Krause.
