It is the case that the wedge product of aspherical spaces is aspherical, so in particular $K(G, 1) \vee K(H, 1) \simeq K(G*H, 1)$.
I know homotopy groups are notoriously poorly behaved with respect to coproducts/pushouts, but is an analogous result true for at least for Eilenberg-MacLane spaces with higher and arbitrary dimensions? Specifically, if $X\simeq K(G, n) \vee K(H, m)$ for $n, m > 1$, then could anyone verify or provide counter-examples to the following claims:
1) If $n = m$ then $X \simeq K(G\oplus H, n)$.
2) If $n \neq m$ then $\pi_n(X) \cong G$, $\pi_m(X)\cong H$ and $\pi_k(X) = 0$ if $k\neq n, m$.
Note that in order to expect 2) to be true for all $G, H$ we need both $n$ and $m$ to be greater than $1$, because if we choose $G$ to have an element of infinite order, $n=1$, and $H$ to be finitely generated then $\pi_m(X)$ will nevertheless be infinitely generated (by considering the universal cover) and so can't be $H$.
For 1) I at least know that $\pi_n(X) \cong G \oplus H$ by the Hurewicz theorem, but I don't know how to address the higher homotopy groups of the wedge. For 2) I know that if $n < m$ then $\pi_n(X) \cong G$ and otherwise $\pi_m(X) \cong H$ but again I don't know how to approach the higher homotopy groups. My feeling is that 2) is much less likely to be true than 1).