@Alessandro So take a presentation $\langle g_i | r_i \rangle$ of $G$. Look at a wedge of $n$-sphere $\bigvee S^n$ indexed over the generators $g_i$. Attach cells $D^{n+1}$ corresponding to relators $r_i$.
@Alessandro In the event that I could make a case for doing grad geometry without officially taking algebraic topology I'd be up for doing some this summer
Indeed, this does not even work for $n = 2$, $G = \Bbb Z$. You have $S^2$ from my construction, which has a lot of higher homotopy groups. $\pi_3$ is nonzero.
The trick is to take this CW complex $X$ I constructed, which might have nonzero $\pi_k(X)$ for $k \geq n+1$, so you take a map $f : S^k \to X$ representing a nontrivial $k$-homotopy class, and glue a cell $D^{k+1}$ along it (that is, glue $\partial D^{k+1}$ by $f$)
Do it for ALL nontrivial homotopy classes in $\pi_k$, for ALL $k$. So that's a big big big space, possibly infinite dimension (usually so), which has $\pi_n = G$ and $\pi_k = 0$ for all $k \neq n$
@AkivaWeinberger Really? Are you, like, absolutely sure?! I mean, as far as English is concerned, punctuation is all around; especially here, and there.
@Alessandro Neat, isn't it? In any case, if you want $\pi_i = G_i$ for $i = 1, 2, \cdots$ then $K(G_1, 1) \times K(G_2, 2) \times \cdots \times K(G_i, i) \times \cdots \times$ is a space with those specified homotopy groups.
