I want to build a finite CW complex such that $\pi\_1$ is non-abelian and $H\_i$ are zero for $i\geq 2.$
From Hatcher for a given group G, one can create an example of a 2-complex $X\_G$ with $\pi\_1(X\_G)=G.$ I also checked from Mayer-Vietoris that if $G$ is cyclic such complex won't have any higher homology for $i\geq 2.$ I tried to take $G=S\_3,$ the symmetric group of order 6 and from Mayer-Vietoris I get $H\_2$ is $Z.$ I believe this was a correct calculation. Or is there a way to get the groups $G$, with $G$ non-abelian, such that we can get $\pi\_1 =G$ and $H\_i=0$ for $i\geq 2.$


Any reference or idea to create such an example? Or is there a way to claim such a finite complex can't exist?



Finite CW complex with finite non-abelian fundamental group and higher homologies zero
**Theorem.** Let $G$ be a group. There exists a finite 3-complex $X\_G$ with $\pi\_1 X\_G = G$ and $H\_i X\_G = 0$ for $i > 1$ if, and only if, $G$ is finitely presentable and has second group homology $H\_2(G) = 0$.


The more interesting question to me is whether this is possible for a *finite 2-complex*, and Jens Reinhold's answer gives us the first step there. This is possible for the group $2I$ mentioned in my comment above.




---


**Lemma:** Let $X$ be any CW complex. Then the cokernel of the Hurewicz map $\pi\_2 X \to H\_2 X$ is isomorphic to $H\_2(\pi\_1 X)$. (Proof: attach cells to make $X$ into a $K(\pi\_1 X, 1)$; doing so kills off precisely $\pi\_2 X$ inside of $H\_2 X$.) This is exercise 23 in Hatcher section 4.2.


**Proof of theorem.** Pick any finite presentation $P$ of $G$ and construct the presentation complex $X\_P$. This is a finite complex with $\pi\_1 X\_P = G$, and with $H\_2(X\_P)$ a free abelian group $\Bbb Z^k$ on a finite number of generators. Because the cokernel of $\pi\_2(X\_P) \to H\_2(X\_P)$ is precisely $H\_2(\pi\_1 X\_P) = H\_2(G) = 0$, it follows that $\pi\_2(X\_P) \to H\_2(X\_P) = \Bbb Z^k$ is surjective. Now choose $k$ maps $\rho\_i: S^2 \to X\_P$ so that these give a basis of the second homology group, and set $$X\_G = X\_P \cup\_{i=1}^k D^3,$$ attaching these 3-cells along the $\rho\_i$. The resulting complex has $H\_i X\_G = 0$ for $i > 1$ by a Mayer-Vietoris argument, while $\pi\_1 X\_G = \pi\_1 X\_P \cong G$ by the van Kampen theorem.


