# Cayley graph is not Eilenberg Maclane space by itself?

Let $$G$$ be a group having presentation $$\langle g_i|r_i\rangle$$ with $$g_i$$ generators and $$r_i$$ relations. One can construct its associated topological space $$X_G$$ by wedging $$S^1$$ and putting $$2$$-cells to each relation that has corresponding words.

$$X_G$$ is path connected, semi local simply connected, local path connected. Thus $$X_G$$ has simply connected universal covering.

$$\textbf{Q:}$$ Why $$X_G$$ itself is not Eilenberg Maclane space?(i.e. It is not $$K(G,1)$$.) Hatcher went through construction of $$BG$$ space and showed this space is $$K(G,1)$$.

Ref. Hatcher, Algebraic Topology Chpt 1, Appendix 1.B Example 1B.7

• You might have misread the construction in Hatcher, it does not stop after attaching 2-cells. Once $X_G$ is constructed, one considers $\pi_2(X_G)$, and attaches 3-cells to kill $\pi_2$. One then considers $\pi_3$, and attaches $4$-cells to kill $\pi_3$. And so on inductively. Commented Jun 15, 2019 at 21:42

Consider $$G=\mathbb{Z}/2$$, it has the presentation $$\{g: g^2=1\}$$. To define $$X_G$$ one attaches a $$2$$-cell along the boundary of $$S^1$$ for the relation $$g^2=1$$ the resulting space is the projective plane $$\mathbb{R}P^2$$ whose universal cover $$S^2$$ is simply connected, but $$\pi_2(S^2)=\mathbb{Z}=\pi_2(\mathbb{R}P^2)$$. We deduce that $$\mathbb{R}P^2$$ is not an Eilenberg McLane space.