Question

The fact that $\mathcal{A}(G)\simeq\mathcal{S}(G)$ is normally proved using Quillen's Theorem A as explained by Dan, but here is another argument that I like better. (I don't think it is in the literature.) For any poset $X$, let $sX$ be the poset of nonempty chains in $X$, ordered by inclusion. There is a homeomorphism $m:|sX|\to |X|$ by barycentric subdivision. There is also a poset map $\max:sX\to X$, and $|\max|$ is homotopic to $m$ (by a linear homotopy) so it is a homotopy equivalence. Next, if $P$ is a nontrivial $p$-group then the group $TP=\{g\in ZP: g^p=1\}$ is nontrivial (by a standard lemma) and elementary abelian. Unfortunately, the map $T:\mathcal{S}(G)\to\mathcal{A}(G)$ is not order-preserving. However, suppose we have a chain $C=(P_0\leq\dotsb\leq P_r)\in s\mathcal{S}(G)$. When $i\leq j$ we note that $TP_i\leq P_j$ and $TP_j$ is central in $P_j$ so $TP_i$ commutes with $TP_j$. It follows that the product $UC=TP_0.TP_1.\dotsb.TP_r$ is again elementary abelian. This defines a poset map $U:s\mathcal{S}(G)\to\mathcal{A}(G)$ and thus a map $|U|:|\mathcal{S}(G)|\simeq|s\mathcal{S}(G)|\to|\mathcal{A}(G)|$. If we let $i\:\mathcal{A}(G)\to\mathcal{S}(G)$ i:\mathcal{A}(G)\to\mathcal{S}(G)$ denote the inclusion then $i\circ U\leq\max$ so $|i|\circ|U|$ is homotopic to $|\max|$ and thus to $m$. We also have $U\circ si=\max:s\mathcal{A}(G)\to\mathcal{A}(G)$. It follows that $|i|$ is a homotopy equivalence as claimed.