# Status of Quillen's conjecture on elementary abelian p-groups

These are questions on D. Quillen's 1978 paper Homotopy properties of the poset of nontrivial p-subgroups of a group.

Let $G$ be a finite group, $p$ a prime number, $\mathcal S(G)$ the poset of non-trivial $p$-subgroups of $G$, and $\mathcal A (G)$ the poset of non-trivial elementary Abelian $p$-subgroups of $G$, both ordered by inclusion.

Question One: Is $\mathcal S(G)$ homotopic or weakly homotopic to $\mathcal A (G)$? (Whichever is true is a theorem of Quillen.) If the latter, can someone give a specific example showing that the two posets are not homotopic?

Quillen also proved that for $G$ solvable, $\mathcal A (G)$ is contractible if and only if $G$ has a non-trivial normal $p$-subgroup.

Conjecture (Quillen): $\mathcal A (G)$ is contractible if and only if $G$ has a non-trivial normal $p$-subgroup.

Question Two: Is this still open (I know it was a few years ago)? What are lines of attack on this problem? Have attempts to prove it led to any a priori unrelated work?

• I'll try to say more later, but you might look at Jesper Grodal's paper in the Annals about higher limits. Quillen's conjecture shows up in there in an interesting way. Sep 16, 2010 at 13:55

To answer your first question, the inclusion $A(G)\to S(G)$ is a homotopy equivalence. This is an application of Quillen's "Theorem A" (aka his "fiber lemma"). See Prop. 2.1 in his paper. To apply the fiber lemma, you just need to show that the fibers, which are those points mapping below a particular P-subgroup in S, are contractible. This means you need to show that the elementary abelian p-subgroups of a P-group form a contractible poset. Well, that's done by a conical contraction: just multiply each subgroup by the maximal elementary abelian subgroup of the center (to slide it up above this characteristic subgroup), and then slide it down to this subgroup.
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 $i:\mathcal{A}(G)\to\mathcal{S}(G)$ 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.