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Let $G$ be a finite group. We define a $G$-simplicial complex $\mathcal{A}(G)$ with set of vertices $G^*:=G-\{e\}$ and the simplices are the abelian subsets of $G^*$. The groupe $G$ act simplicially on $\mathcal{A}(G)$, the action on vertices is given by : $g.s=gsg^{-1}$. The simpicial complex $\mathcal{A}(G)$ is called the Abel complex of $G$.

My question is : for which conditions on $G$, the simplicial complex $\mathcal{A}(G)$ is connected ???

$\textbf{Examples:}$

1) If the center $Z(G)$ is not trivial, then the simplicial complex $\mathcal{A}(G)$ is a cone, because every vertex $s$ of $\mathcal{A}(G)$ is connected to a fixed vertex $s_{0}\in Z(G)-\{e\}$. In particular, if $G$ is abelian, then $\mathcal{A}(G)$ is a cone.

2) If the groupe $G=H\times K$ is the direct product of two finite groups, then $\mathcal{A}(G)$ is connected.

3) $\mathcal{A}(S_{9})$ is connected but $\mathcal{A}(S_{n})$ is not connected for $n\leqslant8$.

$\textbf{Remarks:}$

1) The connectedness of $\mathcal{A}(G)$ is equivalent to the connectedness of another simplicial complex. Indeed, for every proper abelian subgroup $H$ of $G$, we denote $\mathcal{A}(H)$ the Abel complex of $H$. It's clear that $\mathcal{A}(G)=\bigcup\mathcal{A}(H)$, where $H$ runs all proper abelian subgroups of $G$. The nerve of this recouvrement is the simplicial complex $N(G)$ whose vertices the proper abelian subgroups of $G$, and simplices of the form $\sigma=\{H_{0},...,H_{p}\}$ where $H_{0}\cap H_{1}\cap ...\cap H_{p}\neq\{e\}$. since the subcomplexes $\mathcal{A}(H)$ are connected, then $\mathcal{A}(G)$ is connected if and only if $N(G)$ is.

2) Le nerve lemma implies that $\mathcal{A}(G)$ has the same $G$-homotopy type as $N(G)$.

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    $\begingroup$ Usually people just remove the center instead of e. The 1-Skeleton is called the commuting graph. Lots of people look at this $\endgroup$ – Benjamin Steinberg Apr 11 '15 at 2:17

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