论文标题
无限图的基本组的完整群体完成
The profinite completion of the fundamental group of infinite graphs of groups
论文作者
论文摘要
令$（\ Mathcal {G}，γ）$为有限组的抽象图。如果$γ$是有限的，我们可以以自然方式构建组的图表$（\ hat {\ nathcal {g}}，γ）$，其中$ \ hat {\ hat {\ nathcal {g}}}（m）$是$ \ mathcal {g}（g}（g}（m）$ for All $ m m m m \ us，这样做的主要原因是$γ$是有限的，因此它已经是有限的。在本文中，我们通过构造一个profinite图$ \Overlineγ$来处理无限案例，其中$γ$密集地嵌入了$γ$，然后定义了$（\ wideHat {\ Mathcal {g}}}的offinite图，\overlineγ）$。我们还证明，基本组$π_1（\ wideHat {\ Mathcal {g}}}，\Overlineγ）$是$π_1^{abs}（\ Mathcal {g}，γ）$的profinite完成。这本书的《 pifinite图和群体》（Luis Ribes）于2017年出版的书6.7.1的答案是6.7.1。后来，我们概括了路易斯·里伯斯（Luis Ribes）和第二作者的论文主要定理$ \ OVERLINE {n_ {r}（h）} = n _ {\ hat {r}}}（\ oferline {h}）$回答《肋骨书》的打开问题15.11.10。最后，我们概括了希拉·查加斯（Sheila Chagas）和第二作者的论文的主要定理，这表明每个实际上免费的群体都是亚组结合在一起的。这是同一本肋骨书的打开问题15.11.11。
Let $(\mathcal{G},Γ)$ be an abstract graph of finite groups. If $Γ$ is finite, we can construct a profinite graph of groups in a natural way $(\hat{\mathcal{G}},Γ)$, where $\hat{\mathcal{G}}(m)$ is the profinite completion of $\mathcal{G}(m)$ for all $m \in Γ$. The main reason for this is that $Γ$ is finite, so it is already profinite. In this paper we deal with the infinite case, by constructing a profinite graph $\overlineΓ$ where $Γ$ is densely embedded and then defining a profinite graph of groups $(\widehat{\mathcal{G}},\overlineΓ)$. We also prove that the fundamental group $Π_1(\widehat{\mathcal{G}},\overlineΓ)$ is the profinite completion of $Π_1^{abs}(\mathcal{G},Γ)$. This answers Open Question 6.7.1 of the book Profinite Graphs and Groups, published by Luis Ribes in 2017. Later we generalise the main theorem of a paper by Luis Ribes and the second author, proving that if $R$ is a virtually free abstract group and $H$ is a finitely generated subgroup of $R$, then $\overline{N_{R}(H)}=N_{\hat{R}}(\overline{H})$ answering Open Question 15.11.10 of the book of Ribes. Finally, we generalise the main theorem of a paper by Sheila Chagas and the second author, showing that every virtually free group is subgroup conjugacy separable. This answers Open Question 15.11.11 of the same book of Ribes.
