Timeline for Existence of simultaneously normal finite index subgroups
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2011 at 14:13 | comment | added | Colin Reid | As $H_1$ and $H_2$ are topologically finitely generated in Yiftach's example, there are pairs of elements $P = \{x,y\}$ such that $x$ and $y$ individually are contained in open compact subgroups of $G$, but $P$ is not contained in the normaliser of any open compact subgroup. | |
| Jul 29, 2011 at 13:56 | comment | added | Colin Reid | If $H_1,\dots H_n$ are profinite and open in the topological group G, they have arbitrarily small open normal subgroups in common if and only if they generate a residually discrete subgroup of $G$. See Corollary 4.1 of: P-E. Caprace, N. Monod, Decomposing locally compact groups into simple pieces, Math. Proc. Cambridge Philos. Soc. 150 Nr. 1 (2011), 97--128. | |
| Sep 3, 2010 at 20:07 | comment | added | user6976 | The group of Burger and Mozes: Lattices in product of trees, Inst. Hautes Études Sci. Publ. Math. No. 92 (2000), 151--194, gives a counterexample too. It is an amalgamated product $G$ of two free groups over a subgroup $K$ which has finite indexes in both of the free groups, but the group $G$ itself if simple. | |
| Aug 8, 2010 at 14:12 | vote | accept | Terry Tao | ||
| Aug 5, 2010 at 17:03 | comment | added | Ian Agol | Just a comment that when $d=2$, $G$ acts on the Bass-Serre tree of similarity classes of rank 2-$\mathbb{Z}_p$ lattices in $\mathbb{Q}_p^2$, and $H_i$ are stabilizers of distinct vertices of this tree. So this shows that there is an example which comes from automorphism groups of trees, as suggested to Tao. Also, this demonstrates that $G$ is an amalgamated product of conjugates of $H_1$ fixing adjacent vertices, amalgamated over the edge stabilizer. | |
| Aug 5, 2010 at 10:47 | comment | added | BCnrd | More generally, if $k$ is non-arch. local field and $G$ is conn'd semisimple $k$-gp which is $k$-split and simply conn'd (e.g., ${\rm{SL}}_d$, ${\rm{Sp}}_ {2g}$, Spin gps, etc.) then Tits proved proper open subgps of $G(k)$ are compact (since $G(k)$ is gen'td by subgps $U(k)$ for unip. radicals $U$ of Borel $k$-subgps, due to the hypotheses on $G$). So max'l compact open subgps are max'l subgps. Same then holds for $G(k)/Z$, where $Z$ is finite center of $G(k)$, but $G(k)/Z$ is simple. So distinct max'l compact subgps $H_1$ and $H_2$ of $G(k)/Z$ (which always exist) are always a counterex. | |
| Aug 5, 2010 at 9:46 | history | answered | Yiftach Barnea | CC BY-SA 2.5 |