3
$\begingroup$

In this question, which we flag it as a community wiki question, we search for a big list of groups $G$ which can not be isomorphic to a structure mentioned in $i.$ for some $i \in \{1,2,\ldots,6 \}$

  1. The homeomorphism group of a topological space $X$.
  2. The automorphism group of a (not necessarily finite dimensional) Lie algebra.
  3. The automorphism group of a coalgebra.
  4. (Assuming $G$ is a Lie group) The isometry group of a Riemannian manifold.
  5. The automorphism group of another group.
  6. The automorphism group of a ring.
Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ Every group $G$ is the group of homeomorphisms of a topological space $X$. Let $X$ have as point set $\{t_g | g\in G\} \sqcup \{u_{g,h},v_{g,h}, w_{g,h} | (g,h) \in G\times G\}$. Define the topology to be the finest topology satisfying the following specialization conditions: for every $(g,h)\in G\times G$, impose $u_{g,h}$ specializes to $t_g$, $v_{g,h}$ specializes to $u_{g,h}$, $t_g$ and $t_h$, and $w_{g,h}$ specializes to $v_{g,h}$, $u_{g,h}$, $t_g$, $t_h$ and $t_{g\cdot h}$. $\endgroup$
    – Jason Starr
    Commented Sep 4, 2015 at 18:57
  • 1
    $\begingroup$ Do you mean any structure or a structure? $\endgroup$
    – Yiftach Barnea
    Commented Sep 4, 2015 at 19:08
  • $\begingroup$ @YiftachBarnea thank you for your comment. i revise the question. $\endgroup$
    – Ali Taghavi
    Commented Sep 4, 2015 at 20:57
  • 2
    $\begingroup$ @AliTaghavi: "could you pleae more explain on thi topology?" I would like to recommend that you use the spell checker that is built into the MO comments system. A subset $U$ of $X$ is open with respect to the topology I mention if and only if it satisfies all of the following conditions: for every $(g,h)$ in $G\times G$, (a) if $U$ contains $t_g$ then $U$ contains $u_{g,h}$, (b) if $U$ contains $t_h$ or $u_{g,h}$, then $U$ contains $v_{g,h}$, and (c) if $U$ contains $t_{g\cdot h}$ or $v_{g,h}$ then $U$ contains $w_{g,h}$. $\endgroup$
    – Jason Starr
    Commented Sep 4, 2015 at 23:26
  • 1
    $\begingroup$ For (5), $\mathbb{Z}$ works. I think also any cyclic group of odd (prime?) order. The idea is to note that $G/Z(G)$ embeds into $\operatorname{Aut}(G)$, and hence $G$ is abelian (standard undergrad exercise). Such a group always has an automorphism of even order, assuming AOC (less standard exercise, not sure of level), a contradiction. $\endgroup$
    – user1729
    Commented Sep 11, 2015 at 13:32

0

Reset to default

You must log in to answer this question.