Wikipedia:Reference desk/Archives/Mathematics/2009 March 31

= March 31 =

Finite order subgroups of CTG
Let G be a compact topological group, with a distance d. Is it true that for any ε>0 there exists a finite subgroup Gε such that each element of G has a distance at most ε from the closest element of Gε?--79.38.22.37 (talk) 11:44, 31 March 2009 (UTC)


 * The additive group of the p-adic integers Zp is a compact metric abelian group, and it is torsion-free, hence it contains no nontrivial finite subgroup. So the answer is no. — Emil J. 12:35, 31 March 2009 (UTC)
 * Very clear, thanks a lot. --79.38.22.37 (talk) 13:21, 31 March 2009 (UTC)
 * Well, given any metric d of G, one can form a metric of G that induces the same topology, but is bounded by ε > 0 (call this metric dε). So given ε > 0, dε is a feasible metric for G and in this case, the trivial group can be Gε. Then the conclusion of your result is satisfied. As for dε, let dε (x, y) = min {d (x, y), ε}.
 * So indeed the result is true assuming that you allow any metric for G. For a fixed metric d of G however, the result is false as User:EmilJ notes. Note however, that any metric d of G must be bounded (because d is a continuous function with compact domain). -- PS T  01:38, 2 April 2009 (UTC)
 * but this has very little to do with the question. if you truncate or re-normalize the distance of course everything became smaller than epsilon,so what? then, why not to change G itself and choose G to be the trivial group. --131.114.72.122 (talk) 12:45, 2 April 2009 (UTC)
 * Furthermore that is incorrect because G itself might not be finite (as in Emil's answer). Eric.  131.215.158.238 (talk) 19:54, 2 April 2009 (UTC)
 * I never assumed G to be finite. -- PS T  03:47, 5 April 2009 (UTC)
 * It needn't be finite, but it is indeed bounded as PST says (in general, a metric space is compact if and only if it is complete and totally bounded). — Emil J. 14:16, 3 April 2009 (UTC)
 * Very interesting example, anyway; I did not know it and I am glad to learn it. I wonder if it is true the following weaker version of the OP: "given a compact metric topological group $$\scriptstyle(G,\cdot, d)$$ and ε>0, there exists a finite subset of G, Gε, such that each element of G has a distance at most ε from the closest element of Gε (as before), and a group operation * on Gε  such that $$\scriptstyle d(x*y,x\cdot y)\leq\epsilon$$ and $$\scriptstyle d(x^{-1_*},x^{-1_{\cdot}})\leq\epsilon$$ for all x and y in Gε" (if the awful notation is not clear: I just mean that the multiplication and the reciprocal in the two group structures are epsilon close). It seems to me that this can be easily proven true, if G is also abelian, but in general I don't see it clearly. This should be a nice way a tentative of saying in a nice way that "G is almost finite", which is somehow in the spirit of the OP --pma (talk) 16:53, 3 April 2009 (UTC)

Motion Groups
I want to find a reference to learn about motion groups SE(N). I couldn't find anything on wikipedia. Any help will be appreciated. Regards, deeptrivia (talk) 18:28, 31 March 2009 (UTC)
 * We have an article Euclidean group, which briefly mentiones SE(n) under the name E+(n). Algebraist 18:36, 31 March 2009 (UTC)
 * Thanks! deeptrivia (talk) 02:14, 1 April 2009 (UTC)