Mathematics desk
< March 30	<< Feb \| March \| Apr >>	April 1 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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 G_ε such that each element of G has a distance at most ε from the closest element of G_ε?--79.38.22.37 (talk) 11:44, 31 March 2009 (UTC)[reply]

The additive group of the p-adic integers Z_p 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)[reply]

Very clear, thanks a lot. --79.38.22.37 (talk) 13:21, 31 March 2009 (UTC)[reply]

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 d_ε). So given ε > 0, d_ε is a feasible metric for G and in this case, the trivial group can be G_ε. Then the conclusion of your result is satisfied. As for d_ε, let d_ε (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)[reply]

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)[reply]

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)[reply]

I never assumed G to be finite. --PS T 03:47, 5 April 2009 (UTC)[reply]

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)[reply]

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, G_ε, such that each element of G has a distance at most ε from the closest element of G_ε (as before), and a group operation * on G_ε 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 G_ε" (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)[reply]

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)[reply]

We have an article Euclidean group, which briefly mentiones SE(n) under the name E⁺(n). Algebraist 18:36, 31 March 2009 (UTC)[reply]

Thanks! deeptrivia (talk) 02:14, 1 April 2009 (UTC)[reply]