Wikipedia:Reference desk/Archives/Mathematics/2010 May 10

= May 10 =

Group Theory
Hi. I'm doing some work in preparation for my exam and have been asked to explain 'how to make the set of left cosets of H (where H is a normal subgroup of G) into a group'. This is clearly just bookwork but I can't seem to find it in my notes and Google hasn't helped me. Can anyone point me to the answer? Thanks. 128.232.247.9 (talk) 13:55, 10 May 2010 (UTC)
 * See quotient group.—Emil J. 14:20, 10 May 2010 (UTC)


 * (Longer explanation) You make a left coset gH by taking some element g of G and multiplying it in turn by each element h of H. To make the set of left cosets into a group you first need to define a group operation that takes any two left cosets - say gH and kH - and produces another left coset, which we might call gH*kH. There is a natural candidate for this group operation * on left cosets, but you need to make sure it is well-defined - don't forget that aH and bH may be different representations of the same left coset, but the results of the group operation should not depend on how each coset is represented. This is where you use the fact that H is a subgroup of G - and not just any subgroup, but also a normal subgroup. Once you know that * is well-defined, then you need to show that it satisfies the group axioms. But if you make the natural choice of group operation on the left cosets then this isn't hard, as the properties of this group operation are "inhertited" from the original group opration on G. Gandalf61 (talk) 14:24, 10 May 2010 (UTC)

Proving that 0 is the only bounded 1-periodic planar vector field
It seems to me that a planar vector field v which is bounded and such that its flow φt satisfy φ1(x)=x must be zero everywere.

I have found a proof but I'm not really satisfied. I define the lenght function
 * $$L(x)=\int_0^1|v(\phi^t(x))|dt$$

which must be C1, bounded and constant on the orbits. If we take point x where v is nonzero than it must lay on a circular periodic orbit C. I define S=sup{L(x):x is inside an orbit which surrounds C}. Then exist xk such that L(xk)→S, and also xk must be bounded therefore there exist a converging subsequence xk→x0. Since all the orbits φt(xk) surround C by continuity also φt(x0) must surround C. Now we have an absurd: there is an orbit which surrounds C with lenght greater than L(x0): if we consider a small neighbourhood B of x0 then all the points in B will travel near the orbit of x0, therefore outside the orbit there must be another orbit with greater lenght.

Is there a nicer proof?--Pokipsy76 (talk) 20:49, 10 May 2010 (UTC)