Is the hypothesis (formulated in the introduction to the Preprint On geometric constructions of finite groups and Lie algebras
) about the one-to-one correspondence between Abelian and non-Abelian finite groups correct?
Correspondence is ensured with the help of bundles of circles (semicircles). Non-Abelian groups are formed as groups of isomorphisms of bundles, and Abelian groups correspond to the arrangement of nodes of bundles of circles.
If explanations are needed, then you can look into the text with 8 figures. The link is in the head post.
We now use Abelian groups to construct the bundles of semicircles presented here. Suppose we have a one-dimensional lattice $L$, which is a straight line with integer nodes, that is, $L={Z; R}$. Then a bunch of semicircles in the form of an eight is obtained as a result of factorization ${Z/Z}; R/2Z}$, and a bunch in the form of a circle is obtained as a result of factorization ${Z/2Z; R/2Z}$.
An integer $n$-dimensional lattice is not an algebraic space, but a set of integers (points) in an $n$-dimensional real space through which lines parallel to the basis vectors are drawn.
Semicircles are halves of circles, which sometimes (with a certain type of factorization) turn into circles.
And in order not to fantasize, take a look at the drawings.
And how would you answer the question that is indicated in the title?
The edges are continuous, and the elements of the abelian group are discrete. The edges of the lattice are not oriented.
A bundle is a bundle of circles obtained as a result of compactification of the lattice axes.
The group of automorphisms of a bundle of circles is formed due to topological isomorphisms, that is, such transformations of the bundle that allow changing the radii of circles and their location, but prohibit breaks and intersections of circles.
There are more non-commutative groups, since any commutative group can be re-written in "non-commutative" terms, yet the converse does not apply really.
Peter Breuer For a better understanding, I have slightly corrected the text of the preprint
Valentin B. Bura I don't think so. If every noncommutative finite group is isomorphic to some subgroup of the group of monomial substitutions, then with the help of the corresponding bundle of circles we form a commutative group.
Igor, what I am saying is as follows, roughly speaking... Commutative multiplication generates a smaller set.
A.B.C would map into A.B.C, A.C.B, B.C.A, C.B.A., B.A.C., C.A.B. right?
This is because ABC is just one under commutativity...
Valentin, I agree that there will be fewer elements in the commutative group than in the corresponding noncommutative group. However, the question is different, namely, whether there is a corresponding commutative group for every noncommutative group.
I ask a general question and concretize it further by referring to the preprint. What's unethical about that?
Maybe I'm clumsily expressing my thoughts, but the point is that the connection between the groups should be the most natural, geometric, topological.
The group of monomial substitutions is isomorphic to the wreath product of the substitution groups $S_n$ and $S_2$.
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