Mazurov V.D.'s 2-groups with an odd-order automorphism that is the identity PDF

February 14, 2018 | Symmetry And Group | By admin | 0 Comments

By Mazurov V.D.

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S, , so the assertion follows by induction. ]. In the following theorem, the topology is not necessarily Hausdorff. 4 (A. L. S. Corner). Let 6 be a set of subgroups S of the group A and let 6 , consist of all S E 6 withfinite index in A . , the linear topology defined by taking S E 6 , open]. We need only prove the " only if" part. Assume A topological in the closed 6-topology. Given an open set V = A\(a T ) (a E A , T E 6 ) containing 0, there is an open neighborhood U of 0, such that U - U E V.

Because the intersection of two cosets (al + U,) n (a, + U,) is vacuous or a coset mod U , n U,, and U , n U , E D whenever U , , U, E D, all the open sets will be unions of cosets a + U with U E D. The continuity of the group operations is obvious from the simple observation that x - y E a U implies (x U ) - ( y + U ) c a U. We call the arising topology the D-topology of A. Obviously, it is Hausdorfs if and only if + + + + n u=o, UED and discrete exactly if 0 E D. , there is a countable base of open neighborhoods about 0.

Assume that, for every A E d , there is defined a topology t ( A ) under which A is a topological group. We call t = { [ ( A )I A E d }a functorial topology if every homomorphism in d is continuous. In this sense, the Z-adic, the p-adic, the Priifer, and the finite index topologies are all functorial. A more general method of obtaining a functorial topology is to choose a class X of groups and to take the subgroups Ker 4 with 4 : A + X E X as a subbase of open neighborhoods about 0 in A . This will yield a linear topology on A which will be Hausdorff whenever there are sufficiently many homomorphisms 4, in the sense that for every a E A , a # 0, there are X E X and 4 : A + X with &I # 0.

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2-groups with an odd-order automorphism that is the identity on involutions by Mazurov V.D.


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