Center not is surjective endomorphism-balanced: Difference between revisions

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{{sdf subgroup property dissatisfaction|
{{sdf subgroup property dissatisfaction|
sdf = center|
sdf = center|
property = highly strictly characteristic subgroup}}
property = surjective endomorphism-balanced subgroup}}


==Statement==
==Statement==
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==Related facts==
==Related facts==


===Other proofs of similar subgroup-defining functions not being highly strictly characteristic===
===Other proofs of similar subgroup-defining functions not being surjective endomorphism-balanced===


* [[Baer norm not is highly strictly characteristic]]
* [[Baer norm not is surjective endomorphism-balanced]]
* [[Wielandt subgroup not is highly strictly characteristic]]
* [[Wielandt subgroup not is surjective endomorphism-balanced]]


===Proofs of similar subgroup-defining functions being strictly characteristic===
===Proofs of similar subgroup-defining functions being strictly characteristic===

Latest revision as of 23:45, 10 August 2009

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., center) does not always satisfy a particular subgroup property (i.e., surjective endomorphism-balanced subgroup)
View subgroup property satisfactions for subgroup-defining functions

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View subgroup property dissatisfactions for subgroup-defining functions

Statement

Statement with symbols

It is possible to have a group G and a surjective endomorphism σ of G such that the restriction of σ to the center Z(G) is not surjective as an endomorphism of Z(G).

Related facts

Other proofs of similar subgroup-defining functions not being surjective endomorphism-balanced

Proofs of similar subgroup-defining functions being strictly characteristic

Facts used

  1. Isomorphic to inner automorphism group not implies centerless: We can have a group G such that the center Z(G) is nontrivial, but there is an isomorphism between G/Z(G) and G.

Proof

Proof using fact (1)

Let G be an example group for fact (1). Let α:GG/Z(G) and the quotient map and φ:G/Z(G)G be an isomorphism. Then, σ=φα is a surjective endomorphism of G. On the other hand, the restriction of σ to Z(G) is the trivial map, which is not surjective by the assumption that Z(G) is a nontrivial group.