Normal closure-finding problem: Difference between revisions
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===Given data=== | ===Given data=== | ||
Our universe is some group <math>U</math> | Our universe is some group <math>U</math> specified by means of an [[encoding]]. | ||
A group <math>G</math> in <math>U</math> is specified by a generating set <math>A</math>, and a subgroup <math>H</math> of <math>G</math> is specified by a generating set <math>B</math>. (We are given a guarantee that <math>H</math> is a subgroup of <math>G</math>, if not, we can test it using the algorithm for the [[subgroup testing problem]]). | A group <math>G</math> in <math>U</math> is specified by a generating set <math>A</math>, and a subgroup <math>H</math> of <math>G</math> is specified by a generating set <math>B</math>. (We are given a guarantee that <math>H</math> is a subgroup of <math>G</math>, if not, we can test it using the algorithm for the [[subgroup testing problem]]). | ||
Revision as of 20:13, 25 June 2013
This article describes the subgroup operator computation problem for the subgroup operator: [[normal closure]]
Description
Given data
Our universe is some group specified by means of an encoding.
A group in is specified by a generating set , and a subgroup of is specified by a generating set . (We are given a guarantee that is a subgroup of , if not, we can test it using the algorithm for the subgroup testing problem).
Goal
Let be the normal closure of in . We are required to determine a generating set for .
Relation with other problems
Problems it reduces to
- Membership testing problem: Normal closure-finding can be effected using the membership testing problem for all subgroups of . The idea is as follows: let and . At each stage, check whether is normal in (the normality testing problem, which again reduces to membership testing). If not, the algorithm for normality testing will yield a conjugate of an element in that is not in . Add this to the current generating set and hence get a generating set for . This process terminates after a finite number of steps and the stable values give and .