Normal closure-finding problem: Difference between revisions

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==Description==
==Description==


===Given data===
We are given a [[group]] <math>G</math> and a [[subgroup]] <math>H</math> specified by some means (typically using [[generating set]]s, though we may also specify them using membership tests). The goal is to obtain a description of the [[normal closure]] <math>K = H^G</math> of <math>H</math> in <math>G</matH> using the same language (so if <math>G</math> and <math>H</math> are described using generating sets, we want a generating set for <math>K</math>, whereas if <math>G</math> and <math>H</math> are described using membership tests, then <math>K</math> is also to be described using a membership test.
 
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]]).
 
===Goal===
 
Let <math>K</math> be the [[normal closure]] of <math>H</math> in <math>G</math>. We are required to determine a generating set <math>C</math> for <math>K</math>.


==Relation with other problems==
==Relation with other problems==

Revision as of 20:27, 25 June 2013

This article describes the subgroup operator computation problem for the subgroup operator: normal closure

Description

We are given a group G and a subgroup H specified by some means (typically using generating sets, though we may also specify them using membership tests). The goal is to obtain a description of the normal closure K=HG of H in G using the same language (so if G and H are described using generating sets, we want a generating set for K, whereas if G and H are described using membership tests, then K is also to be described using a membership test.

Relation with other problems

Problems it reduces to