Free quotient group admits a section: Difference between revisions

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Then, <math>N</math> is a [[fact about::complemented normal subgroup]] of <math>G</math>. In other words, there exists a [[retract]] <math>B</math> of <math>G</math> with [[normal complement]] <math>N</math>, i.e., <math>B</math> is a subgroup of <math>G</math> such that <math>G</math> is the [[fact about::internal semidirect product]] <math>N \rtimes B</math>. Explicitly, <math>N \cap B</math> is trivial and <math>NB = G</math>.
Then, <math>N</math> is a [[fact about::complemented normal subgroup]] of <math>G</math>. In other words, there exists a [[retract]] <math>B</math> of <math>G</math> with [[normal complement]] <math>N</math>, i.e., <math>B</math> is a subgroup of <math>G</math> such that <math>G</math> is the [[fact about::internal semidirect product]] <math>N \rtimes B</math>. Explicitly, <math>N \cap B</math> is trivial and <math>NB = G</math>.


A normal subgroup <math>N</math> such that <math>G/N</math> is a free group is termed a [[fact about::free-quotient group]].
A normal subgroup <math>N</math> such that <math>G/N</math> is a free group is termed a [[fact about::free-quotient subgroup]].


==Related facts==
==Related facts==


* [[Retract of free group is free on fewer generators]]
* [[Retract of free group is free on fewer generators]]

Revision as of 20:24, 19 May 2010

Statement

Suppose N is a normal subgroup of a group G such that the quotient group G/N is a Free group (?).

Then, N is a Complemented normal subgroup (?) of G. In other words, there exists a retract B of G with normal complement N, i.e., B is a subgroup of G such that G is the Internal semidirect product (?) NB. Explicitly, NB is trivial and NB=G.

A normal subgroup N such that G/N is a free group is termed a Free-quotient subgroup (?).

Related facts