Macdonald polynomials: Difference between revisions

From formulasearchengine
Jump to navigation Jump to search
en>ClueBot NG
m Reverting possible vandalism by 75.221.134.104 to version by 98.110.175.194. False positive? Report it. Thanks, ClueBot NG. (1718418) (Bot)
en>Trappist the monk
m References: replace mr template with mr parameter in CS1 templates; using AWB
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
In [[group theory]], a [[group (mathematics)|group]] <math>G</math> is said to be '''free-by-cyclic''' if it has a [[free group|free]] [[normal subgroup]] <math> F</math> such that the [[quotient group]]
Nice to meet you, my title is Araceli Oquendo but I don't like when people use my full name. I've always loved residing in Idaho. What she loves doing is taking part in croquet and she is attempting to make it a occupation. I am a production and distribution officer.<br><br>My web page: [http://Www.Jvawest.com/UserProfile/tabid/152/userId/2157/Default.aspx Www.Jvawest.com]
 
:<math> G/F</math>
 
is [[cyclic group|cyclic]]. 
 
In other words, <math>G</math> is free-by-cyclic if it can be expressed as a [[group extension]] of a free group by a cyclic group (NB there are two conventions for 'by').
 
If <math>F</math> is a [[finitely generated group]] we say that <math>G</math> is ''(finitely generated free)-by-cyclic'' (or (f.g. free)-by-cyclic). 
 
==References==
* A. Martino and E. Ventura (2004), [http://www.crm.es/Publications/04/pr574.pdf ''The Conjugacy Problem for Free-by-Cyclic Groups'']. Preprint from the Centre de Recerca Matemàtica, Barcelona, Catalonia, Spain.
 
[[Category:Infinite group theory]]
 
 
{{Abstract-algebra-stub}}

Latest revision as of 01:30, 25 September 2014

Nice to meet you, my title is Araceli Oquendo but I don't like when people use my full name. I've always loved residing in Idaho. What she loves doing is taking part in croquet and she is attempting to make it a occupation. I am a production and distribution officer.

My web page: Www.Jvawest.com