Macdonald polynomials: Difference between revisions
Jump to navigation
Jump to search
en>Helpful Pixie Bot m ISBNs (Build KE) |
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) |
||
| 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]] | |||
:<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}} | |||
Revision as of 21:03, 30 August 2013
In group theory, a group is said to be free-by-cyclic if it has a free normal subgroup such that the quotient group
is cyclic.
In other words, 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 is a finitely generated group we say that is (finitely generated free)-by-cyclic (or (f.g. free)-by-cyclic).
References
- A. Martino and E. Ventura (2004), The Conjugacy Problem for Free-by-Cyclic Groups. Preprint from the Centre de Recerca Matemàtica, Barcelona, Catalonia, Spain.