Difference between revisions of "Group action"

From Maths
Jump to: navigation, search
m (Fixed markup, marked as dire page)
m
 
Line 1: Line 1:
{{Dire page|This was supposed to be something Alex Clark did, it's in urgent need of content!}}
+
{{Stub page|grade=A|msg=Needs fleshing out and neatening up, I'd like to introduce right group actions in a different way to left, however in my current attempt they're the same length!}}
==Definition==
+
__TOC__
A ''group action'' of a group {{M|G}} on a set {{M|A}} is a map from {{M|G\times A \to A}} usually written as {{M|g\cdot a}} for all {{M|g\in G}} and {{M|a\in A}}, that satisfies the following two properties:
+
==Defintion==
 +
A (left) ''group action'' of a [[group]] {{M|(G,*)}} on a [[set]] {{M|X}} is a [[mapping]]{{rAAPAG}}:
 +
* {{M|(\cdot):G\times X\rightarrow X}}<ref group="Note">I have written {{M|(\cdot):G\times X\rightarrow X}} rather than the usual {{M|\cdot:G\times X\rightarrow X}} notation for [[function|functions]] to make it clearer that there is a dot there; this notation isn't new or different, it's just because a lone {{M|\cdot}} looks out of place.</ref> defined by {{M|(\cdot):(g,x)\mapsto g\cdot x}} such that:
 +
** {{M|1=\forall x\in X[1\cdot x=x]}} (where {{M|1}} is the identity element of {{M|(G,*)}} group) and
 +
** {{M|1=\forall g,h\in G\ \forall x\in X[g\cdot(h\cdot x)=(g*h)\cdot x]}}
 +
Notations for {{M|g\cdot x}} include {{M|gx}} and {{M|{}^gx}}  
  
* {{M|1=g_1 \cdot(g_2\cdot a) =(g_1g_2)\cdot a}} for all {{M|g_1,g_2\in G,a\in A}}
 
  
 +
A ''right group action''<ref name="AAPAG"/> is almost exactly the same, just the other way around; defined by {{M|(\cdot):X\times G\rightarrow X}} given by {{M|(\cdot):(x,g)\mapsto x\cdot g}} which must satisfy {{M|1=\forall x\in X[x\cdot 1=x]}} and {{M|1=\forall g,h\in G\ \forall x\in X[(x\cdot g)\cdot h=x\cdot(g*h)]}}.
 +
 +
Notations for {{M|x\cdot g}} include {{M|xg}} and {{M|x^g}}
 +
==See also==
 +
* '''Examples:'''
 +
** [[Every group acts on itself by multiplication]]
 +
** [[Every subgroup of a group acts on the group by multiplication]]
 +
** [[The symmetric group on a set acts on the set by evaluation]]
 +
* [[Any group action can be thought of as a permutation]]
 +
==Notes==
 +
<references group="Note"/>
 +
==References==
 +
<references/>
 +
{{Group theory navbox|plain}}
 +
{{Abstract algebra navbox}}
 
{{Definition|Abstract Algebra|Group Theory}}
 
{{Definition|Abstract Algebra|Group Theory}}

Latest revision as of 23:28, 21 July 2016

Stub grade: A
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Needs fleshing out and neatening up, I'd like to introduce right group actions in a different way to left, however in my current attempt they're the same length!

Defintion

A (left) group action of a group [ilmath](G,*)[/ilmath] on a set [ilmath]X[/ilmath] is a mapping[1]:

  • [ilmath](\cdot):G\times X\rightarrow X[/ilmath][Note 1] defined by [ilmath](\cdot):(g,x)\mapsto g\cdot x[/ilmath] such that:
    • [ilmath]\forall x\in X[1\cdot x=x][/ilmath] (where [ilmath]1[/ilmath] is the identity element of [ilmath](G,*)[/ilmath] group) and
    • [ilmath]\forall g,h\in G\ \forall x\in X[g\cdot(h\cdot x)=(g*h)\cdot x][/ilmath]

Notations for [ilmath]g\cdot x[/ilmath] include [ilmath]gx[/ilmath] and [ilmath]{}^gx[/ilmath]


A right group action[1] is almost exactly the same, just the other way around; defined by [ilmath](\cdot):X\times G\rightarrow X[/ilmath] given by [ilmath](\cdot):(x,g)\mapsto x\cdot g[/ilmath] which must satisfy [ilmath]\forall x\in X[x\cdot 1=x][/ilmath] and [ilmath]\forall g,h\in G\ \forall x\in X[(x\cdot g)\cdot h=x\cdot(g*h)][/ilmath].

Notations for [ilmath]x\cdot g[/ilmath] include [ilmath]xg[/ilmath] and [ilmath]x^g[/ilmath]

See also

Notes

  1. I have written [ilmath](\cdot):G\times X\rightarrow X[/ilmath] rather than the usual [ilmath]\cdot:G\times X\rightarrow X[/ilmath] notation for functions to make it clearer that there is a dot there; this notation isn't new or different, it's just because a lone [ilmath]\cdot[/ilmath] looks out of place.

References

  1. 1.0 1.1 Abstract Algebra - Pierre Antoine Grillet