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Difference between revisions of "Abelian group"
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Commutativity
 
Commutativity
 
           For all <math>a,b \in S</math>, <math>a \bullet b = b \bullet a</math>.
 
           For all <math>a,b \in S</math>, <math>a \bullet b = b \bullet a</math>.
 +
 +
A simple example of an abelian group is <math>\mathbb{Z}</math> under addition. It is simple to show that it meets all the requirements.
 +
 +
Closure
 +
          For all <math>a,b \in \mathbb{Z} , a+b \in \mathbb{Z}</math>.
 +
Associativity
 +
          For all <math>a,b,c \in \mathbb{Z} , (a+b)+c = a+(b+c)</math>.
 +
Identity Element
 +
          For all <math>a \in \mathbb{Z} , a+0 = 0+a = a</math>.
 +
Inverse Element
 +
          For all <math>a \in \mathbb{Z} , a+ -a = 0</math>.
 +
Commutativity
 +
          For all <math>a,b \in \mathbb{Z} , a+b = b+a</math>.
 +
 +
Seeing as <math>\mathbb{Z}</math> meets all of these requirements under addition, we can say that <math>\mathbb{Z}</math> is abelian under addition.
  
 
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Revision as of 18:50, 12 August 2015

An abelian group is a group in which the group operation is commutative. For a group to be considered "abelian", it must meet several requirements.

Closure 

         For all $a,b$ $\in$ $S$, and for all operations $\bullet$, $a\bullet b \in S$.

Associativity 

         For all $a,b,c$ $\in$ $S$ and all operations $\bullet$, $(a\bullet b)\bullet c=a\bullet(b\bullet c)$.

Identity Element 

         There exists some $e \in S$ such that $a \bullet e = e \bullet a = a$.

Inverse Element 

         For all $a \in S$, there exists some $a^{-1}$ such that $a \bullet a^{-1} = e$

Commutativity 

         For all $a,b \in S$, $a \bullet b = b \bullet a$.

A simple example of an abelian group is $\mathbb{Z}$ under addition. It is simple to show that it meets all the requirements.

Closure 

         For all $a,b \in \mathbb{Z} , a+b \in \mathbb{Z}$.

Associativity 

         For all $a,b,c \in \mathbb{Z} , (a+b)+c = a+(b+c)$.

Identity Element 

         For all $a \in \mathbb{Z} , a+0 = 0+a = a$.

Inverse Element 

         For all $a \in \mathbb{Z} , a+ -a = 0$.

Commutativity 

         For all $a,b \in \mathbb{Z} , a+b = b+a$.

Seeing as $\mathbb{Z}$ meets all of these requirements under addition, we can say that $\mathbb{Z}$ is abelian under addition.

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