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IB Mathematics HL Option: Sets, Relations and groups (Abelian Groups- Cayley table)

How can we show that the set under the addition modulo 3 is an Abelian group? and what is the order of each element?

Thanks

How can we show that the set under the addition modulo 3 is an Abelian group? and what is the order of each element?

Thanks

- elizabeth
**Posts:**0**Joined:**Mon Jan 28, 2013 8:09 pm

IB Mathematics HL Option Sets, Relations and Groups - Cayley table

The basic principles about Abelian Groups are:

First of all I remind you the definition of Group (G,*)

Let G be a non-empty set on which a binary operation * is defined . We say G is a Group under this operation if each of the following four properties hold:

1. G is closed under * , i.e., for every

2. The operation * is associative on G,

i.e., for every

3. The operation * has an identity element in G,

i.e., for every

4. Each element of G has an inverse under *,

i.e., for every

A group {G,*} is called Abelian if for every

Now concerning your question

First we must construct the Cayley table in order to determine closure and the existence of identity and inverse elements.

the set under the addition modulo 3

Closure: The addition modulo 3 on the set is closed

i.e. for every

Associative: The addition on is Associative

i.e. x+(y+z)=(x+y)+z for every

Identity: For every element there is an element, the zero, which is the identity element in under addition.

i.e. x+0=0+x=0

Inverse: The identity element {0} appears once in every row and every column, so every element has an iverse.

Commutative: The addition on is Commutative

i.e. x+y=y+x for every

As for the order you have to know the following

The order of a group {G,*} is the number of elements in G

The order of an element x of a group {G,*} is the smallest positive integer n for which with e the identity element.

A group {G,*} is said to be finite if it has a finite order and is said to be infinite if it has infinite elements.

So in you exercise the identity element has order 1 since 0=0

the element 1 has order 3 since 1+1+1=0

and finally the element 2 has also order 3 since 2+2+2=0

Hope these help

The basic principles about Abelian Groups are:

First of all I remind you the definition of Group (G,*)

Let G be a non-empty set on which a binary operation * is defined . We say G is a Group under this operation if each of the following four properties hold:

1. G is closed under * , i.e., for every

2. The operation * is associative on G,

i.e., for every

3. The operation * has an identity element in G,

i.e., for every

4. Each element of G has an inverse under *,

i.e., for every

A group {G,*} is called Abelian if for every

Now concerning your question

First we must construct the Cayley table in order to determine closure and the existence of identity and inverse elements.

the set under the addition modulo 3

Closure: The addition modulo 3 on the set is closed

i.e. for every

Associative: The addition on is Associative

i.e. x+(y+z)=(x+y)+z for every

Identity: For every element there is an element, the zero, which is the identity element in under addition.

i.e. x+0=0+x=0

Inverse: The identity element {0} appears once in every row and every column, so every element has an iverse.

Commutative: The addition on is Commutative

i.e. x+y=y+x for every

As for the order you have to know the following

The order of a group {G,*} is the number of elements in G

The order of an element x of a group {G,*} is the smallest positive integer n for which with e the identity element.

A group {G,*} is said to be finite if it has a finite order and is said to be infinite if it has infinite elements.

So in you exercise the identity element has order 1 since 0=0

the element 1 has order 3 since 1+1+1=0

and finally the element 2 has also order 3 since 2+2+2=0

Hope these help

- miranda
**Posts:**268**Joined:**Mon Jan 28, 2013 8:03 pm

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