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Group Theory – Determine if (S,*) is a group


(Phils)










































Abstract Algebra - Group Theory

Determine if (S,*) is a group (verify the properties of a group are satisfied).

    a. S= The set of real numbers; the binary operation * is defined by a*b= (a+b)/2, a,b are element of S

    b. S= The set of integers Z; the binary operation * is the ordinary operation subtraction.

    c. S= The set 3Z = {3n| n E Z};the binary operation * is the ordinary operation addition.

    d. S= The set G={0,1,2}; the binary operation * is defined by a*b = |a-b|, a and b are element of S.

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I know that in order for the set S to be a group, it must satisfy the four condition( closure, associative, identity, and inverse)

for Qa

1.it satisfies condition 1 because for every a*b, there is always c that is an element of S but i can't prove conditions 2,3 and 4

For Qb.

1. yes it is closed under subtraction because for evry a*b, there is c that is belong to the set S.

2. Subtraction is not associative on Z
condition 3 and 4 confuse me because accrding to the defintion of identity a*e=a. Based on this definition, the identity is 0. But if the identity is zero then 0 must be the difference of any integer and its inverese. This does not happen (ex. 3-(-3) =6 not 0
Does this mean there is no identity and inverse?

I don't know how to start question c

Qd.

yes it is closed because for every A*b, there is c thetis an element of S

Idon't know how to show that it is associative

yes it has an identity. It's 0

the inverses are:

the inverse of 0 is 0
the inverse of 1 is 1
the inverse of 2 is 2


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Aug 16, 2012
Determine if (S,*) is a group
by: Staff


Answer:
Part I

• Group: A group (S,*) is a nonempty set S and a binary operation * on S, where the following four conditions are true:

    (1) Closure: If a and b are elements of set S (a,b∈S) the element a*b is also an element of S.

    (2) Associativity: If a, b, and c are elements of set S (a,b,c∈S), the following is true for every a, b, and c: a*(b*c) = (a*b)*c.

    (3) Identity: If a is an element of set S (a∈S), there exists a single identity element e which is also an element of set S (e∈S), such that e*a=a and a*e=a for every a (a∈S).

    (4) Inverses : If a is an element of set S (a∈S), there also exists an inverse element of set S (a-1∈S), such that a*a-1 = e and a-1*a = e.


a. S= The set of real numbers; the binary operation * is defined by a*b= (a+b)/2, a,b are elements of S

***>>>(S,*) is NOT a GROUP

     (1) Closure: If a and b are elements of set S (a,b∈S) the element a*b is also an element of S.

         True.

         >>> The operation * is CLOSED on set S for the set of real numbers.


     (2) Associativity: If a, b, and c are elements of set S (a,b,c∈S), the following is true for every a, b, and c: a*(b*c) = (a*b)*c.

         a*b= (a+b)/2

         (a*b)*c

         (a*b)*c = [(a+b)/2 + C]/2 = a/4 + b/4 + c/2

         a*(b*c) = [A + (b+c)/2]/2 = a/2 + b/4 + c/4

         a*(b*c) ≠ (a*b)*c


         >>> The operation * is NOT ASSOCIATIVE on set S.


     (3) Identity: There exists an identity element e∈S such that e*a=a and a*e=a for all a∈S.

         e*a= (e+a)/2

         a*e= (a+e)/2

         (e+a)/2 = (a+e)/2

         solve for “e”

         e = a

         since the identity element depends upon the value of “a”, NO single IDENTITY ELEMENT EXISTS for all of set S.


         >>> The operation * does NOT have an IDENTITY ELEMENT in set S.


     (4) Inverses : For each a∈S there exists an inverse element a-1∈S such that a*a-1=e and a-1*a=e.

         since NO single IDENTITY ELEMENT EXISTS for all the elements of set S, NO INVERSE EXISTS.

         >>> The operation * does NOT have an INVERSE ELEMENT in set S.


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Aug 16, 2012
Determine if (S,*) is a group
by: Staff


------------------------------------------------

Part II


b. S= The set of integers Z; the binary operation * is the ordinary operation subtraction.

the binary operation * is defined by a*b = a - b

***>>>(S,*) is NOT a GROUP

     (1) Closure: If a and b are elements of set S (a,b∈S) the element a*b is also an element of S.

         True.

         >>> The operation * is CLOSED on set S for the set of integers (Z).


     (2) Associativity: If a, b, and c are elements of set S (a,b,c∈S), the following is true for every a, b, and c: a*(b*c) = (a*b)*c.

         True.

         >>> The operation * is ASSOCIATIVE on set S.


     (3) Identity: There exists an identity element e∈S such that e*a=a and a*e=a for all a∈S.

         e*a= e - a

         a*e= a – e

         e - a ≠ a - e


         e*a ≠ a*e

         >>> The operation * does NOT have an IDENTITY ELEMENT in set S.


     (4) Inverses : For each a∈S there exists an inverse element a-1∈S such that a*a-1=e and a-1*a=e.

         since NO single IDENTITY ELEMENT EXISTS for all the elements of set S, NO INVERSE EXISTS.

         >>> The operation * does NOT have an INVERSE ELEMENT in set S.


c. S = The set 3Z = {3n| n ∈ Z};the binary operation * is the ordinary operation addition.

the binary operation * is defined by a*b = a + b


the meaning of 3Z = {3n| n ∈ Z}


Set S = {3n | n ∈ ℤ}


{} curly brackets surround the expression
∈ = element of a set
| and : can be used interchangeably. Both notations are separators which mean “where” or “such that”

3n: the first element in the “output function”, shown as = {3n |
n: the second “n” is the “variable”, shown as | n ∈ ℤ,
ℤ (the set of all integers) is the“input set”


Reading from left to right: “S” is the set of all numbers “3n” {3n | … , …} where “n” is an element of the set of integers ℤ { … | n ∈ ℤ, … }.


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Aug 16, 2012
Determine if (S,*) is a group
by: Staff


------------------------------------------------

Part III


***>>>(S,*) is a GROUP

     (1) Closure: If a and b are elements of set S (a,b∈S) the element a*b is also an element of S.

         True.

         >>> The operation * is CLOSED on set S for the set of integers (Z).


     (2) Associativity: If a, b, and c are elements of set S (a,b,c∈S), the following is true for every a, b, and c: a*(b*c) = (a*b)*c.

         True.

         >>> The operation * is ASSOCIATIVE on set S.


     (3) Identity: There exists an identity element e∈S such that e*a=a and a*e=a for all a∈S.

         e*a= e + a

         a*e= a + e

         e + a = a + e

         0 + a = a + 0

         e*a = a*e

         >>> The operation * does have an IDENTITY ELEMENT in set S.


     (4) Inverses : For each a∈S there exists an inverse element a-1∈S such that a*a-1=e and a-1*a=e.

         a*a-1 = e

         a*a-1 = a + (-a) = 0, the identity element

         a-1*a = -a + a = 0, the identity element




         >>> The operation * does have an INVERSE ELEMENT in set S.



d. S= The set G={0,1,2}; the binary operation * is defined by a*b = |a - b|, a and b are elements of S.

the binary operation * is defined by a*b = |a - b|

Set S = {0,1,2}

***>>>(S,*) is NOT a GROUP


     (1) Closure: If a and b are elements of set S (a,b∈S) the element a*b is also an element of S.

         True.

         >>> The operation * is CLOSED on set S.


     (2) Associativity: If a, b, and c are elements of set S (a,b,c∈S), the following is true for every a, b, and c: a*(b*c) = (a*b)*c.

a*(b*c) = (a*b)*c


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Aug 16, 2012
Determine if (S,*) is a group
by: Staff


------------------------------------------------

Part IV


You can easily try out the possible permutations

6 possible Permutations without repetition

{0,1,2} {0,2,1} {1,0,2} {1,2,0} {2,0,1} {2,1,0}

--------------------------------------
a*(b*c) for {0,1,2}
|1-2| = |-1| = 1
|0-1| = |-1| = 1
a*(b*c) = 1

(a*b)*c for {0,1,2}
|0-1| = |-1| = 1
|1-2| = |-1| = 1
(a*b)*c = 1

a*(b*c) = (a*b)*c , OK

--------------------------------------

a*(b*c) for {0,2,1}
|2-1| = |1| = 1
|0-1| = |-1| = 1
a*(b*c) = 1

(a*b)*c for {0,2,1}
|0-2| = |-1| = 1
|1-1| = |0| = 0
(a*b)*c = 0

a*(b*c) ≠ (a*b)*c , ERROR

--------------------------------------

There is no need for any other calculations.


         >>> The operation * is NOT ASSOCIATIVE on set S.



     (3) Identity: There exists an identity element e∈S such that e*a=a and a*e=a for all a∈S.

         e*a= |e - a|

         a*e= |a - e|

         |e - a|= |a - e|

         |0 + a| = |a + 0|

         e*a = a*e

         >>> The operation * does have an IDENTITY ELEMENT in set S. The identity element is 0. 0 is an element of the set {0,1,2}.


     (4) Inverses : For each a∈S there exists an inverse element a-1∈S such that a*a-1=e and a-1*a=e.

         a*a-1 = e

         a*a-1 = |a - (-a)| ≠ 0


         >>> The operation * does NOT have an INVERSE ELEMENT in set S.






• Summary:

    Question a: (S,*) is NOT a GROUP

    Question b: (S,*) is NOT a GROUP

    Question c: (S,*) is a GROUP

    Question d: (S,*) is NOT a GROUP






Thanks for writing.

Staff
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