You are not logged in.
#1 2006-08-03 03:28:20
- Ricky
- Moderator
Offline
Group Theory
To be used with Introduction to Groups
1. Show that the set {0} with addition is a group.
2. Show that the set {0} with multiplication is a group.
3. Show that the set {1} with addition is not a group.
4. Show that the set {1} with multiplication is a group.
5. Show that the set {-1, 1} is a group under multiplication, but not addition.
6. Name the multiplicative inverse for -1 in the group {-1, 1} under multiplication.
7. Show that if abab = aabb, then it must be that ab = ba.
8. Show that the matrix [1 2][2 1] * [3 2][1 2] does not equal [3 2][1 2] * [1 2][2 1] (It may be noted that matrices of integers are groups). This would mean that matrices are not abelian.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#4 2011-01-27 21:26:57
- bobbym
- Administrator
Online
Re: Group Theory
Hi dom pong;
Welcome to the forum! Why did you post here instead of Introductions?
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
#5 2011-03-04 00:29:20
- Reek
- Member
Offline
Re: Group Theory
1. Show that the set {0} with addition is a group.
For any elements a and b of {0}, (a+b) is an element of {0}. The closure law has been followed.
For any a,b,c of {0}; a+(b+c) = (a+b)+c. The associative law has been followed.
For any a of {0} i+a=a, where i is a particular element in {0}.The left identity element i is 0 here.
For any a of {0} the equation x+a=i has a solution known as the left inverse of a.0 is the only element in {0} and the left inverse of 0 is 0.
All these properties are followed by this set that is closed under addition.
Therefore, {0} is a group with respect to addition.
Last edited by Reek (2011-03-22 21:56:38)
#6 2011-03-04 00:41:31
- Reek
- Member
Offline
Re: Group Theory
2.Show that the set {0} with multiplication is a group.
For any elements a and b of {0}, (a*b) is an element of {0}. The closure law has been followed.
For any a,b,c of {0}; a*(b*c) = (a*b)*c. The associative law has been followed.
For any a of {0} i*a=a, where i is a particular element in {0}.The left identity element i is 0 here.
For any a of {0} the equation x*a=i has a solution known as the left inverse of a.0 is the only element in {0} and the left inverse of 0 is 0.
All these properties are followed by this set that is closed under multiplication.
Therefore, {0} is a group with respect to multiplication.
Last edited by Reek (2011-03-22 21:57:16)
#7 2011-03-04 00:49:17
- Reek
- Member
Offline
Re: Group Theory
3. Show that the set {1} with addition is not a group.
For any elements a and b of {1}, (a+b) is not an element of {1}. 1+1=2. It is enough to show only one rule-break to prove that {1} is not a group with respect to addition.
We conclude {1} is not a group with respect to addition.
Last edited by Reek (2011-03-04 00:58:20)
#8 2011-03-04 00:57:20
- Reek
- Member
Offline
Re: Group Theory
4. Show that the set {1} with multiplication is a group.
For any elements a and b of {1}, (a*b) is an element of {1}. The closure law has been followed.
For any a,b,c of {1}; a*(b*c) = (a*b)*c. The associative law has been followed.
For any a of {1} i*a=a, where i is a particular element in {1}.The left identity element i is 1 here.
For any a of {1} the equation x*a=i has a solution known as the left inverse of a.1 is the only element in {1} and the left inverse of 1 is 1.
All these properties are followed by this set that is closed under multiplication.
Therefore, {1} is a group with respect to multiplication.
Last edited by Reek (2011-03-22 21:57:49)
#9 2011-03-04 01:10:15
- Reek
- Member
Offline
Re: Group Theory
5. Show that the set {-1, 1} is a group under multiplication, but not addition.
For any elements a and b of {-1, 1}, (a+b) is not an element of {-1, 1}. It is enough to show only one rule-break to prove that {-1, 1} is not a group with respect to addition.
We conclude {-1, 1} is not a group with respect to addition.
------------------------------------------------------------------------------------------------------------------------
For any elements a and b of {-1,1}, (a*b) is an element of {1,-1}. The closure law has been followed.
For any a,b,c of {1,-1}; a*(b*c) = (a*b)*c. The associative law has been followed.
For any a of {1,-1} i*a=a, where i is a particular element in {1,-1}.The left identity element i is 1 here.
For any a of {1,-1} the equation x*a=i has a solution known as the left inverse of a.
All these properties are followed by this set that is closed under multiplication.
Therefore, {1,-1} is a group with respect to multiplication.
Last edited by Reek (2011-03-22 21:58:38)
#10 2011-05-03 21:27:58
- JaneFairfax
- Legendary Member
Offline
Re: Group Theory
Ricky wrote:
(It may be noted that matrices of integers are groups).
Are you sure? Or have I misunderstood something here? 