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#1 2006-08-03 03:28:20
- Ricky
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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
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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.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#5 2011-03-04 00:29:20
- Reek
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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
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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
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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
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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
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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.
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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
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Re: Group Theory
Ricky wrote:
(It may be noted that matrices of integers are groups).
Are you sure? Or have I misunderstood something here? 