You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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..."

Offline

**xueli****Member**- Registered: 2010-10-18
- Posts: 1

im still not understand....

Offline

**vichet love****Member**- Registered: 2011-01-26
- Posts: 2

hello my name dom pong

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,927

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.**

**If it ain't broke, fix it until it is.**

Offline

**Reek****Member**- Registered: 2010-12-22
- Posts: 11

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 (2011-03-21 22:56:38)*

Offline

**Reek****Member**- Registered: 2010-12-22
- Posts: 11

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 (2011-03-21 22:57:16)*

Offline

**Reek****Member**- Registered: 2010-12-22
- Posts: 11

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 (2011-03-03 01:58:20)*

Offline

**Reek****Member**- Registered: 2010-12-22
- Posts: 11

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 (2011-03-21 22:57:49)*

Offline

**Reek****Member**- Registered: 2010-12-22
- Posts: 11

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 (2011-03-21 22:58:38)*

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Ricky wrote:

(It may be noted that matrices of integers are groups).

Are you sure? Or have I misunderstood something here?

Offline

**sobia****Member**- Registered: 2014-03-07
- Posts: 1

i still dont don't understand basics of groups and how to solve probems....:(

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,922

JaneFairfax wrote:

Ricky wrote:(It may be noted that matrices of integers are groups).

Are you sure? Or have I misunderstood something here?

I don't think they are a group under multiplication, because not all the matrices in the set have inverses...

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,016

hi sobia,

Welcome to the forum.

There's a very simple introduction at

http://www.mathsisfun.com/sets/groups-introduction.html

The link at the end takes you back to this post. If you have a particular question that you need help with, post it here.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

Great!!

Why didn't I see that before?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'But our love is like the wind. I can't see it but I can feel it.' -A Walk to remember

Offline

**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,646

I don't know why, but recently I have started to really enjoy abstract algebra. Taking a graduate level group theory course right now!

Offline

What is abstract algebra like?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'But our love is like the wind. I can't see it but I can feel it.' -A Walk to remember

Offline

**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,646

It's pretty abstract and broad. You start learning about algebraic structures by them selves, like groups, rings, fields etc. You start learning why things actually work (certain properties). I preferred analysis over it for quite a bit, but I'm starting to like it more nowadays.

Offline

**eigenguy****Member**- Registered: 2014-03-18
- Posts: 78

The greatest tool in the mathematician's tool chest is abstraction. I cannot count all the times when I've done some long, horrific calculation or proof, and some time later came across an abstraction that made the whole thing almost trivial.

Example: prove that det(AB) = det(A)det(B). If you try to do that for higher dimensions by looking at the coordinate formula, you'll be gibbering before very long (2D isn't bad, 3 is a pain, 4 is awful, much higher, and you might as well throw the whole thing in). But if you develop the concept of vector spaces, and introduce the wedge product, then suddenly, the determinant pops up in such a way that the det(AB) = det(A)det(B) result is a trivial consequence.

I find analysis more interesting myself, but abstract algebra shows up in pretty much all other fields, so it is a very good thing to master.

"Having thus refreshed ourselves in the oasis of a proof, we now turn again into the desert of definitions." - Bröcker & Jänich

Offline

Pages: **1**