<![CDATA[Math Is Fun Forum / Group Theory]]> 2014-03-19T02:13:36Z FluxBB http://www.mathisfunforum.com/viewtopic.php?id=4056 <![CDATA[Re: Group Theory]]> 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.

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http://www.mathisfunforum.com/profile.php?id=204521 2014-03-19T02:13:36Z http://www.mathisfunforum.com/viewtopic.php?pid=305256#p305256
<![CDATA[Re: Group Theory]]> 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.

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http://www.mathisfunforum.com/profile.php?id=97578 2014-03-08T18:00:36Z http://www.mathisfunforum.com/viewtopic.php?pid=304180#p304180
<![CDATA[Re: Group Theory]]> What is abstract algebra like?

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http://www.mathisfunforum.com/profile.php?id=95904 2014-03-08T17:44:21Z http://www.mathisfunforum.com/viewtopic.php?pid=304172#p304172
<![CDATA[Re: Group Theory]]> I don't know why, but recently I have started to really enjoy abstract algebra. Taking a graduate level group theory course right now!

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http://www.mathisfunforum.com/profile.php?id=97578 2014-03-08T17:07:37Z http://www.mathisfunforum.com/viewtopic.php?pid=304146#p304146
<![CDATA[Re: Group Theory]]> Great!!

Why didn't I see that before?

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http://www.mathisfunforum.com/profile.php?id=95904 2014-03-08T13:46:20Z http://www.mathisfunforum.com/viewtopic.php?pid=304076#p304076
<![CDATA[Re: Group Theory]]> 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

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http://www.mathisfunforum.com/profile.php?id=67694 2014-03-08T11:34:17Z http://www.mathisfunforum.com/viewtopic.php?pid=304044#p304044
<![CDATA[Re: Group Theory]]> 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...

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http://www.mathisfunforum.com/profile.php?id=118786 2014-03-08T10:51:04Z http://www.mathisfunforum.com/viewtopic.php?pid=304026#p304026
<![CDATA[Re: Group Theory]]> i still dont don't understand basics of groups and how to solve probems....:(

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http://www.mathisfunforum.com/profile.php?id=204272 2014-03-08T10:24:47Z http://www.mathisfunforum.com/viewtopic.php?pid=304025#p304025
<![CDATA[Re: Group Theory]]> Ricky wrote:

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

Are you sure? Or have I misunderstood something here? ]]>
http://www.mathisfunforum.com/profile.php?id=6777 2011-05-03T11:27:58Z http://www.mathisfunforum.com/viewtopic.php?pid=173201#p173201
<![CDATA[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.

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http://www.mathisfunforum.com/profile.php?id=87949 2011-03-03T14:10:15Z http://www.mathisfunforum.com/viewtopic.php?pid=166943#p166943
<![CDATA[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.

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http://www.mathisfunforum.com/profile.php?id=87949 2011-03-03T13:57:20Z http://www.mathisfunforum.com/viewtopic.php?pid=166942#p166942
<![CDATA[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.

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http://www.mathisfunforum.com/profile.php?id=87949 2011-03-03T13:49:17Z http://www.mathisfunforum.com/viewtopic.php?pid=166940#p166940
<![CDATA[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.

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http://www.mathisfunforum.com/profile.php?id=87949 2011-03-03T13:41:31Z http://www.mathisfunforum.com/viewtopic.php?pid=166938#p166938
<![CDATA[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.

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http://www.mathisfunforum.com/profile.php?id=87949 2011-03-03T13:29:20Z http://www.mathisfunforum.com/viewtopic.php?pid=166937#p166937
<![CDATA[Re: Group Theory]]> Hi dom pong;

Welcome to the forum! Why did you post here instead of Introductions?

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http://www.mathisfunforum.com/profile.php?id=33790 2011-01-27T10:26:57Z http://www.mathisfunforum.com/viewtopic.php?pid=163893#p163893