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.

]]>Why didn't I see that before?

]]>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

]]>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...

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

Are you sure? Or have I misunderstood something here?

]]>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.

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.

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.]]>

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.

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.

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

]]>