I have several questions here: Define F: R^2 to R^2 by F(a,b) = (-a, -b). Show t
ID: 2943941 • Letter: I
Question
I have several questions here:Define F: R^2 to R^2 by F(a,b) = (-a, -b). Show that G is an automorphism for this group under vector addition. He has NOT shown this in class, it's not in the book.
Define G: R^2 to R^2 by G(a,b) = (a,-b). Show G is an automorphism.
Isn't any group automatically automorphic to itself? What's the deal with defining F and referencing G in the first question?
Lastly: Let G = (a + b sqrt 2), where a, b are natural numbers not both 0. Prove G is a group under ordinary multiplication.
Thank you, once again, super math people. Thank goodness you are out there in cyberspace willing to explain something my teachers (whom I pay to teach me) will NOT.
Explanation / Answer
The word 'automorphism' can have different meanings depending upon the context so you will have to check with your textbook but most often it means that the mapping preserves the group operation. i.e., the mapping F satisfies the property
F( x + y) = F(x) + F(y) where x and y are ordered pairs in 2-space.
For the first part, let (a,c) and (b,d) be two points in R^2.
Then F{(a,c) + (b,d)} = F{(a+b, c+d)} = {-(a+b), -(c+d)}= (-a-b, -c-d) = (-a,-c) + (-b,-d) = F(a.c) + F(b,d)
Your mapping G works exactly the same way so you try that.
For the set whose elements are of the form a + b2 (I think you mean a,b real numbers not both zero) you need to show closure under multiplication and the existence of an inverse.
For closure, (a + b2)(c + d2) = ac + ad2 + bc2 + 2bd = (ac + 2bd) + (ad + bc)2 - the desired format.
The (multiplicative) identity here is 1 + 02. So to find the inverse element consider the equaion in x and y,
(a + b2)(x + y2) = (1 + 02).
Multiplying and equating like terms will give you:
x = -a/(2b^2 - a^2)
y = b/(2b^2 - a^2) (This is why a and b can not both be zero).