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Showing Two Groups are Isomorphic 
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#19
Aug304, 08:25 AM

P: 867

Sorry about that. You have to show whether or not it is a group homomorphism and decide if it is an isomorphism.



#20
Aug304, 08:27 AM

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P: 9,396

diag{1,x} and diag{x, 1/x}



#21
Aug304, 08:27 AM

P: 867

Matt, how did you see that it isn't injective? I am having some difficulty working this out :C



#22
Aug304, 08:39 AM

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what's the determinant of a diagonal matrix with entries x and 1/x for any nonzero x?



#23
Aug304, 08:44 AM

P: 867

The determinant of a diagonal matrix is just the product of the entries on the main diagonal. So a 2x2 diagonal matrix with diagonal entries x is x². And the determinant of the 2x2 diagonal matrix with entries 1/x is 1/x². Is this what you mean?



#24
Aug304, 08:53 AM

P: 696

The determinant of a 2x2 diagonal matrix with elements x and 1/x (x nonzero) would be 1, not 1/x^2... I.e. there are many matrices which are not equal to one another, but have the same determinant.
Also, you can prove that no isomorphism can exist between those two groups ((R\{0}, x) is abelian but GL_2(R) isn't). 


#25
Aug304, 09:55 AM

P: 867

I see, I misunderstood. The 2x2 diagonal matrix has entry(11)= x and entry(22) = 1/x with entries(12,21) = 0. Then I agree, the determinant is 1.
But, as you said, many matrices have determinant 1 so the mapping between the two groups is not bijective. Hence it is not isomorphic? You said that because one group is abelian and the other is not indicates that the two groups are not isomorphic, is this another property that I do not know about? Could you explain. 


#26
Aug304, 10:22 AM

P: 696




#27
Aug304, 10:38 PM

P: 867

Why did you choose the diagonal entries to be x and 1/x?
I understand that the identity is mapped to itself det(I) = 1 And that the group operation is preserved. I can show this by proving det(A*B) = detA X detB not equal to 0. Hence the det is a group homomorphism between the two groups. Thanks for the help guys. 


#28
Aug404, 04:23 AM

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I chose x and 1/x because that was the most obvious way of showing that the map is not injective. Always make your life easier on yourself; diagonal and upper triangular matrices have determinants and traces that are easy to read off.



#29
Aug504, 02:57 AM

P: 867

I have another quick question.
If I were asked to prove that a certain operation * was a binary operation on a given set S. Would I simply prove that it obeys closure? That is, take x, y in S and show that x*y is back in S. If so, then it is a binary operator on S. 


#30
Aug504, 04:11 AM

P: 696

See the definition of a binary operator (you have to prove that given two elements x and y, x*y is uniquely determined as well).



#31
Aug504, 05:53 AM

P: 867

So if my set S was the set of all real numbers excluding 1. And (S, *) was a group where x*y = x + y + xy. How would I start proving that * is a binary operation?



#32
Aug504, 06:04 AM

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you would show it satisfied all the axioms defining a binary operation, that's all. It's a "just do it" proof. Clearly given two inputs there is a unique output, how about closure? Note, since you've called (S,*) a group, then * must be a binary operation, or it isn't a group.



#33
Sep2304, 11:31 AM

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The hints here have been outstanding. I wanted to add something though, and I can only think of this summary of what has been said:
1. to show two groups are isomorphic, you must find an isomorphism between them, 2. to show two groups are not isomorphic, you must show there cannot be any isomorphism, not just that one particular attempt fails. this is harder, as your argument has to apply to all potential isomorphisms. hence you need to find a property of groups that would be preserved by all isomorphisms, and yet which your two groups do not share, such as being commutative (which is called "abelian" for groups, in honor of Niels Abel). In general the search for properties that would be preserved by all isomorphisms is a deep and fundamental one in every area, sometimes called the search for "invariants". for example in algebraic curve theory, to show the projective plane curve x^3 + y^3 = z^3, is not rationally isomorphic to the line, can be done by outright cleverness, but is most efficiently done by producing the invariant called the genus. I.e. topoloogically the cubic is a doughnut and the "line" is a sphere. the proof of the fundamental theorem of algebra in topology, is done by finding some way of discerning the difference between the punctured plane and the plane itself, which eventually becomes the first homology group. i.e. you have to show why the unit circle cannot be pulled away from the origin without passing through the origin. this is usually done by computing the integral of dtheta, and applying greens theorem from calculus. in number theory one uses reduction "mod n" which says that any solution of an equation in integers would also yield a solution mod every n. Hence, since after division by 4, the equation x^2 = 2 has no solution (the left side always has remainder 0 or 1 after division by 4,) hence the equation x^2 = 204,840,962 also has no solution. 


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