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Topic: Isomorphism between groups (Read 4278 times) 

knightfischer
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Isomorphism between groups
« on: Jun 8^{th}, 2009, 6:09pm » 
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I am working my way through an Abstract Algebra text (selfstudy) and I'm trying to solve the following problem. Group G = {R  {1}} , where a, b from G have the binary operation: a*b = a + b +ab. Show this groups is isomorphic to R^{x} (nonzero reals). Hint: to find the function Phi, try to map identity to identity. I know the identity in G is 0, and the identity in R^{x} is 1. I think the inverse of a in G is a/(1a). The inverse in R^{x} is the multiplicative inverse. I cannot figure out the function Phi. I thought of some constant C raised to a power, like Phi(x) = C^{x}, which would map 0 in G to 1 in R^{x}. But I could not figure out how that would map inverses to inverses, or how to show that this function defines an isomorphism. I'm a bit lost on this problem and any help would be appreciated. Thanks.


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Obob
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Re: Isomorphism between groups
« Reply #1 on: Jun 8^{th}, 2009, 6:30pm » 
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Phi(x) = C^x couldn't possibly work, because its image is only the positive reals. You need the image to be all nonzero reals. Hint: what's the absolute simplest function you know that takes 0 to 1, aside from phi(x) = 1 for all x? That function will work.

« Last Edit: Jun 8^{th}, 2009, 6:38pm by Obob » 
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knightfischer
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Re: Isomorphism between groups
« Reply #2 on: Jun 9^{th}, 2009, 11:23am » 
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I now see Phi(x) = X + 1. Thanks for your help. Is my reasoning and wording correct below: To show it is an isomorphism, I need to show: 1) Phi(a*b) = Phi(a)Phi(b) 2) Phi(x) is a bijection For 1), Phi(a*b) = Phi(a+b+ab) = a+b+ab+1 = (a+1)(b+1) = Phi(a)Phi(b). For 2), Phi(x) = 1, implies x = 0, so Phi(x) is injective. Then for any a in R^{x}, there exists an x in G such that a = x+1, so x =a1. Therefore Phi(x) is surjective. I'm not clear how to formally word the surjective part. What is the correct mathematical wording of this?


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Obob
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Re: Isomorphism between groups
« Reply #3 on: Jun 9^{th}, 2009, 1:07pm » 
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1) is exactly correct. For 2), the injectivity proof is fine (although it is slightly nontrivial that being injective is the same thing as saying that only the identity maps to the identity; this requires 1), for instace). For surjective, here is how I would write it if I were being pedantic: Let a be an element of R^{x}. Since a is not 0, a  1 is not 1, so a  1 is in G. Furthermore, Phi(a1) = a. Therefore Phi is surjective. Your explanation for surjectivity is the correct way to find the proof. When you actually write the proof, though, you don't need to go through the derivation of finding x. You also need to note that the x you found is actually an element of the group. Alternately, to show Phi is a bijection, you could define Psi(x) = x  1. Then observe that Psi Phi is the identity map of R^{x} and Phi Psi is the identity map of G. Finally, for one last proof that it is a bijection, you could say that as a map R > R it is a bijection (which should be obvious, although technically the proof is more or less the same as the above) and it takes 0 to 1. Thus the restriction to R^{x} > G is a bijection.

« Last Edit: Jun 9^{th}, 2009, 1:10pm by Obob » 
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knightfischer
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Re: Isomorphism between groups
« Reply #4 on: Jun 9^{th}, 2009, 1:40pm » 
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Thanks for your help with this problem.


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Eigenray
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Re: Isomorphism between groups
« Reply #5 on: Jun 9^{th}, 2009, 5:21pm » 
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If you already know that G is a group, then to show phi is injective it is enough to show that only the identity maps to the identity. But otherwise it doesn't follow. That is, suppose X has a binary operation on it, and we have a surjective function f : X > G such that f(a*b) = f(a)f(b), and the preimage of the identity of G consists of a single point. Then f need not be injective unless X is a group. For example, X could be G with a 'doubled point', say X = {0,1,1'}, G = Z/(2Z), f(0)=0, f(1)=f(1')=1, and 0*1 = 1, 0*1' = 1', 1*1' = 0. But on the other hand, if f is injective, then X must be a group (and f an isomorphism), because we have just taken the group structure on G and 'renamed' the elements (using f) to be elements of X. In fact, this would be the easiest way to show that G is in fact a group, if you did not already have this information. That is, phi(x) = x+1 is a bijection because x x1 is its inverse, and the operation on G is the same as that induced by f from the operation on R^{*}.


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