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zepdrix
 one year ago
Abstract Algebra
Showing that an isomorphism goes both ways.
zepdrix
 one year ago
Abstract Algebra Showing that an isomorphism goes both ways.

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Recall that if \(\large\rm f:A\to B\) is a onetoone function mapping A onto B, then \(\large\rm f^{1}(b)\) is the unique \(\large\rm a\in A\) such that \(\large\rm f(a)=b\). Prove that if \(\large\rm \phi:S\to S'\) is an isomorphism of \(\large\rm \left<S,*\right>\) with \(\large\rm \left<S',*'\right>\), then \(\large\rm \phi^{1}\) is an isomorphism of \(\large\rm \left<S',*'\right>\) with \(\large\rm \left<S,*\right>\).

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Teacher actually did this one in class... I'm just having trouble with the notation I think.. hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2His notes for showing onetoone: \(\large\rm \phi^{1}\) is onetoone as \(\large\rm \phi^{1}(a)=\phi^{1}(b)\) implies that \[\large\rm a=\phi(\phi^{1}(a))=\phi(\phi^{1}(b))=b\] I'm just not seeing how that makes use of the original relationship being isomorphic. hmm

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0showing isomorphic means to show it's homorphism (bijective)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So since \(\large\rm \phi\) is a onetoone mapping, I'm able to immediately say that \(\large\rm \phi^{1}\) must also be onetoone? Is that where I'm starting from? :o

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0hmm with some twist xD

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2What he wrote for onto: \(\large\rm \phi^{1}\) is onto as \(\large\rm \phi^{1}(\phi(x))=x\) for all \(\large\rm x\in S\). Again, I just don't understand how this makes use of phi mapping S onto S' or whatever. I think I can understand the isomorphic property that I have to show after that. I'm just getting confused on these bijective properties.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0the twist only a need to make sure phi^1 fits the origin domain :3

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0f: A>B mapping unique 11 also onto thus its isomorphic on (s,*) (we call this line phi) u need to show f^1: B>A isomorphic on (s',*') means prove that phi^1 isomorphic by showing f^1 is 11 and onto from B to A notations are clear now @zepdrix ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Since \(\phi\) is an isomorphism, it is also a homomorphism, so we have \(\phi(xy) = \phi(x)\phi(y)\) for \(x,y\in S\) let \(x = \phi^{1}(a)\) and \(y = \phi^{1}(b)\), plug them above : \(\phi( \phi^{1}(a) \phi^{1}(b)) = \phi(\phi^{1}(a))\phi (\phi^{1}(a))\) \( \implies \phi^{1}(a) \phi^{1}(b) = ab\) \( \implies \phi^{1}(\phi^{1}(a) \phi^{1}(b)) = \phi^{1}(ab)\) that proves the homomorhism part for inverses

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yah, that definitely helps with that portion of it :) Buhaha, you keep editing XD making little corrections

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2@zepdrix I think what you are missing here is a common thing when talking about inverse relations, that is, we need to prove it is a function. To prove that \(\phi^{1} :S'\rightarrow S\) is a function you must show that it is 1) Well defined i.e. \(a=b\implies f(a) = f(b)\) and 2) defined everywhere i.e. \(\forall \ x\in S' \ \exists \ y\in S \ \text{ s.t. } f(x) = y\) Now the fact that \(\phi\) is onto, gives that \(\phi^{1}\) is defined everywhere and the fact that \(\phi\) is onetoone gives that \(\phi^{1}\) is well defined.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2@zepdrix I think what you are missing here is a common thing when talking about inverse relations, that is, we need to prove it is a function. To prove that \(\phi^{1} :S'\rightarrow S\) is a function you must show that it is 1) Well defined i.e. (a=b\implies \phi^{1}(a) = \phi^{1}(b)\) and 2) defined everywhere i.e. \(\forall \ x\in S' \ \exists \ y\in S \ \text{ s.t. } \phi^{1}(x) = y\) Now the fact that \(\phi\) is onto, gives that \(\phi^{1}\) is defined everywhere and the fact that \(\phi\) is onetoone gives that \(\phi^{1}\) is well defined.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2woops lil boo boo on part 1 :3 lol

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2bah, now I dont have in my clip board and I am to lazy lol

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2if you did not use the fact that the original function was well defined to show that the inverse is one to one and if you did not use the fact that the original function was defined everywhere to show that the inverse is onto Then it is wrong. again, I am to lazy to read :)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2For the isomorphism part you want to start with two arbitrary elements in \(S'\), say \(a,b\in S'\) and we want to show \(\phi^{1}(ab)=\phi^{1}(a)\phi^{1}(b)\). Now \(\exists x,y\in S\) s.t. \(\phi(x) = a, \phi(y)=b\) So we have \(\phi^{1}(ab)=\phi^{1}(\phi(x)\phi(y))=\phi^{1}(\phi(xy))=xy=\phi^{1}(\phi(x))\phi^{1}(\phi(y))=\\\phi^{1}(a)\phi^{1}(b)\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2@zzr0ck3r If you get bored, maybe you can read this and see if my wording and steps make sense :) I doubt you'd get THAT bored though lol.  Since \(\phi:S\to S'\) is an isomorphism of \(<S,*>\) to \(<S',*'>\), this implies that \(\phi\) is a bijective function. Since \(\phi\) is onetoone, \(\phi^{1}\) is well defined. Meaning, that if \(\phi^{1}(a)=\phi^{1}(b)\) then \(a=b\) for all \(a,b\in S'\). Since \(\phi\) is onto, \(\phi^{1}\) is everywhere defined. Meaning, that for all \(x\in S\) there exists a unique element \(y\in S'\) such that \(\phi^{1}(x)=y\). For \(a,b\in S'\), let \(a=\phi(x)\) and \(b=\phi(y)\). Then \[\phi^{1}(a*'b)\\=\phi^{1}(\phi(x)*'\phi(y))\\=\phi^{1}(\phi(x*y))\\=x*y\\=\phi^{1}(a)*\phi^{1}(b)\] I used the word "since" a lot, it doesn't read very nicely. Just trying to see if the steps are logical and sufficient.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2`I guess I should have inserted`: Therefore \(\phi^{1}\) is onetoone `and` Therefore \(\phi^{1}\) is onto \(S\) `somewhere into there :\` `woops`

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2you need to prove those two things

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\(f:S\rightarrow S'\) is an isomorphism. We want to show \(f^{1}:S'\rightarrow S\) is a bijective homomorphism. 1) Since \(f\) is onto it follows that \(f^{1}\) is defined everywhere and since \(f\) is onetoone it follows that \(f^{1}\) is well defined. So \(f^{1}\) is a function. 2) Let \(s\in S\), then \(f(s)\in S'\) that is \(f(s) = s'\) for some \(s'\in S'\) (since \(f\) is defined everywhere). Now \(f(s) = s'\implies f^{1}(f(s))=f^{1}(s)\implies s=f^{1}(s)\). So \[\forall y\in S \ \exists \ x \in S' \text{ s.t. } f^{1}(x)=y\] that is \(f^{1}\) is surjective. 3) Suppose \(f^{1}(a)= f^{1}(b)\) then, since \(f\) is well defined, \(f(f^{1}(a))=f^{1}(f(b))\implies a=b\), showing \(f^{1}\) is injective. 1), 2), and 3) show that \(f^{1}\) is a bijection. 4) Suppose \(a,b\in S'\). Then, since \(f\) is onto, \(\exists \ x,y \in S\) such that \(\phi(x) = a, \phi(y)=b\). So \[\phi^{1}(ab)=\phi^{1}(\phi(x)\phi(y))=\phi^{1}(\phi(xy))=xy\\=\phi^{1}(\phi(x))\phi^{1}(\phi(y))=\phi^{1}(a)\phi^{1}(b).\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Thanks for the help broski! I should get to sleep while I still can >.<

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2fo sure. no I love this stuff. :)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2gn and may you dream of symmetries of the square
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