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zepdrix

  • one year ago

Abstract Algebra Showing that an isomorphism goes both ways.

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  1. zepdrix
    • one year ago
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    Recall that if \(\large\rm f:A\to B\) is a one-to-one 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 \(\large\rm \left<S,*\right>\).

  2. zepdrix
    • one year ago
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    @ganeshie8

  3. zepdrix
    • one year ago
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    Teacher actually did this one in class... I'm just having trouble with the notation I think.. hmm

  4. zepdrix
    • one year ago
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    His notes for showing one-to-one: \(\large\rm \phi^{-1}\) is one-to-one 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

  5. ikram002p
    • one year ago
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    showing isomorphic means to show it's homorphism (bijective)

  6. zepdrix
    • one year ago
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    So since \(\large\rm \phi\) is a one-to-one mapping, I'm able to immediately say that \(\large\rm \phi^{-1}\) must also be one-to-one? Is that where I'm starting from? :o

  7. ikram002p
    • one year ago
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    yes exactly :)

  8. ikram002p
    • one year ago
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    hmm with some twist xD

  9. zepdrix
    • one year ago
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    What 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.

  10. ikram002p
    • one year ago
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    the twist only a need to make sure phi^-1 fits the origin domain :3

  11. ikram002p
    • one year ago
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    f: A-->B mapping unique 1-1 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 1-1 and onto from B to A notations are clear now @zepdrix ?

  12. ganeshie8
    • one year ago
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    Since \(\phi\) 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

  13. zepdrix
    • one year ago
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    Yah, that definitely helps with that portion of it :) Buhaha, you keep editing XD making little corrections

  14. zzr0ck3r
    • one year ago
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    @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 one-to-one gives that \(\phi^{-1}\) is well defined.

  15. zzr0ck3r
    • one year ago
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    @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 one-to-one gives that \(\phi^{-1}\) is well defined.

  16. zzr0ck3r
    • one year ago
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    fixed a few typos...

  17. zepdrix
    • one year ago
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    woops lil boo boo on part 1 :3 lol

  18. zzr0ck3r
    • one year ago
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    bah, now I dont have in my clip board and I am to lazy lol

  19. zepdrix
    • one year ago
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    haha

  20. zzr0ck3r
    • one year ago
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    if 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 :)

  21. zzr0ck3r
    • one year ago
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    For 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 \(\phi^{-1}(ab)=\phi^{-1}(\phi(x)\phi(y))=\phi^{-1}(\phi(xy))=xy=\phi^{-1}(\phi(x))\phi^{-1}(\phi(y))=\\\phi^{-1}(a)\phi^{-1}(b)\)

  22. zepdrix
    • one year ago
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    @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 one-to-one, \(\phi^{-1}\) 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.

  23. zepdrix
    • one year ago
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    `I guess I should have inserted`: Therefore \(\phi^{-1}\) is one-to-one `and` Therefore \(\phi^{-1}\) is onto \(S\) `somewhere into there :\` `woops`

  24. zzr0ck3r
    • one year ago
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    you need to prove those two things

  25. zepdrix
    • one year ago
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    o :c hah

  26. zzr0ck3r
    • one year ago
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    \(f:S\rightarrow S'\) 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 one-to-one 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 \[\phi^{-1}(ab)=\phi^{-1}(\phi(x)\phi(y))=\phi^{-1}(\phi(xy))=xy\\=\phi^{-1}(\phi(x))\phi^{-1}(\phi(y))=\phi^{-1}(a)\phi^{-1}(b).\]

  27. zepdrix
    • one year ago
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    Thanks for the help broski! I should get to sleep while I still can >.<

  28. zzr0ck3r
    • one year ago
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    fo sure. no I love this stuff. :)

  29. zzr0ck3r
    • one year ago
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    gn and may you dream of symmetries of the square

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