zepdrix
  • zepdrix
Abstract Algebra Showing that an isomorphism goes both ways.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zepdrix
  • zepdrix
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 \(\large\rm \left\) with \(\large\rm \left\), then \(\large\rm \phi^{-1}\) is an isomorphism of \(\large\rm \left\) with \(\large\rm \left\).
zepdrix
  • zepdrix
@ganeshie8
zepdrix
  • zepdrix
Teacher actually did this one in class... I'm just having trouble with the notation I think.. hmm

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zepdrix
  • zepdrix
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
ikram002p
  • ikram002p
showing isomorphic means to show it's homorphism (bijective)
zepdrix
  • zepdrix
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
ikram002p
  • ikram002p
yes exactly :)
ikram002p
  • ikram002p
hmm with some twist xD
zepdrix
  • zepdrix
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.
ikram002p
  • ikram002p
the twist only a need to make sure phi^-1 fits the origin domain :3
ikram002p
  • ikram002p
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 ?
ganeshie8
  • ganeshie8
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
zepdrix
  • zepdrix
Yah, that definitely helps with that portion of it :) Buhaha, you keep editing XD making little corrections
zzr0ck3r
  • zzr0ck3r
@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.
zzr0ck3r
  • zzr0ck3r
@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.
zzr0ck3r
  • zzr0ck3r
fixed a few typos...
zepdrix
  • zepdrix
woops lil boo boo on part 1 :3 lol
zzr0ck3r
  • zzr0ck3r
bah, now I dont have in my clip board and I am to lazy lol
zepdrix
  • zepdrix
haha
zzr0ck3r
  • zzr0ck3r
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 :)
zzr0ck3r
  • zzr0ck3r
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 \(\phi^{-1}(ab)=\phi^{-1}(\phi(x)\phi(y))=\phi^{-1}(\phi(xy))=xy=\phi^{-1}(\phi(x))\phi^{-1}(\phi(y))=\\\phi^{-1}(a)\phi^{-1}(b)\)
zepdrix
  • zepdrix
@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 \(\) to \(\), 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.
zepdrix
  • zepdrix
`I guess I should have inserted`: Therefore \(\phi^{-1}\) is one-to-one `and` Therefore \(\phi^{-1}\) is onto \(S\) `somewhere into there :\` `woops`
zzr0ck3r
  • zzr0ck3r
you need to prove those two things
zepdrix
  • zepdrix
o :c hah
zzr0ck3r
  • zzr0ck3r
\(f:S\rightarrow S'\) 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 \[\phi^{-1}(ab)=\phi^{-1}(\phi(x)\phi(y))=\phi^{-1}(\phi(xy))=xy\\=\phi^{-1}(\phi(x))\phi^{-1}(\phi(y))=\phi^{-1}(a)\phi^{-1}(b).\]
zepdrix
  • zepdrix
Thanks for the help broski! I should get to sleep while I still can >.<
zzr0ck3r
  • zzr0ck3r
fo sure. no I love this stuff. :)
zzr0ck3r
  • zzr0ck3r
gn and may you dream of symmetries of the square

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