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

1. 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 $$\large\rm \left<S,*\right>$$ 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

@ganeshie8

3. zepdrix

Teacher actually did this one in class... I'm just having trouble with the notation I think.. hmm

4. 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

5. ikram002p

showing isomorphic means to show it's homorphism (bijective)

6. 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

7. ikram002p

yes exactly :)

8. ikram002p

hmm with some twist xD

9. 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.

10. ikram002p

the twist only a need to make sure phi^-1 fits the origin domain :3

11. 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 ?

12. 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

13. zepdrix

Yah, that definitely helps with that portion of it :) Buhaha, you keep editing XD making little corrections

14. 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.

15. 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.

16. zzr0ck3r

fixed a few typos...

17. zepdrix

woops lil boo boo on part 1 :3 lol

18. zzr0ck3r

bah, now I dont have in my clip board and I am to lazy lol

19. zepdrix

haha

20. 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 :)

21. 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 $$\phi^{-1}(ab)=\phi^{-1}(\phi(x)\phi(y))=\phi^{-1}(\phi(xy))=xy=\phi^{-1}(\phi(x))\phi^{-1}(\phi(y))=\\\phi^{-1}(a)\phi^{-1}(b)$$

22. 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 $$<S,*>$$ 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

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

you need to prove those two things

25. zepdrix

o :c hah

26. 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 $\phi^{-1}(ab)=\phi^{-1}(\phi(x)\phi(y))=\phi^{-1}(\phi(xy))=xy\\=\phi^{-1}(\phi(x))\phi^{-1}(\phi(y))=\phi^{-1}(a)\phi^{-1}(b).$

27. zepdrix

Thanks for the help broski! I should get to sleep while I still can >.<

28. zzr0ck3r

fo sure. no I love this stuff. :)

29. zzr0ck3r

gn and may you dream of symmetries of the square