Origional Isomorphisms

  1. If an group G is isomorphic to a group G(prime) then they are only equal or approximately equal. If this can continue on, i.e. G(prime) is isomorphic to a group W... then we are not sure if G is the origional group. Is it possible to find a group that is an origional isomorphism? Does this question even make sense to ask? thanks.
     
  2. jcsd
  3. matt grime

    matt grime 9,395
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    Equal and isomorphic are not the same, and origional isn't even a word.

    In general it only makes sense to say two things in maths are equal if they are 'exactly the same thing', thus if we take the group of symmetries of the n-gon where n is even, eg the square or hexagon, then the stabilzer of one vertex equals the stabilizer of the opposite vertex (and is the reflection through those vertices plus the identity), whereas the stabilizers of two arbitrary vertices are merely isomorphic but not equal.
     
    Last edited: Dec 16, 2005
  4. JasonRox

    JasonRox 2,327
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    I have a question myself.

    Isomorphism does not mean equal, but let's say you have an isomorphism between G and H, with the isomorphic function being f, which is known.

    My thoughts are that if you are working with a group, and just hate doing multiplication (or whatever the binary operation might be) in group G, but doing it in H isn't as bad. Wouldn't it just make sense to use f to map the necessary numbers to H, and then map them back to G with the solution?

    Just curious.
     
  5. Hurkyl

    Hurkyl 16,089
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    It could. Or, you might hate applying f or f^-1 even more than you hate multiplying in G. :smile: (IOW, it depends)
     
  6. HallsofIvy

    HallsofIvy 40,946
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    This doesn't make a whole lot of sense. If G is isomorphic to G', that means that they are, in fact, the same group but with different notation. I don't see what it could mean to say two groups are "approximately equal". Certainly that could have nothing to do with "isomorphic".

    Similarly, it makes no sense to talk about "a group that is an original isomorphism". A group is not an isomorphism- an isomorphism is a function between two groups.
     
  7. JasonRox

    JasonRox 2,327
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    That's very true, but you never know.
     
  8. matt grime

    matt grime 9,395
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    f is an isomorphism not 'an isomorphic function', and just because you know two groups are isomorphic does not mean you know of any explicit isomorphism between them.

    numbers? what numbers?

    [quoteto H, and then map them back to G with the solution?
    [/QUOTE]
     
  9. JasonRox

    JasonRox 2,327
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    Did I not say f is known?
     
  10. matt grime

    matt grime 9,395
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    And I am saying that finding an isomorphism between two groups is genuinely hard (in the computational sense as well).
     
  11. JasonRox

    JasonRox 2,327
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    I already know that it's not easy. That wasn't my question though.
     
  12. matt grime

    matt grime 9,395
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    There are certainly some presentations of an abstract group that are easier to work with than others, if that's what you mean. Though I have no idea what 'hard composition' might mean, and unless you're going to tell me *how* you're going to give me two elements, and what kind of identification you want me to make from the answer to some preferred set of descriptions, then I can't offer any opinion. After all if you give me a and b the composition is ab. But perhaps you're asking me: given a and b and another element c is ab the same as c? As a simple case in S_3 the composition of (12) and (23) is (12)(23), it is also (123).
     
    Last edited: Dec 17, 2005
  13. JasonRox

    JasonRox 2,327
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    Here is what I'm saying.

    Let's say you have a and b in group G, but for some reason it's pain in the ass to find ab. If there is an isomorphism f to H, where finding f(a)f(b), wouldn't it just be easier to map it to H, find f(ab) is much easier, then map it back to G.

    But like Hurkyl said, applying f or f^-1 will probably be just as bad or worse than just simply finding ab.
     
  14. Hurkyl

    Hurkyl 16,089
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    I didn't say "probably", I said "might". :tongue2: Transforming a problem into a more efficient format is a common optimization theme. For example, when doing linear algebra on a computer, matrices are usually stored as a vector. (And for certain operations, are sometimes even converted into different things that can be manipulated more efficiently for that operation)
     
  15. matt grime

    matt grime 9,395
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    I know what you're saying and as far as i can tell it is full of mights, maybes, and perhapses. Try and find an example instead of speculating wildly.
     
  16. Jason: Yes, the method you suggest can be employed for calculating integral transforms (convolving is harder than multiplying). f and f^-1 may be hard, but there are tables for them. Note that these contexts are not purely group-theoretic, but the idea is the same.

    Matt grime: what's your deal? When some one asks a question, why do you speak of unrelated things? Someone says "I have two groups and an isomorphism", and respond with "isomorphisms are hard to find"? You claim that you knew what he was asking, so why did you change the subject? It seems like you're being deliberately obtuse.
     
  17. matt grime

    matt grime 9,395
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    If some one asks about using isomorphisms to make things easier then surely it is reasonable to point out that finding an isomorphism is a (computationally) hard thing to do? Or perhaps I misread the first post of Jason's (which I believe I did). Very rarely do you have two groups and an explicit isomorphism in any practical situation. Since the question is about practical considerations surely that is a reasonable point to raise.

    Certainly if you know a set of matrices of size 10^6 square is isomorphic to C_n as a group you ain't going to multiply together two large matrices, are you? So the answer is obviously, yes, isomorphisms can save you a lot of work. I can't believe this question even needs to be asked.

    However, determining the structure of even known abelian groups is hard (eg points on elliptic curves over finite fields).

    Since obtuse means blunt (or even insensitive) then it is probably exactly right to say I am being deliberately obtuse. You might also want to consider the fact that Jason raised a question unrelated to the original point of the thread if you're going to accuse people of bringing up unrelated things.
     
    Last edited: Dec 21, 2005
  18. JasonRox

    JasonRox 2,327
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    matt's a good guy. Just some misunderstandings.

    Just like matt grime said, the answer is obvious. I should have never of asked. I just wanted some certainty.
     
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