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Math Help - An Isomorphism Question!

  1. #1
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    An Isomorphism Question!

    Suppose that (A,*), (B,&) and (C,@) are sets with prescribed binary operations.

    Prove that if (A,*) is isomorphic to (B,&) and (B,&) is isomorphic to (C,@), then (A,*) is isomorphic to (C,@).
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  2. #2
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    By the given condition you have an isomorphism f:A\to B and g:B\to C.

    What is the obvious choice if you want to make an isomorphism from A to C?

    Claim this obvious choice is an isomorphism and prove it, meaning you need to show that this map is a 1-1, onto, homomorphism.
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  3. #3
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    Quote Originally Posted by watchmath View Post
    By the given condition you have an isomorphism f:A\to B and g:B\to C.

    What is the obvious choice if you want to make an isomorphism from A to C?

    Claim this obvious choice is an isomorphism and prove it, meaning you need to show that this map is a 1-1, onto, homomorphism.
    Should i use the funtion f(x)=x, and g(x)=x, then f(g(x))=x. Is that what u ment by the obvious choice?
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  4. #4
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    We don't know what is the explicit formula for f and g . But now if you have a function from A to B and from B to C, how would you make a function from A to C (by using f and g)?
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  5. #5
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    g o f?
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  6. #6
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    Quote Originally Posted by Lipticboven View Post
    g o f?
    Exactly!!
    Now your job is to verify that this function is a homomorphism, 1-1 and onto.
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  7. #7
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    Quote Originally Posted by watchmath View Post
    Exactly!!
    Now your job is to verify that this function is a homomorphism, 1-1 and onto.
    To prove it is a homomorhism would it be ok to say:
    since(A,*) is isomorphic to (B,&) and (B,&) is isomorphic to (C,@) then,
    f(a*b) = f(a) & f(b)
    g(a&b) = g(a) @ g(b)

    So

    g(f(a*b)) = g(f(a) & f(b)) = g(f(a)) @ g(f(b))

    Therefore (A,*) is isomorphic to (C,@).
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  8. #8
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    Quote Originally Posted by Lipticboven View Post
    To prove it is a homomorhism would it be ok to say:
    since(A,*) is isomorphic to (B,&) and (B,&) is isomorphic to (C,@) then,
    f(a*b) = f(a) & f(b)
    g(a&b) = g(a) @ g(b)

    So

    g(f(a*b)) = g(f(a) & f(b)) = g(f(a)) @ g(f(b))

    Therefore (A,*) is isomorphic to (C,@).
    You just proved that g 0 f is a homomorphism. You need now to prove that this function is 1-1 and onto
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  9. #9
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    and to prove it is one-to-one would it be ok to say:

    Since (A,*) is isomorphic to (B,&) it means,
    if f(a) = f(b) then a = b

    Similarly since (B,&) is isomorphic to (C,@) it means,
    if g(a) = g(b) then a = b

    Therefore if g(f(a)) = g(f(b) it means f(a) = f(b), which in turn means a = b therefore it shows (A,*) and (C,@) are one to one.

    I am a bit stuck proving it is onto thou
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  10. #10
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    Yest, thats OK!

    For onto, take arbitrary element c in C. You show that this is onto if you can find x in A such that g 0 f (x)= g(f(x))=c.
    You need to use the onto-ness of f and g.

    Since g is onto there is an element b in B such that g(b)=c.

    So -----> g(f(x))=c <---------- what you want

    --------> g(b)=c <------------------------ what you have

    So now how to produce x so that from what you have you get what you want
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