Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

7 iso and hom questions

  1. Nov 22, 2004 #1
    1-3) determine whether the given map (theta) is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not?

    1) <Z,+> with <Z,+> where theta(n)= -n for n elements in Z

    2) <Z,+> with <Z,+> where theta(n)= 2n for n elements in Z

    3) <Z,+> with <Z,+> where theta(n)= n+1 for n elements in Z

    4) Let alpha: <Z4,+> -> <Z#5,X> by alpha(0)=1, alpha(1)=2, alpha(2)=4, alpha(3)=3. Prove alphais an isomorphism of groups. (verify 16 equations).

    5) Let gamma:<C*,X> -> <R+,X> by gamma(a+bi)= a^2 + b^2. Prove gamma is a homomorphism.

    6) Let theta: <Z,+> -> <{+/-1,+/-1},X> by theta(k)=i^k. Prove theta is homomorpism.

    7) Let G be a group. Prove that alpha:G->G by alpha(a)=a^2 is a homomorphism if and only if G is abelian.
  2. jcsd
  3. Nov 22, 2004 #2
    1-3) I dont know what "first binary structure with the second" means, but the first two are isomorphisms. The third isnt: try theta(a+b)=theta(a)+theta(b)...
  4. Nov 22, 2004 #3
    5) gamma(1) = 1

    gamma( (a+bi)*(c+di) ) = ... = gamma(a+bi)*gamma(c+di)
    I didn't work this out so I'm not sure if it works...

    (a+bi)' = 1/(a+bi) = a/(a^2+b^2) - bi/(a^2+b^2)
    gamma( a/(a^2+b^2) - bi/(a^2+b^2) ) = ... = 1/(a^2 + b^2)
    Again, I didn't work this out so I'm not sure if it works...

    Also I'm not sure what C* means so I'm not sure what affect this has on the problem...
    Last edited: Nov 22, 2004
  5. Nov 22, 2004 #4
    i think by binary structure it's just the structure <Z,+> where + is the binary operation, mapping it to itself by the function f(n)=-n , thanks again. I can see the answers but i am not good with proofs, for these three

    the 2nd one is not, the ans in the book reads
    No, because theta does not map Z onto Z'. theta(n) is not equal to 1 for all n in Z.
  6. Nov 22, 2004 #5
    thanks again, it's very helpful,...i'll have 4 more in an hour
    thanks alot, appreciate it.
  7. Nov 22, 2004 #6
    I'm not sure what that means, but:

    f(x) = 2x

    injective: f(x) = y & f(x') = y -> x = x' -> 2x = 2x' ->...-> x=x'
    surjective: you know that if y is an even integer then it is equal to 2x for some x, where x is an integer...

    the last part is showing that f(x+y) = f(x) + f(y), pretty straight forward...
  8. Nov 22, 2004 #7
    ok thanks again
  9. Nov 23, 2004 #8
    The second on is not an iso. since it is not surjective.
    Take, for example, 1 in Z. If (2) was surjective, then we could find an n in Z such that 2n = 1. In otherwords we need n=0.5 which is not in Z. So not surjective and so not an iso.

    (7) is quite straight forward.
    Let a, b be in G. Then alpha(a)alpha(b) = (a^2)(b^2) and alpha(ab) = (ab)^2 =abab. Clearly this only equals (a^2)(b^2) (and hence meaning alpha is a homomorphism) if you can "swap" the order of multiplication of the elements around, ie, if the group is abelain.
    Last edited: Nov 23, 2004
  10. Nov 23, 2004 #9

    surjective: for every y of B, there exists an x of A such that f(x)=y

    Your example doesn't fit this criteria as 1 is not in B. However, you know that if y is an even integer then it is equal to 2x for some x, where x is an integer...

    I may be missing something or misreading the notation, but under normal circumstances, f(x)=2x is an isomorphism for integers under addition.
  11. Nov 23, 2004 #10
    I don't know what this means, but I get the impression that it is not asking for normal isomorphisms, so what I previously posted may not apply...
  12. Nov 23, 2004 #11
    Unless I'm the one to have missed something, then you are the one that is wrong. In (2) we a function f: Z ---> Z where f(n) = 2n. For this map to be an isomorphism from Z to Z we need it to be surjective. That is fro any integer y in Z we have an x in Z such that f(x) = y. Since f maps only to the even integers (i.e. the image of f is the even integers), there is no such x that maps to the odd integers. As such the map is not surjective.

    If I'm right you are making the mistake of trying to show the map is surjective onto the image of the map, which clearly is the case, but this is not what the questions wants. Instead you need to show f maps ontot he whole of Z.
  13. Nov 23, 2004 #12
    Yes, thats what I was doing... I see my mistake now, I was trying to prove it for Z->2Z.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook