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Is this number isomorphic?

  1. Nov 19, 2006 #1
    Show that Z3 X Z5 is isomorphic to Z15 (where Zn is like the intergers mod n)

    i m not sure if i m proving it right.. if i first right out Z3 x Z5 ={(1x1), (1X2),(1x3),(1x4), (1X5), (2X1),(2x2),(2x3), (2x4), (2X5), (3x1), (3x3),(3x3),(3x4), (3X5)}
    Z15={1,2,3....15}

    both of which have 15 elements... there for they are in the same form?? therefore isomorphic??

    i would appreciate any help
    thanks
     
  2. jcsd
  3. Nov 21, 2006 #2
    To show that Z3 X Z5 is isomorphic to Z15 you have to come up with a one-to-one onto function between the two. I think something like this might work if (a,b) is in Z3 X Z5 then the mapping p^a*q^b where p and q are unique primes is one-to-one by the unique factorization thm. However, it's not onto but we do know it's well ordered so we have elements a_1<a_2<....<a_15 then you can take a_n to n and this would make it onto.
     
  4. Nov 21, 2006 #3
    To show they're isomorphic
    You also need to show that if F is the mapping between them, * is the operation on Z3xZ5, and # is the operation on Z15

    that
    F(a*b) = F(a)#F(b)
    where a and b are elements of Z3xZ5 if the function maps Z3xZ5 to Z15.
     
  5. Nov 21, 2006 #4
    Sorry, I forgot to say that the mapping also has to be a homomorphism.
     
  6. Nov 21, 2006 #5

    matt grime

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    The number of elements does not determine the group. This is clear, since abelian groups cannot be isomorphic to non-abelian groups.

    It is easier to map elements of Z15 to the other group, as it happens, from the description you've given.

    Of course, if you can show that some element of Z3xZ5 has order 15 you're also done, but do you understand why?
     
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