Results 1 to 7 of 7

Math Help - Direct Product Isomorphism

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    225

    Question Direct Product Isomorphism

    Could anyone please explain clearly, why is it that we have \mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{3^2} \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5, but \mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 \ncong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5?

    So, why is it that \mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 is isomorphic to the first direct product but not to the second one?

    I mean, we can break down \mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 to the following (which is the same as the first direct product):

    \mathbb{Z}_3 \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3^2}\oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}

    And we further to the following (which is the same as the second direct product):

    \mathbb{Z}_3 \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}

    So is it not isomorphic to the latter?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by demode View Post
    Could anyone please explain clearly, why is it that we have \mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{3^2} \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5, but \mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 \ncong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5?

    So, why is it that \mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 is isomorphic to the first direct product but not to the second one?

    I mean, we can break down \mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 to the following (which is the same as the first direct product):

    \mathbb{Z}_3 \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3^2}\oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}

    And we further to the following (which is the same as the second direct product):

    \mathbb{Z}_3 \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}

    So is it not isomorphic to the latter?
    In effect, you are asking why \mathbb{Z}_{3^2} is not isomorphic to \mathbb{Z}_{3} \oplus \mathbb{Z}_{3}. One answer would be that the first of those groups has elements of order 9, but all the elements (except the identity) in the second group have order 3.

    In general, it is true that \mathbb{Z}_{mn} is isomorphic to \mathbb{Z}_{m} \oplus \mathbb{Z}_{n} if m and n are co-prime, but not otherwise.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    225
    I see, but here's another example I don't understand:

    " \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}. But \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \ncong \mathbb{Z}_{60}."

    So for the first one we know that \mathbb{Z}_{30} \cong \mathbb{Z}_3 \oplus \mathbb{Z}_{10}, but can we write

    \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{10}?

    My notes say that if m=n_1n_2...n_k, then Z_m is isomorphic to Z_{n_1} \oplus ... \oplus Z_{n_k} iff n_i and n_j are coprime when i \neq j.

    But in this case 2 and 10 are not coprime. Furthermore in \mathbb{Z}_6 \oplus \mathbb{Z}_{10}, 6 and 10 are not coprime either.

    So how did they conclude \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by demode View Post
    I see, but here's another example I don't understand:

    " \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}. But \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \ncong \mathbb{Z}_{60}."

    So for the first one we know that \mathbb{Z}_{30} \cong \mathbb{Z}_3 \oplus \mathbb{Z}_{10}, but can we write

    \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{10}?

    My notes say that if m=n_1n_2...n_k, then Z_m is isomorphic to Z_{n_1} \oplus ... \oplus Z_{n_k} iff n_i and n_j are coprime when i \neq j.

    But in this case 2 and 10 are not coprime. Furthermore in \mathbb{Z}_6 \oplus \mathbb{Z}_{10}, 6 and 10 are not coprime either.

    So how did they conclude \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}?
    It requires you to note two things, that gcd(10, 3)=1 and gcd(2, 3)=1, so you get,

    \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{10} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2009
    Posts
    225
    Quote Originally Posted by Swlabr View Post
    It requires you to note two things, that gcd(10, 3)=1 and gcd(2, 3)=1, so you get,

    \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{10} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}.
    Why is it that you don't require that gcd(2,10)=1? Those two are not co-prime! I thought they ALL should be co-prime with each other, at least that's what my notes say...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    You only need that the one's you're actually manipulating be coprime: Notice that \mathbb{Z}_2 never interacts directly with \mathbb{Z}_{10}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by demode View Post
    Quote Originally Posted by Swlabr View Post
    It requires you to note two things, that gcd(10, 3)=1 and gcd(2, 3)=1, so you get,

    \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{10} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}.
    Why is it that you don't require that gcd(2,10)=1? Those two are not co-prime! I thought they ALL should be co-prime with each other, at least that's what my notes say...
    It might make it more obvious if you write it like this:

    \mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \bigl(\mathbb{Z}_3 \oplus \mathbb{Z}_{10}\bigr) \cong \bigl(\mathbb{Z}_2 \oplus \mathbb{Z}_3\bigr) \oplus \mathbb{Z}_{10} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}.

    That makes it clear that only co-prime numbers are being used at each step. It also brings out a hidden point in the proof, namely that the direct sum operation is associative.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Direct Product
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 16th 2011, 06:57 PM
  2. Direct Product
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 24th 2009, 08:10 PM
  3. Direct product
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 7th 2009, 08:41 PM
  4. External Direct Product isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 1st 2007, 11:10 AM
  5. Isomorphism in external direct products
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 30th 2007, 03:38 AM

Search Tags


/mathhelpforum @mathhelpforum