Results 1 to 7 of 7

Math Help - Direct Products

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Direct Products

    How do I do...

    Z x Z/<(1,2)> isomorphic to Z

    and

    Z x Z x Z/<(1,1,1)> isomorphic to Z x Z

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: Direct Products

    Quote Originally Posted by JSB1917 View Post
    How do I do...

    Z x Z/<(1,2)> isomorphic to Z

    and

    Z x Z x Z/<(1,1,1)> isomorphic to Z x Z

    Thanks.
    What have you done my friend? Without any context, I'm not sure how you would even want to proceed by this? The obvious thing to do, for the first example you want a mapping \mathbb{Z}^2\to\mathbb{Z} whose kernel is a(1,2). How about something like (1,0)\mapsto 2, and (0,1)\mapsto -1. Then, (a,b)\mapsto 2a-b which....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Re: Direct Products

    Quote Originally Posted by Drexel28 View Post
    What have you done my friend? Without any context, I'm not sure how you would even want to proceed by this? The obvious thing to do, for the first example you want a mapping \mathbb{Z}^2\to\mathbb{Z} whose kernel is a(1,2). How about something like (1,0)\mapsto 2, and (0,1)\mapsto -1. Then, (a,b)\mapsto 2a-b which....
    I have nothing really because I'm not sure what I'm supposed to exactly do.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: Direct Products

    Quote Originally Posted by JSB1917 View Post
    I have nothing really because I'm not sure what I'm supposed to exactly do.
    Clearly you are going to want to construct an epimorphism \mathbb{Z}^2\twoheadrightarrow\mathbb{Z} with \ker=\langle(1,2) and then apply the FIT. Now, since \mathbb{Z}^2 has basis \{(1,0),(0,1)\} (i.e. it's free) you need only specify what you want your homomorphism to do to those guys, and then just extend by linearity. Now, how could you define a map on them, so that your kernel is what you want?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,392
    Thanks
    759

    Re: Direct Products

    Drexel28 has already given you a mapping which has the property that f(2,1) = 0. all you have to do is show it is an onto homomorphism....
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Re: Direct Products

    I have f(a,b)=2a-b for the first showed it's a homomorphism, showed the kern f = <(1,2)>, just need the onto.

    Similarly, for the other one I have f(a,b,c)=(a-c,b-c), showed it's a homomorphism, kern f = <(1,1,1)>, just need the onto.

    I just need help with onto.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,392
    Thanks
    759

    Re: Direct Products

    pick a k in Z. now we need to find a pair of integers (a,b) such that f(a,b) = 2a-b = k. there could be a LOT of pairs with this property, we only need to find ONE. so let's make it easy on ourselves, and see if we can find a pair of the form (0,b) such that f(0,b) = - b = k.

    well, it appears that b = -k, will do the trick. so

    (we could have decided to let (a,b) = (a,k) (that is, choose b = k). then, f(a,k) = 2a - k = k implies that 2a = 2k, and we could choose a = k. and sure enough f(2k,k) = 2k-k = k. there's literally infinite ways we could have gone about them, and all are equally good. the trick is to pick "some" value for either a, or b, and then find out how we have to choose "the other part of the pair". for example, suppose we choose a = m. then f(a,b) = 2m - b. if 2m - b = k, then b has to be 2m - k. and sure enough, f(m,2m-k) = 2m - (2m - k) = 2m - 2m + k = k. see how that works?).

    a similar logic applies to the second problem. we want to pick a,b,c so that a-c = k, b-c = m, for any arbitrary pair (k,m). one possible choice is (k,m,0). do you see the similarity here with solving a system of linear equations?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Group and Direct Products
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: December 17th 2009, 07:03 AM
  2. direct products
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 4th 2009, 11:57 AM
  3. direct products
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: November 9th 2008, 07:42 PM
  4. external direct products
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 5th 2008, 06:00 AM
  5. external direct products
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: March 3rd 2008, 08:26 PM

Search Tags


/mathhelpforum @mathhelpforum