Results 1 to 4 of 4

Math Help - Isomorphism Question

  1. #1
    xyz
    xyz is offline
    Banned
    Joined
    Oct 2009
    Posts
    24

    Isomorphism Question




    G has 16 elements and H has 4 elements....so G/H = 4?

    Now I have what seems like a corollary in my book, but not stated as...it says "a group of order 4 is isomorphic to Z4 or Z2+Z2 (direct prodcut)"...so would the first part be yes? Maybe I am over complicating things...
    and then on the second part
    k = <(1,2)>
    How do you find out how many elements a generating function like K has?
    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
    Quote Originally Posted by xyz View Post



    G has 16 elements and H has 4 elements....so G/H = 4?

    Now I have what seems like a corollary in my book, but not stated as...it says "a group of order 4 is isomorphic to Z4 or Z2+Z2 (direct prodcut)"...so would the first part be yes? Maybe I am over complicating things...
    and then on the second part
    k = <(1,2)>
    How do you find out how many elements a generating function like K has?
    I'm sorry. But do you know what G/H means?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    xyz
    xyz is offline
    Banned
    Joined
    Oct 2009
    Posts
    24
    Let G be a group and let H be a normal subgroup of G.
    The set G/H = {aH | a ∈ G} is a group under the operation (aH)(bH) = abH. I mentioned that the order of G/H is 4.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    what makes \mathbb{Z}_2 \oplus \mathbb{Z}_2 and \mathbb{Z}_4 different is that \forall x \in \mathbb{Z}_2 \oplus \mathbb{Z}_2: \ 2x=0, which is not true in \mathbb{Z}_4. now we have K=\{(0,0),(1,2),(2,0),(3,2) \}. so if you let x=(3,1) + K \in G/K, then

    2x=(2,2) + K \neq 0, because (2,2) \notin K. thus G/K \cong \mathbb{Z}_4. such situation does not happen in G/H, which means G/H \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Isomorphism Question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 6th 2011, 02:26 PM
  2. Isomorphism question
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: May 24th 2010, 03:40 AM
  3. Replies: 4
    Last Post: February 14th 2010, 04:05 AM
  4. isomorphism question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 22nd 2009, 03:19 PM
  5. Question about Isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 31st 2009, 12:42 PM

Search Tags


/mathhelpforum @mathhelpforum