Results 1 to 5 of 5
Like Tree2Thanks
  • 1 Post By HallsofIvy
  • 1 Post By HallsofIvy

Thread: i need help with this isomorphic

  1. #1
    Newbie
    Joined
    Feb 2017
    From
    greece
    Posts
    3

    i need help with this isomorphic

    I have a hard time to show the following isomorphism anyone can help me with the type f : G-> Q^* .Show that the group

    1. G=Q{1/2} with binary operation x∗y=x+y+2xy is isomorphic to Q{0} with multiplication.
    2. Find a subgroup HGH≤G so that G/HZ2G/H≅Z2 (the group with two elements)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,115
    Thanks
    2803

    Re: i need help with this isomorphic

    Okay, do you know what an "isomorphism" is?

    An isomorphism, from group A to group B, is a mapping from, f, A to B that
    1) is both "one-to-one" and "onto" B. "one to one" means that every member of A is mapped to exactly one member of B and "onto" B means that for every member of B there is a member of A that is mapped to that member of B.

    2) If a*b= c in group A then f(a)*f(b)= f(c) in group B.
    (From that it follows that if e is the identity in A then f(e) is the identity in B and that is a is the inverse of b in A then f(a) is the identity of f(b) in B.

    Here A is the set of all rational numbers except -1/2 with operation a*b= a+ b+ 2ab and Q is the set of all rational numbers except 0 with operation the usual multiplication.

    I would start by looking at the identity of each group. Obviously 1 is the identity of B. is a is the identity in A then, for any b in A, we must have a*b= a+ b+ 2ab= b. Subtracting b from both sides a+ 2ab= (1+ 2b)a= 0. That will be true for any b if and only if a= 0. So the identity in A is 0. Any isomorphism would have to map 0 to 1. If a and b are inverse to each other in A then we must have a+ b+ 2ab= 0. We can write that as a+ 2ab= a(1+ 2b)= -b so that a= -b/(1+ 2b) which exists for every b except -1/2. (Equivalently b+ 2ab= b(1+ 2a)= -a so b= -a/(1+ 2a).) For example, 1 is in A and its inverse is -1/(1+ 2)= -1/3: 1- 1/3+ 2(1)(-1/3)= 0. So any isomorphism must map a to -a/(1+ 2a). That tells us what the isomorphism is.
    Thanks from tredy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2017
    From
    greece
    Posts
    3

    Re: i need help with this isomorphic

    ty for your answer i know what isomorphick is..i know that the e=0 and for every x ε A , x^-1 =-x/1+2x i found it also, because i had to show that A is group..but i cant link them with the proper type of f
    Last edited by tredy; Feb 24th 2017 at 11:50 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,115
    Thanks
    2803

    Re: i need help with this isomorphic

    Did you not understand my post? I told you exactly what the isomorphism, f, should be for the first question.
    Thanks from tredy
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2017
    From
    greece
    Posts
    3

    Re: i need help with this isomorphic

    i understand everything except the last line you wrote...sry
    "So any isomorphism must map a to -a/(1+ 2a). That tells us what the isomorphism is"
    do you mean that f(a)=-a/1+2a ?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Not isomorphic
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 22nd 2012, 11:06 PM
  2. D2n Isomorphic To Dn X Z2
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 19th 2010, 12:27 PM
  3. Isomorphic
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Oct 13th 2008, 08:55 PM
  4. Isomorphic or not?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 13th 2008, 04:41 PM
  5. Isomorphic?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 15th 2007, 04:42 PM

Search Tags


/mathhelpforum @mathhelpforum