Results 1 to 4 of 4

Math Help - Abstract Algebra: Groups, Tables, Isomorphisms

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    2

    Abstract Algebra: Groups, Tables, Isomorphisms

    I need help with these problems.

    1. Let G be a group and f:G-->G defined by f(x)=x*x (group operation). Determine whether f is injective, surjective, bijective, or neither.

    2. Let A={x element of R| x does not equal 0 or 1}. Write the table for G where G is a group of permutations of A consisting of {e(x), f(x), g(x), h(x), k(x), m(x)} defined by:
    e(x)=x, f(x)=1-x, g(x)=1/x h(x)=1/(1-x) k(x)=(x-1)/x and m(x)=x/(x-1)

    3. If a, b, and c are cycles of length m, n, and r, respectively, show that abc is even or odd depending on whether m + n + r - 3 is even or odd.

    4. Show that the two groups G and H are isomorphic:

    G=< Z2 x Z2 x Z2, +> and H={e, a, b, c, ab, ac, bc, abc} where a^2=b^2=c^2=e and (ab)^2=(bc)^2=(ac)^2=e.
    [IMG]file:///C:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtml1/01/clip_image002.gif[/IMG]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    what have you tired?
    If you haven't tried anything i will start by giving you a hint for question 1. apply the function on a few examples of groups that you know.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2010
    Posts
    2
    I have done problems similar to this before, I was just confused about how I should approach it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    okay lets start on question 1 then. for each part you must prove that the function must always have a property, or give a counter example to show that it doesn't have that property.

    So lets assume to begin with that the function is injective, that means if we f(x) = f(y) if and only if x = y. trivially if x = y then f(x) = f(y), suppose f(x) = f(y) does that always imply that x = y. Can you think of a group G and two elements x , y such that x*x =y*y but x and y are distinct ?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Abstract Algebra: Completing coset Multiplication Tables
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 4th 2010, 03:41 AM
  2. Abstract Algebra-Groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 30th 2009, 01:39 PM
  3. Isomorphisms in Abstract Algebra
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 9th 2008, 07:06 PM
  4. Abstract Algebra: Groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 5th 2008, 04:13 AM
  5. Abstract Algebra: Groups
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: February 11th 2008, 10:39 AM

Search Tags


/mathhelpforum @mathhelpforum