Results 1 to 3 of 3

Math Help - Group Problem

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    62

    Group Problem

    Let G be a group. Suppose that
    (\alpha\ast\beta)^3=\alpha^3\ast\beta^3 -----(1)
    for all \alpha, \beta \in G.
    (i)Show that for all \alpha, \beta \in G, then
    (\beta\ast\alpha)^2=\alpha^2\ast\beta^2 and \alpha^3\ast\beta^2=\beta^2\ast\alpha^3.
    (ii)If G satisfies (1) and the condition that the identity element e in G, is the only element in G satisfying x^3=e, show that for all \alpha, \beta \in G,
    \alpha^2\ast\beta=\beta\ast\alpha^2
    and conclude that G is abelian.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Mar 2010
    Posts
    62
    Proving part (i) is okay. The difficult portion is part (ii).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Markeur View Post
    Let G be a group. Suppose that
    (\alpha\ast\beta)^3=\alpha^3\ast\beta^3 -----(1)
    for all \alpha, \beta \in G.
    (i)Show that for all \alpha, \beta \in G, then
    (\beta\ast\alpha)^2=\alpha^2\ast\beta^2 and \alpha^3\ast\beta^2=\beta^2\ast\alpha^3.
    (ii)If G satisfies (1) and the condition that the identity element e in G, is the only element in G satisfying x^3=e, show that for all \alpha, \beta \in G,
    \alpha^2\ast\beta=\beta\ast\alpha^2
    and conclude that G is abelian.
    Hint 1. By condition (1), the map given by f(\alpha) = \alpha^3 is a homomorphism from the group to itself.

    Hint 2. If the kernel of f consists solely of the identity, then f is injective. In particular, if (\alpha^2\ast\beta)^3 = (\beta\ast\alpha^2)^3 then \alpha^2\ast\beta=\beta\ast\alpha^2.

    Hint 3. Use the properties in (i) to show that (\alpha^2\ast\beta)^3 = (\beta\ast\alpha^2)^3.

    To conclude that G is abelian, notice that (\alpha\ast\beta)^3 = \alpha^3\ast\beta^3 = \alpha^2\ast\beta^2\ast\alpha\ast\beta, from which (\alpha\ast\beta)^2 = \alpha^2\ast\beta^2, and it easily follows that \beta\ast\alpha = \alpha\ast\beta.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Difficult group problem
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 15th 2010, 09:16 AM
  2. Replies: 1
    Last Post: November 4th 2009, 09:52 AM
  3. GROUP problem please help??
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 14th 2009, 11:39 AM
  4. group problem...
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 21st 2009, 09:27 PM
  5. Group Problem
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: August 25th 2007, 04:09 PM

Search Tags


/mathhelpforum @mathhelpforum