Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By SlipEternal

Math Help - Generating a subgroup of D(2n)

  1. #1
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,852
    Thanks
    321
    Awards
    1

    Generating a subgroup of D(2n)

    First, just to make sure we are all working with the same notation: D(2n) is the Dihedral group order 2n, where (in a physical representation) n is the number of points of a regular plane figure, r is the usual rotation of the plane element rotating the figure by one point in the clockwise direction, and s is the inversion element. The presentation I'm using is <r, s| r^n = s^2 = 1, rs = s r^{-1} >.

    Okay. Here's my problem. I am constructing a set X(2n) from D(2n) by selecting all the x elements from D(2n) that are not powers of r and have the property rx = xr^{-1}. The problem is that my text claims that X(2n) is a group. But the identity does not belong to X(2n): re = er \neq er^{-1} unless |r| = 2, which is not general. Not only that, the multiplication in X(2n) is not closed as it contains elements of the sort s(sr) = (ss)r = er = r, which is not in X(2n) by definition.

    I could take this all as a typo, but the very next problem in the set does the same thing with another kind of sub"group" of D(2n).

    Am I going bonkers?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,810
    Thanks
    698

    Re: Generating a subgroup of D(2n)

    You are not going bonkers. The set X(2n) you are describing is not a group. The smallest subgroup of D(2n) containing X(2n) is D(2n). But, X(2n) = \{sr^k \mid k \in \mathbb{Z}, 0\le k < n\}. So, it is just the group of rotations multiplied by s. So, the set sX(2n) is a group.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 2nd 2011, 08:07 PM
  2. characterisitic subgroup implies normal subgroup
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 8th 2010, 03:13 PM
  3. Centralizer of a subgroup is a subgroup of the main group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 1st 2010, 07:16 AM
  4. subgroup of a normal subgroup...
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 23rd 2009, 08:06 AM
  5. Normal subgroup interset Sylow subgroup
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 10th 2008, 12:21 AM

Search Tags


/mathhelpforum @mathhelpforum