Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Rigid Motions

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    225

    Rigid Motions

    Let SM denote the subset of orientation-preserving motions of the
    plane. Prove that SM is a normal subgroup of M, and determine its
    index in M.


    I know if I can show this has an index 2, that means it is normal, but I don't know how to prove this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,371
    Thanks
    740

    Re: Rigid Motions

    you should tell us what "M" is, too. we don't have your text available for ready reference.

    my assumption is that M is the set of all rigid motions (isometries) of the euclidean plane. how many orientations can the plane have?

    see if you can find a surjection from M to {-1,1} (under the operation of multiplication for the latter set).

    is this surjection a group homomorphism? if so, what is its kernel?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2008
    Posts
    225

    Re: Rigid Motions

    Oops. You're right, M is the group of all rigid motions of the plane.
    I'm confused about this but I think we can use the determinant map for the rotation matrix, and this will go to 1 if the orientation of the plane is not flipped.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,371
    Thanks
    740

    Re: Rigid Motions

    ok, you have a map in mind, det:M --->{1,-1}.

    is this surjective?

    what does it mean to say det is a homomorphism?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2008
    Posts
    225

    Re: Rigid Motions

    This map is surjective. For det to be a homomorphism, det (AB)=det(A)det(B)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,371
    Thanks
    740

    Re: Rigid Motions

    and, is det(AB) = det(A)det(B) a true statement (if it is, you're almost done!)?

    what is ker(det)? (remember, ker(det) = {T in M: det(T) = 1}, since 1 is the identity of {-1,1}).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2008
    Posts
    225

    Re: Rigid Motions

    Because detA=detB=1, the statement is true.
    So I think the kernel would be all the matrices with det =1
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,371
    Thanks
    740

    Re: Rigid Motions

    Quote Originally Posted by veronicak5678 View Post
    Because detA=detB=1, the statement is true.
    So I think the kernel would be all the matrices with det =1
    no, not ALL transformations have a determinant of 1. for example, if:

    T(x,y,z) = (2x,2y,z), det(T) = 4.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Aug 2008
    Posts
    225

    Re: Rigid Motions

    I think if the transformation is a reflection, the determinant is -1. So if it is an orthogonal rotation matrix, the determinant is 1.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,371
    Thanks
    740

    Re: Rigid Motions

    that is correct, however, you should understand that det(AB) = det(A)det(B) is true for any 2 nxn matrices.

    so det is a (group) homomorphism from GL(n,R) to R, no matter what n is (you are dealing with the special case of n = 2, and det(A) = 1, which is one way of defining M).

    so if ker(M) = SM, how does that show SM is normal?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Aug 2008
    Posts
    225

    Re: Rigid Motions

    We have proved in my class that the kernel of a group is always normal to it.
    How do I find the index of the group?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,371
    Thanks
    740

    Re: Rigid Motions

    what is the order of M/SM (which, by the first isomorphism theorem (big important theorem, don't leave home without it!), is isomorphic to det(M) = ____)?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    Aug 2008
    Posts
    225

    Re: Rigid Motions

    Does M/SM mean the left cosets of SM in M? I tried looking up the first iso theorem, but I don't really understand it...
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,371
    Thanks
    740

    Re: Rigid Motions

    it does. for a finite group, with a finite normal subgroup: [G:H] = |G/H| = |G|/|H|.

    the trouble is, here, we have an infinite group, and an infinite subgroup.

    what the first isomorphism theorem says is this: for a group homomorphsim f: G-->G'

    1) f(G) is a subgroup of G'
    2) ker(f) is a normal subgroup of G
    3) the set of cosets of ker(f), is also a group, G/ker(f), and is isomorphic to f(G).

    the part that is useful to you is (3). it tells you that the number of cosets of SM in M is the number of possible determinant values elements of M can have,

    and this number IS the index of SM in M.

    (by the way, since SM is normal, the left cosets and the right cosets of SM in M are the same).
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    Aug 2008
    Posts
    225

    Re: Rigid Motions

    Can you explain how part 3 is telling me that?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 8th 2010, 06:16 PM
  2. two rigid motions agree on three non-colinear points
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 2nd 2010, 08:50 AM
  3. Rigid motions
    Posted in the Geometry Forum
    Replies: 4
    Last Post: February 26th 2010, 06:52 AM
  4. group of motions
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 26th 2009, 01:05 PM
  5. [SOLVED] Concerning Rigid Motions
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: February 27th 2008, 08:05 PM

Search Tags


/mathhelpforum @mathhelpforum