Results 1 to 5 of 5

Math Help - Basic Group Question

  1. #1
    Junior Member
    Joined
    Dec 2011
    Posts
    27
    Thanks
    1

    Basic Group Question

    Let S be a non-empty set with a binary operation that satisfies the following:

    1) the operation is associative
    2) there exists a right sided identity
    3) there exists a right sided inverse

    Is S a group?

    It seems like this should be simple but I can neither produce a counter nor adequately show it is true (the couple of 'proofs' I made seemed a bit hand-wavy)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,277
    Thanks
    669

    Re: Basic Group Question

    short answer: yes.

    but why? why does "one-sided" (in this case, right) identity and inverse, imply "two-sided"?

    first of all, it's important to realize that we need both a right-identity AND a right-inverse for this to work.

    let's call our (right) identity, e'. so all we know, so far, is that for any a in S, a*e' = a.

    but...we also have a right-inverse for a, that is, some b for which a*b = e'.

    now, suppose for a moment that a*a = a (for some particular a). let b be a right inverse for a.

    then (a*a)*b = a*b = e', so
    a*(a*b) = e' (by associativity)
    a*e' = e' (since a*b = e')
    a = e'. remember this fact.

    now, since every element has a right-inverse, so does e', that is, there is some f, for which:

    e'*f = e'.

    but then:

    f = f*e' = f*(e'*f) = (f*e')*f = f*f, so f = e' (from our "remembered fact").

    so e' is its own right inverse.

    again, let's chose a and b so that a*b = e'.
    now b*a = (b*e')*a (since e' is a right-identity)
    = (b*(a*b))*a (since a*b = e')
    = ((b*a)*a)*a = (b*a)*(b*a), by associativity (twice).

    thus b*a = a*b = e'.

    now, if b is such that b*a = a*b = e', then:

    e'*a = (a*b)*a = a*(b*a) = a*e' = a, so e' is also a left-identity.

    is this two-sided identity unique?

    suppose we had another, called e".

    then e' = e'*e" = e" (since both are two-sided identities).

    so our two-sided identity is unique, and we can just call it "e".

    is the element b for which b*a = a*b = e unique?

    suppose that b' is another such element. so b'*a = a*b' = e, as well.

    then b' = b'*e = b'*(a*b) = (b'*a)*b = e*b = b.

    so inverses are unique as well.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2011
    Posts
    27
    Thanks
    1

    Re: Basic Group Question

    By the way here are the proofs I roughed out (e is the identity and z is the inverse):

    For identity:
    a \star e = a (known)
    (a \star e) \star a = a \star a
    a \star (e \star a) = a \star a (associative)
    Because a=a, this implies e \star a = a (possible hand waving)

    For inverse:
    a \star z = e (known)
    (a \star z) \star a = e \star a
    (a \star z) \star a = a \star e (by previous hand waving)
    a \star (z \star a) = a \star e (associative)
    Because a=a, this implies z \star a = e (more possible hand waving)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,277
    Thanks
    669

    Re: Basic Group Question

    Quote Originally Posted by wsldam View Post
    By the way here are the proofs I roughed out (e is the identity and z is the inverse):

    For identity:
    a \star e = a (known)
    (a \star e) \star a = a \star a
    a \star (e \star a) = a \star a (associative)
    Because a=a, this implies e \star a = a (possible hand waving)

    For inverse:
    a \star z = e (known)
    (a \star z) \star a = e \star a
    (a \star z) \star a = a \star e (by previous hand waving)
    a \star (z \star a) = a \star e (associative)
    Because a=a, this implies z \star a = e (more possible hand waving)
    with the first one, you can't "cancel a on the left" until you know every a has a left-inverse. the same applies to the second "proof".
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: Basic Group Question

    I would like to comment, that while this theorem may seem very stupid, in the sense that it isn't useful there is a canonical application of this theorem (most particularly the weakening to existence of only left inverses): a unital ring D is a division ring if and only if the only left ideals of D are (0),(1) (same goes for right).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] basic group theory Q
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: October 9th 2011, 10:31 AM
  2. Replies: 3
    Last Post: September 21st 2011, 07:21 PM
  3. Group Theory - Basic question on existence of sub-groups
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: August 15th 2009, 11:36 PM
  4. basic group theory questions
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: December 5th 2008, 08:18 AM
  5. Group Theory Question, Dihedral Group
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: March 4th 2008, 10:36 AM

Search Tags


/mathhelpforum @mathhelpforum