Results 1 to 3 of 3

Math Help - theorem

  1. #1
    Super Member
    Joined
    Aug 2009
    Posts
    639

    theorem

    im trying to prove that the subgroup of a cyclic group is always cyclic but im wondering what is wrong with my proof.

    what i did:

    let G=<g>={g^n l n is an integer} and let H be the subgroup with order d. then d is the least positive integer such that g^d=e.
    thus d is a divisor of n.
    since the order of g^n =G is the order of H =d then d=n and hence H=<g>

    im wondering what is wrong with my understanding because in my notes, it states that to proof this theorem, i need to let H=<g^d> ={(g^d)^q} and im wondering why i need to do it that way.

    buy letting H =<g^d>, it still means that it has an order d right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by alexandrabel90 View Post
    buy letting H =<g^d>, it still means that it has an order d right?
    No. The order of H is another name for its cardinality. By the way, writing H=<g^d> would mean that H is cyclic, which is what you are asked to prove.

    For instance (there is a simpler proof that I can't remember), you can easily see that \langle g^m,g^n\rangle = \langle g^{\gcd(m,n)}\rangle for any m,n\in\mathbb{Z}, so that, by induction, the subgroup spanned by any number of elements is always spanned by a single element, i.e. is cyclic. In particular, H is spanned by its elements, so that it is cyclic. ** this works if G is finite only...

    To prove the above identity, just write Bézout's identity am+bn=d where d=\gcd(m,n). Thus g^d=(g^m)^a(g^n)^b\in \langle g^m,g^n\rangle. And the reverse inclusion comes from d|m and d|n.

    --
    A general proof: the function \phi:\mathbb{Z}\to G defined by \phi(n)=g^n is a group morphism. Thus, \phi^{-1}(H) is a subgroup of \mathbb{Z}, hence there is a\in\mathbb{Z} such that \phi^{-1}(H)=a\mathbb{Z}, hence H=\phi(a\mathbb{Z})=\langle g^a\rangle is cyclic. This uses the knowledge of the subgroups of \mathbb{Z}.
    If H is a nontrivial subgroup of \mathbb{Z}, let h be the smallest positive element of H, then for every element a\in H, Euclidean division gives a=qh+r with 0\leq r<h, and we have r=a-qh\in H hence r=0 by minimality of h, and thus a=qh\in h\mathbb{Z}. qed.
    Last edited by Laurent; March 14th 2010 at 12:04 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2009
    Posts
    639
    then in this case, how do i find out the order of H=<g^d>?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: January 10th 2011, 08:51 AM
  2. Replies: 3
    Last Post: May 14th 2010, 10:04 PM
  3. Prove Wilson's theorem by Lagrange's theorem
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: April 10th 2010, 01:07 PM
  4. Replies: 2
    Last Post: April 3rd 2010, 04:41 PM
  5. Replies: 0
    Last Post: November 13th 2009, 05:41 AM

Search Tags


/mathhelpforum @mathhelpforum