Results 1 to 6 of 6

Math Help - Group Theory - order of element

  1. #1
    Banned
    Joined
    May 2009
    Posts
    471

    Group Theory - order of element

    Determine the order of a^m where a is an element of order n in a group G

    I don't really know what they book is looking for here. It seems to me this will break down into a bunch of cases depending on what m and n are, but maybe I'm wrong.

    a^n=1 (the identity)

    and yeah that's pretty much all I got so far
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by artvandalay11 View Post
    Determine the order of a^m where a is an element of order n in a group G

    I don't really know what they book is looking for here. It seems to me this will break down into a bunch of cases depending on what m and n are, but maybe I'm wrong.

    a^n=1 (the identity)

    and yeah that's pretty much all I got so far
    We kind of need this lemma

    Lemma: Let g\in G with |g|=n. If g^{m}=e then n|m.

    Proof: By the division algorithim m=qn+r\quad q\in\mathbb{Z},0\le r<n. So then g^m=g^{qn+r}=\left(g^n\right)^qg^r=g^r=e and since 0\le r<n and n is the least positive integer such that g^k=e we must have that r=0. Therefore m=qn\quad q\in\mathbb{Z} and the conclusion follows. \blacksquare

    So basically we need the least number k such that n|mk, i.e. we need \left|a^m\right|=\text{lcm}(m,n)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Apr 2009
    Posts
    677
    Quote Originally Posted by Drexel28 View Post
    We kind of need this lemma

    Lemma: Let g\in G with |g|=n. If g^{m}=e then n|m.

    Proof: By the division algorithim m=qn+r\quad q\in\mathbb{Z},0\le r<n. So then g^m=g^{qn+r}=\left(g^n\right)^qg^r=g^r=e and since 0\le r<n and n is the least positive integer such that g^k=e we must have that r=0. Therefore m=qn\quad q\in\mathbb{Z} and the conclusion follows. \blacksquare

    So basically we need the least number k such that n|mk, i.e. we need \left|a^m\right|=\text{lcm}(m,n)
    Hi - Is your final result correct?
    \left|a^m\right|=\text{lcm}(m,n)
    I trust \left|a^m\right|=n/(m,n) where (m,n) is gcd(m,n)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie lepton's Avatar
    Joined
    Sep 2009
    Posts
    10
    \text{lcm}(m,n)=(mn)/\gcd(m,n)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Apr 2009
    Posts
    677
    Quote Originally Posted by lepton View Post
    \text{lcm}(m,n)=(mn)/\gcd(m,n)
    Right. But what I'm saying is that order is n/\gcd(m,n)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by aman_cc View Post
    Right. But what I'm saying is that order is n/\gcd(m,n)
    Thanks, typo. It was supposed to be \frac{\text{lcm}(m,n)}{m}=\frac{n}{(m,n)}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 3rd 2010, 10:42 AM
  2. order of an element of a group
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 30th 2010, 12:55 PM
  3. Finding the order of a group element
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: November 13th 2009, 07:48 AM
  4. group element order
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 1st 2009, 07:23 AM
  5. order of an element in mulitpicative group
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 15th 2009, 12:25 PM

Search Tags


/mathhelpforum @mathhelpforum