Results 1 to 2 of 2

Math Help - Finite and Infinite order

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    3

    Finite and Infinite order

    an element a of a group is the smallest positive integer m such that a^m = e (where e denotes the identity element of the group, and a^m denotes the product of m copies of a). If no such m exists, we say that a has infinite order. All elements of finite groups have finite order.

    can some1 give me an easy examples using the above facts please? thanks
    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 student17989 View Post
    an element a of a group is the smallest positive integer m such that a^m = e (where e denotes the identity element of the group, and a^m denotes the product of m copies of a). If no such m exists, we say that a has infinite order. All elements of finite groups have finite order.

    can some1 give me an easy examples using the above facts please? thanks
    Here are two groups. One that has infinite order and one that doesn't. While the one that doesn't may seem overcomplicated there is a reassuringly easy way to prove that it's infinite.

    Finite: S_n=\left\{\sigma:\{1,\cdots,n\}\to\{1,\cdots n\}\mid \sigma\text{ is bijective}\right\} with the binary operation of function composition. To prove that each element has finite order we use the following theorem:

    Theorem: Let G be a finite group, then for each g\in G we have that |g|<\infty.

    Proof: Suppose not. Then, \left\{e,g,g^2,g^3,\cdots\right\} must all be distinct, for to suppose that g^{k}=g^{\ell}\implies g^{|k-\ell|}=e contradicting the assumption that |g|=\infty. Thus, since G is closed under multiplication it follows that \{e,g,g^2,g^3,\cdots\}\subseteq G and so |G|=\infty. Contradiction.


    Infinite: Let \Omega be an infinite set and S_\Omega be defined as above. Then, |S_\Omega|=\infty. To see this suppose not. Then, |S_\Omega|=k\in\mathbb{N}. Select elements \omega_1,\cdots,\omega_k\in S_\Omega and consider K=\left\{\sigma\in S_n:\sigma(m)=m,\text{ }\forall m\notin\{\omega_1,\cdots,\omega_k\}\right\}. Clearly K\leqslant S_\Omega and \theta:S_k\hookrightarrow S_\Omega given by the obvious mapping is an embedding and \theta(S_k)=K. Thus, |K|=k!. But, by our assumption |S_\Omega|=k<\infty and so Lagrange's theorem implies that k!\mid k. Contradiction.



    Does that answer your question? There are infinite groups whose elements all have finite order though. For example \bigoplus_{n=1}^\infty Z_n where each Z_n is a copy of \mathbb{Z}_2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finite and infinite sets
    Posted in the Discrete Math Forum
    Replies: 11
    Last Post: August 6th 2011, 03:53 PM
  2. finite or infinite?
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: September 3rd 2010, 09:39 PM
  3. finite or infinite order of matrices
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: October 17th 2009, 02:03 PM
  4. infinite/finite limits
    Posted in the Calculus Forum
    Replies: 9
    Last Post: March 10th 2009, 10:36 AM
  5. Finite>Infinite
    Posted in the Advanced Math Topics Forum
    Replies: 9
    Last Post: February 20th 2006, 11:37 AM

Search Tags


/mathhelpforum @mathhelpforum