Results 1 to 2 of 2

Thread: finite subgroup of C*

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    5

    finite subgroup of C*

    Let $\displaystyle G$ be a finite subgroup of $\displaystyle \mathbb{C}^*$. For example, $\displaystyle \{ 1,-1,i,-i \}$ is a subgroup of order 4.
    (a) Prove that $\displaystyle G$ must be a subset of $\displaystyle T=\{ z \in \mathbb{C}^* | \text{ } |z|=1 \}.$
    (b) Prove that $\displaystyle G$ is cyclic and is generated by an element of form $\displaystyle e^{\frac{2 \pi i}{n}}$.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by numbertheory12 View Post
    (a) Prove that $\displaystyle G$ must be a subset of $\displaystyle T=\{ z \in \mathbb{C}^* | \text{ } |z|=1 \}.$
    (b) Prove that $\displaystyle G$ is cyclic and is generated by an element of form $\displaystyle e^{\frac{2 \pi i}{n}}$.
    Finite subfields of the multiplicative group of a field must be cyclic subgroups.
    From here it should be clear why (a) and (b) follows.


    But you can also see (a) because if $\displaystyle re^{i\theta} \in G$ then it means that $\displaystyle r^k e^{ik\theta} \in G$ for all $\displaystyle k>0$. But if $\displaystyle r\not = 1$ then $\displaystyle r^k \to \infty$ if $\displaystyle r>1$ or $\displaystyle r^k \to 0$ if $\displaystyle r<1$. Therefore $\displaystyle r=1$ and so the complex number in $\displaystyle G$ must lie on the unit circle. To show (b) let $\displaystyle G = \{ e^{i\theta_1}, ... , e^{i\theta_m} | \theta_j \in [0,2\pi)\}$. By Lagrange's theorem we have $\displaystyle (e^{i\theta_j})^m = 1 \implies \theta_j m = 2\pi r, r\in \mathbb{Z}^+$. Therefore, $\displaystyle \theta_j = \tfrac{2\pi r i}{m}$ for some $\displaystyle r\in \mathbb{Z}^+$. Define $\displaystyle H = \left< e^{2\pi i/m} \right>$, this subgroup is cyclic and contains $\displaystyle G$, but, $\displaystyle |G| = |H|$, from here it follows that $\displaystyle G=H$. Thus, $\displaystyle G$ must be a cyclic subgroup.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finite subgroup of an infinite Group
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Dec 13th 2011, 03:26 AM
  2. Prove that no proper nontrivial subgroup of Z is finite.
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jul 18th 2011, 12:11 AM
  3. proof that subgroup has finite index
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Mar 10th 2010, 10:33 PM
  4. Finite group with identity subgroup.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Feb 18th 2010, 09:17 AM
  5. Finite subgroup of Q/Z
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Dec 8th 2008, 02:29 AM

Search Tags


/mathhelpforum @mathhelpforum