Results 1 to 4 of 4
Like Tree2Thanks
  • 1 Post By HallsofIvy
  • 1 Post By johng

Thread: Surjective homomorphism on roots of unity

  1. #1
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,158
    Thanks
    734
    Awards
    1

    Surjective homomorphism on roots of unity

    Let $\displaystyle G = \{ z \in \mathbb{C} |z^n = 1 \text{ for some n } \in \mathbb{Z} ^+ \}$. Prove that for any fixed integer k > 1 the map from G to itself defined by $\displaystyle z \mapsto z^k$ is a surjective homomorphism but is not injective.
    So let
    $\displaystyle \phi _k : G \to G : g \mapsto g^k$.

    Proving the homomorphism is trivial. It's the surjection part that's getting to me. (I haven't tackled the injection yet. Please leave that out of the replies for now.) What I am looking at is the following...I want to show that there is at least one pair of x, y in G such that $\displaystyle x^k = y^k$ is true when x is not equal to y. My other question about this is do we need to find an x, y for each k? For example, $\displaystyle (-1)^k = (1)^k$ is only true for even k.

    I'm just not wrapping this one around in my head today.

    Thanks!
    -Dan
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,791
    Thanks
    3035

    Re: Surjective homomorphism on roots of unity

    Quote Originally Posted by topsquark View Post
    So let
    $\displaystyle \phi _k : G \to G : g \mapsto g^k$.

    Proving the homomorphism is trivial. It's the surjection part that's getting to me. (I haven't tackled the injection yet. Please leave that out of the replies for now.) What I am looking at is the following...I want to show that there is at least one pair of x, y in G such that $\displaystyle x^k = y^k$ is true when x is not equal to y.
    No, you don't! That would be showing this is a NOT an injection. It would not be showing this is a surjection. To show this is a surjection you need to show that if z is a complex number such that $\displaystyle z^n= 1$ for some n then there exist a complex number w such that $\displaystyle w^n= 1$ such that w^k= z for this specific z.

    My other question about this is do we need to find an x, y for each k? For example, $\displaystyle (-1)^k = (1)^k$ is only true for even k.
    For z= -1, take w= -1.

    I'm just not wrapping this one around in my head today.

    Thanks!
    -Dan
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    1,143
    Thanks
    476

    Re: Surjective homomorphism on roots of unity

    1. Surjective: if $\displaystyle z^n=1$ then $\displaystyle z=e^{2mi\pi/n}$ for some $\displaystyle m$ with $\displaystyle 0\leq m<n$. Can't you find $\displaystyle w$ with $\displaystyle w^{kn}=1$ and $\displaystyle w^k=z$?

    2. Injective: Think about the kernel of the homomorphism.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,158
    Thanks
    734
    Awards
    1

    Re: Surjective homomorphism on roots of unity

    Sorry I didn't get back sooner. I just finished the problem, both surjective and injective. I appreciate the help.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Well-defined, surjective ring homomorphism.
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Nov 16th 2011, 09:02 PM
  2. Surjective Ring Homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Mar 2nd 2010, 11:44 PM
  3. proving surjective homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Nov 29th 2009, 07:04 PM
  4. surjective homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 6th 2009, 12:27 PM
  5. Isomorphism of a surjective group homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Dec 7th 2008, 08:51 PM

Search Tags


/mathhelpforum @mathhelpforum