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Is this a group homomorphism?

  1. Mar 19, 2016 #1
    1. The problem statement, all variables and given/known data
    [itex] ℤ_n → D_n [/itex] sending z modn → [itex]g^z[/itex] where g is rotation by an nth of a turn.

    2. Relevant equations
    Group homomorphism imply [itex] θ(g_1*g_2)=θ(g_1)*θ(g_2) [/itex]

    3. The attempt at a solution
    Before anything, I'd like to know if Group homomorphism imply [itex] θ(g_1+g_2)=θ(g_1)[/itex]x[itex]θ(g_2) [/itex] I've seen [itex] θ(g_1[/itex]x[itex]g_2)=θ(g_1)×θ(g_2) [/itex] and [itex] θ(g_1+g_2)=θ(g_1)+θ(g_2) [/itex] but not [itex] θ(g_1+g_2)=θ(g_1)[/itex]x[itex]θ(g_2) [/itex]. Can the operator * be different on the left and right side?

    Attempt:
    [itex] θ(z_1+z_2)[/itex]=g[itex]z_1+z_2 modn [/itex]
    I'm not sure how to go from here. I'm sure I need to use the fact that [itex]g^n=e[/itex] but I don't know how to proceed.
    Any help will be appreciated!
     
    Last edited: Mar 19, 2016
  2. jcsd
  3. Mar 19, 2016 #2

    LCKurtz

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    What is ##D_n##? And what is a "turn" and an nth of a turn?

    Yes, they are usually different groups. Look up the definition of group homomorphism.
     
  4. Mar 19, 2016 #3

    Office_Shredder

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    If you are having trouble with the intuition, you should try picking a value of n, say n=3 to start, and just explicitly write out every possible pair of elements in [itex]\mathbb{Z}_n[/itex] and how they add up and what they would get mapped to in [itex]D_n[/itex].
     
  5. Mar 22, 2016 #4

    vela

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    Yes, the operations generally are different, even if you happen to use the same symbol for them. When you wrote ##\theta(g_1+g_2) = \theta(g_1)+\theta(g_2)##, the + on the lefthand side isn't the same as the + on the righthand side because ##g## and ##\theta(g)## are from two different sets.
     
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