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Isomorphic groups

  1. Mar 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the group [itex](\mathbb Z_4,_{+4})[/itex] is isomorphic to [itex](\langle i\rangle,\cdot)[/itex]?

    2. Relevant equations
    -Group isomorphism

    3. The attempt at a solution

    Let [itex]\mathbb Z_4=\{0,1,2,3\}[/itex].
    [itex](\mathbb Z_4,_{+4})[/itex] can be represented using Cayley's table:
    [tex]
    \begin{array}{c|lcr}
    {_{+4}} & 0 & 1 & 2 & 3 \\
    \hline
    0 & 0 & 1 & 2 & 3 \\
    1 & 1 & 2 & 3 & 0 \\
    2 & 2 & 3 & 0 & 1 \\
    3 & 3 & 0 & 1 & 2 \\
    \end{array}
    [/tex]

    What is the set [itex]\langle i\rangle[/itex]?
    How to define [itex](\langle i\rangle,\cdot)[/itex]?
     
  2. jcsd
  3. Mar 16, 2016 #2

    micromass

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    The subgroup of ##(\mathbb{C},\cdot)## generated by ##i##.
     
  4. Mar 16, 2016 #3
    What should be the order of that subgroup, and how to represent it using Cayley's table?
     
  5. Mar 16, 2016 #4

    Orodruin

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    Why don't you try figuring it out? What is ##i^2##?
     
  6. Mar 16, 2016 #5

    Mark44

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    Should this be ##(\mathbb{Z_4}, +)##?
    A group is defined by a set of elements of the group, together with an operation.

     
  7. Mar 18, 2016 #6
    Could someone explain this problem (using Cayley's tables - easier)? How to form Cayley's table for the group [itex](\langle i\rangle,\cdot)[/itex]?

    One method to show the groups are isomorphic is to create Cayley's tables and compare them (that is only useful for small groups).
    I don't understand the method which requires finding the function (isomorphism) between these groups
     
  8. Mar 18, 2016 #7

    Mark44

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    Why don't you try what Orodruin suggested -- find i2, i3, and so on. This is not a hard problem.
     
  9. Mar 18, 2016 #8
    Let [itex]f:\mathbb Z_4\rightarrow \langle i\rangle=\{i^0,i^1,i^2,i^3\}=\{1,i,-1,-i\}[/itex] where [itex]f[/itex] is an isomorphism.
    From here, how to explicitly define a function [itex]f[/itex]?
     
  10. Mar 18, 2016 #9

    Orodruin

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    What do you think? There are only four possibilities of defining a homomorphism (it is fully defined by specifying how f acts on the group generator). Two of them give isomorphisms!
     
  11. Mar 18, 2016 #10
    [itex]f(x)=e^x[/itex] is one isomorphism.
     
  12. Mar 18, 2016 #11

    Orodruin

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    Not between the given groups.
     
  13. Mar 18, 2016 #12
    [itex]f(x)=e^{2\pi x i}[/itex]?
     
  14. Mar 18, 2016 #13

    micromass

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    If I define ##f(0) = 1## and if I say ##f## is a homomorphism, can you figure out ##f(1)##, ##f(2)## and ##f(3)##? That is, can you describe ##f## completely??
     
  15. Mar 18, 2016 #14
    [itex]f(0)=1,f(1)=i,f(2)=-1,f(3)=-i[/itex].

    Using Lagrange interpolation polynomial on points [itex](0,1),(1,i),(2,-1),(3,-i)[/itex] gives
    [itex]f(x)=-\frac{(x-1)(x-2)(x-3)}{6}+i\frac{x(x-2)(x-3)}{2}+\frac{x(x-1)(x-3)}{2}-i\frac{x(x-1)(x-2)}{6}[/itex].

    But [itex]f(x)[/itex] is not one to one.

    What is the actual method for describing an isomorphism, without taking a guess?
     
    Last edited: Mar 18, 2016
  16. Mar 20, 2016 #15

    LCKurtz

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    @gruba I'm guessing that the lack of further replies is caused by your last reply. It appears to me that you need more help than can be provided under the rules of this forum. My suggestion to is that you need to schedule a personal meeting with your teacher to clear up your misunderstandings on this topic.
     
  17. Mar 20, 2016 #16

    Orodruin

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    It is one-to-one on the relevant sets. You have specified f(x) for all elements of ##\mathbb Z_4##, there is absolutely no need to express it in terms of a polynomial (why would you think there was?).
     
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