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Group Theory

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Let ##G## be a cyclic group of order ##7,## that is, ##G## consists of all ##a^i##, where ##a^7 = e.## Why is the mapping ##\phi:a^i\to a^{2i}## an automorphism of ##G## of order ##3##?

    The attempt at a solution

    I know the group ##G## is formed by the elements ##\{ e, a, a^2, a^3, a^4, a^5, a^6 \}##. Now under the mapping of ##\phi##, the order of elements is changed to: ##e, a^2, a^4, a^6, a, a^3, a^5 ##. Thus, ##\phi## is both one-one and onto. What I don't understand is why ##G## is of order ##3## by the map ##\phi##?
     
  2. jcsd
  3. Oct 30, 2013 #2

    Dick

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    Play around with it. Order 3 means ##\phi^3## is the identity. Is it? Is ##\phi^3(a)=\phi(\phi(\phi(a)))=a##?
     
  4. Oct 30, 2013 #3

    pasmith

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    The automorphisms of [itex]G[/itex] are themselves a group, [itex]\mathrm{Aut}(G)[/itex], under the operation of composition of functions. An automorphism [itex]\phi \in \mathrm{Aut}(G)[/itex] being of order 3 means that [itex]\phi^3[/itex] is the identity function on [itex]G[/itex], [itex]\mathrm{Id}_G \in \mathrm{Aut}(G)[/itex].

    So what's [itex]\phi^3(a^i)[/itex]?
     
  5. Oct 30, 2013 #4
    But how did they get ##\phi^3##? How did they know ##\phi## has order 3? For example, does it have order ##8## or ##9##?
     
  6. Oct 30, 2013 #5

    Dick

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    The order of the automorphism, n, is the smallest number such that ##\phi^n## is the identity automorphism. They didn't just know that it is 3, they figured it out. And they are asking you to do the same. You figured out the mapping for ##\phi## pretty easily, now do ##\phi^3##.
     
  7. Oct 30, 2013 #6
    Ohh, thank you! I understand now. I thought they came up with it out of the blue.

    But is there an easy way to verify the order of the automorphism?
     
  8. Oct 30, 2013 #7

    Dick

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    Sure, there's an easier way than working out the whole group mapping. I think you should work that out by playing with it. Once you've done that and given it some thought, do you see that if ##\phi^n(a)=a## where a is a generator then ##\phi^n(x)=x## for all elements of the group?
     
  9. Oct 31, 2013 #8
    Dick - No I cant see that. Why is that?
     
  10. Oct 31, 2013 #9

    Dick

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    Think about it. Every element of the group is a power of a and ##\phi^n(a)=a## and ##\phi## is a homomorphism. Put those things together.
     
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