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Showing that ##Aut ~ Z_p \simeq Z_{p-1}##

  1. Mar 29, 2017 #1
    1. The problem statement, all variables and given/known data

    I am trying to show that ##Aut~ Z_p## is isomorphic to ##Z_{p-1}##, where ##Z_p## denotes the congruence class of integers ##\mod p## and ##p## a prime.

    2. Relevant equations


    3. The attempt at a solution

    I have shown ##Aut~ Z_p## consists of ##p-1## elements, using the fact that a homomorphism is uniquely determined by how it maps ##[1]_p##, the generator of ##Z_p##; the only element it cannot be mapped to is ##[0]_p##, otherwise ##[1]_p \rightarrow [k]_p \neq [0]_p## extends to an automorphism.


    Now I am trying to show that ##Aut~ Z_p## is a cyclic, and show that mapping the generator of ##Aut~ Z_p## to ##Z_{p-1}## defines an isomorphism. However, I am having trouble showing this. I could use some hints.

    I just want to add that this is an early exercise in Hungerford, so, for example, the order of an element hasn't even been defined yet.
     
  2. jcsd
  3. Mar 30, 2017 #2

    andrewkirk

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    In this exercise is ##Z_p## considered as a group under the operation of addition, or as a field?
     
  4. Mar 30, 2017 #3

    andrewkirk

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    Having thought about this some more, I concluded that we need to interpret ##Z_p## as a group under addition rather than a field.

    With that interpretation, your first sentence in section 3 above is correct.
    Given that result, I don't think you need to ask about cyclicality do you (the subject of your next para)? In fact you have already done most of the proof. The remaining part is just to state the isomorphism between ##Aut\ Z_p## and ##Z_{p-1}##.

    First make a guess as to what map ##\theta## from ##Z_{p-1}## to ##Z_p\times Z_p## (the set of maps from ##Z_p## to itself) might be the required isomorphism, then check that
    1. its image is always in ##Aut\ Z_p##
    2. ##\theta## is a group homomorphism
    3. ##\theta## is one-to-one
    4. ##\theta## is onto

    The hardest part is guessing a form for ##\theta##. We need to specify a function ##f:Z_{p-1}\times Z_p\to Z_p## such that
    $$\theta([k]_{p-1})([r]_p)=[f(k,r)]_p$$
    The properties we require for ##f## are
    1. ##f(0,r)=r## (##\theta## maps ##[0]_{p-1}## to the identity automorphism on ##Z_p##)
    2. ##[f(k,r)]_p\neq [f(k,s)]_p## if ##k\neq 0##, ##r\neq s## and ##0\leq r,s<p## (##\theta(k)## must be injective)
    3. for any ##k##, ##[f(k,r+s)]_p=[f(k,r)]_p+[f(k,s)]_p## (##\theta(k)## must be a homomorphism)
    You can expect that ##f## will be fairly simple, although it is not immediately obvious what it should be.
     
  5. Mar 31, 2017 #4

    Dick

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    Bashyboy has already spelled out what the automorphisms are by considering the possible images of ##1##. That's the straightforward part. The hard part is showing the automorphism group is cyclic. This is a standard result, it's showing that there is a primitive root. But I don't know of any proofs that are simple enough that I would assign them as an exercise before I discussed what 'order of an element' means.
     
  6. Mar 31, 2017 #5
  7. Mar 31, 2017 #6
    Yes, you are right in construing it as a additive group. I am sorry I didn't respond sooner. Also, I forgot to mention that I have already proved that ##Z_p## is multiplicative group whenever ##p## is prime.

    So, do you happen to know what ##\theta## and #f## look like? I'm still going to need to think about it.
     
  8. Mar 31, 2017 #7

    Dick

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    You already know what the automorphisms look like. You said it in the first post. If you know where ##1## is mapped you know where everything is mapped. It's a generator.
     
  9. Mar 31, 2017 #8

    andrewkirk

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    Whoa, that partial quote facility really doesn't play well with latex does it? :smile:

    Try ##f(k,s)=2^ks##
     
  10. Apr 1, 2017 #9

    Dick

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    That gives you a homomorphism into ##Aut Z_p##. Are you sure it's an isomorphism? Look at ##Z_7##.
     
  11. Apr 1, 2017 #10

    andrewkirk

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    Great Caesar's Ghost, you're right - it doesn't work!

    And I felt so sure of that mapping.
     
  12. Apr 2, 2017 #11

    Dick

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    Right. You can't just use 2 for all values of p. You need a primitive root. There's a number of proofs that one exists. None of them give you a formula for finding one.
     
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