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Math Help - Automorphism

  1. #1
    mms
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    Automorphism

    let p be a prime. Show that <br />
Aut\left( {\mathbb{Z}_p } \right) \simeq \mathbb{Z}_{p - 1} <br /> <br />

    and show that <br />
Aut\left( {\mathbb{Z}_8 } \right) \simeq \mathbb{Z}_2 <br /> <br />
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    In general, \mbox{Aut } \mathbb{Z}_n \cong \mathbb{Z}_n^\times.

    It's not trivial at all that \mathbb{Z}_p^\times is cyclic. It's not very difficult to show once you know it, but you are probably not expected to come up with the proof. Instead you are probably expected to use this as a fact.

    The second one is false - you have \mbox{Aut } \mathbb{Z}_8 \cong \mathbb{Z}_8^\times \cong \mathbb{Z}_2 \times \mathbb{Z}_2 .
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  3. #3
    mms
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    thanks for taking the time for explaining...
    but when you write the x on top of Zn you mean its the multiplicative group?

    also, how could you prove that aut(Z8) is isomorphic to Z2 X Z2?
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Yes that is what I mean! Sorry for not being more clear.

    To see that <br />
\mbox{Aut } \mathbb{Z}_8 \cong \mathbb{Z}_8^\times \cong \mathbb{Z}_2 \times \mathbb{Z}_2<br />
, just notice that every one of the four elements of \mathbb{Z}_8^\times has order 2. There is only one group of order 4 in which every element has order 2; it is the Klein 4-group.
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  5. #5
    mms
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    i see.. thank you

    i can't prove the first problem though :/
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  6. #6
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    Notice that an homomorphism f:G \rightarrow H when G is cyclic is completely determined once you know the image of a generator. Knowing this g: \mathbb{Z}_p \rightarrow \mathbb{Z}_p is determined by g(1), but if g is to be an automorphism then g(1)=a must be a generator ie. a nonzero element, but we have p-1 such elements. Now take g,h : \mathbb{Z}_p \rightarrow \mathbb{Z}_p automorphisms then if a=h(1), b=g(1) then gh(1)=ba so we can identify an automorphism g with it's value at 1 and this defines an isomorphism between Aut( \mathbb{Z}_p ) \cong \mathbb{Z}_p^{\times }
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  7. #7
    MHF Contributor Bruno J.'s Avatar
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    Yes, that is how one goes about showing that <br />
\mbox{Aut } \mathbb{Z}_p \cong \mathbb{Z}_p^\times<br />
. However, it is showing that \mathbb{Z}_p^\times \cong \mathbb{Z}_{p-1} which is difficult (and which I doubt has to be proven by mms - it is probably to be assumed).
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  8. #8
    mms
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    thanks for your time...

    i understand how you show that you have p-1 automorphisms, but i dont get hoy you define the isomorphism -.- plz help this dumb man
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  9. #9
    MHF Contributor Bruno J.'s Avatar
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    Jose explained it, but I can explain some more.

    For every a \in \mathbb{Z}_p^\times, let \sigma_a :  \mathbb{Z}_p \rightarrow  \mathbb{Z}_p be the automorphism x \mapsto ax (prove that it is an automorphism). Then the map \pi : \mathbb{Z}_p^\times \rightarrow \mbox{Aut }\mathbb{Z}_p defined by a \mapsto \sigma_a is the isomorphism you are looking for. (Prove it!)
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  10. #10
    mms
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    i think i understand now...thank you
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