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Thread: Automorphism

  1. #1
    mms
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    Automorphism

    let p be a prime. Show that $\displaystyle
    Aut\left( {\mathbb{Z}_p } \right) \simeq \mathbb{Z}_{p - 1}

    $

    and show that $\displaystyle
    Aut\left( {\mathbb{Z}_8 } \right) \simeq \mathbb{Z}_2

    $
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    In general, $\displaystyle \mbox{Aut } \mathbb{Z}_n \cong \mathbb{Z}_n^\times$.

    It's not trivial at all that $\displaystyle \mathbb{Z}_p^\times$ 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 $\displaystyle \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 $\displaystyle
    \mbox{Aut } \mathbb{Z}_8 \cong \mathbb{Z}_8^\times \cong \mathbb{Z}_2 \times \mathbb{Z}_2
    $, just notice that every one of the four elements of $\displaystyle \mathbb{Z}_8^\times$ 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 $\displaystyle f:G \rightarrow H$ when $\displaystyle G$ is cyclic is completely determined once you know the image of a generator. Knowing this $\displaystyle g: \mathbb{Z}_p \rightarrow \mathbb{Z}_p$ is determined by $\displaystyle g(1)$, but if $\displaystyle g$ is to be an automorphism then $\displaystyle g(1)=a$ must be a generator ie. a nonzero element, but we have $\displaystyle p-1$ such elements. Now take $\displaystyle g,h : \mathbb{Z}_p \rightarrow \mathbb{Z}_p$ automorphisms then if $\displaystyle a=h(1), b=g(1)$ then $\displaystyle gh(1)=ba$ so we can identify an automorphism $\displaystyle g$ with it's value at $\displaystyle 1$ and this defines an isomorphism between $\displaystyle Aut( \mathbb{Z}_p ) \cong \mathbb{Z}_p^{\times }$
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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 $\displaystyle
    \mbox{Aut } \mathbb{Z}_p \cong \mathbb{Z}_p^\times
    $. However, it is showing that $\displaystyle \mathbb{Z}_p^\times \cong \mathbb{Z}_{p-1}$ 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 $\displaystyle a \in \mathbb{Z}_p^\times$, let $\displaystyle \sigma_a : \mathbb{Z}_p \rightarrow \mathbb{Z}_p$ be the automorphism $\displaystyle x \mapsto ax$ (prove that it is an automorphism). Then the map $\displaystyle \pi : \mathbb{Z}_p^\times \rightarrow \mbox{Aut }\mathbb{Z}_p$ defined by $\displaystyle a \mapsto \sigma_a$ 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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