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Math Help - Automorphism

  1. #1
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    Automorphism

    Let G = Aut(\mathbb{Z}_{315}) and let n be the order of the largest cyclic subgroup of G. What is n? And what would be the number of distinct cyclic subgroups of order n in G?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    You should start by thinking about what this automorphism group looks like. Essentially, this comes down to `where can the generator, 1, be sent to by an automorphism?'. (e.g. \phi: 1 \rightarrow 2 is an automorphism. Can you think why this is?)

    So, where can the generator be sent to?
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    You should start by thinking about what this automorphism group looks like. Essentially, this comes down to `where can the generator, 1, be sent to by an automorphism?'. (e.g. \phi: 1 \rightarrow 2 is an automorphism. Can you think why this is?)

    So, where can the generator be sent to?
    The generator is sent to U(315), as Aut(\mathbb{Z}_{315}) is isomorphic to U(315)?

    I think the order of the largest cyclic subgroup of G is

    For an element \alpha of Aut(\mathbb{Z}_{315}) we know \alpha(1), we'll know \alpha(k) fora any k, because

    \alpha(k) = \alpha \underbrace{(1+1+...+1)}_\textrm{k terms} = \underbrace{\alpha(1)+...+\alpha(1)}_\textrm{k terms}=k\alpha(1)

    I think the property that isomorphisms preserve orders would tell us that |\alpha(1)|=315, then I think members of U(315) would be the candidates for \alpha(1).

    The largest member of U(315) is 314. But I don't know if this helps...
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  4. #4
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    Oh, I meant since \phi is an automorphism we have \phi(n)=dn for some d \in U(315) and any n \in Z_{315}.

    But how does this help me to prove the question?
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Yes, you need to find out where you can send the generator too. Your choices are precisely those numbers which generate \mathbb{Z}_{315}. So, which elements of your group generate it? How many of them are there?

    This will give you the order of your Automorphism group (why?).
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  6. #6
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    Thanks for the reply. Since |a^k| = \frac{315}{gcd(315,k)}, G = \left\langle a \right\rangle \iff k \in U(315), where

    U(315) = \{1,2,4,6,...,314 \}

    Therefore the generators of \mathbb{Z}_{315} are the elements

    \{ a,a^2,a^4,a^6...,a^{314} \}

    Is this right? If so how do I know how many elements there are in U(315)? (there are too many to count)
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  7. #7
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    Okay I know that if G is a cyclic group of order n, then Aut (G) \cong U(n).

    In our case if we have \mathbb{Z}_{315}= \left\langle a \right\rangle

    Aut(\mathbb{Z}_{315}) = \{ a_i : \mathbb{Z}_{315} \to \mathbb{Z}_{315}, a \mapsto ia, i \in U(315) \} \cong U(315)

    315 = 5.7.3^2

    U(315) \cong \mathbb{Z}_5 \oplus \mathbb{Z}_7 \oplus \mathbb{Z}_9

    Does this mean the order of the largest cyclic subgroup of G is 63? Because 5 \times7 < 5 \times 9 < 7 \times 9 = 63.

    Am I right?
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by demode View Post
    Okay I know that if G is a cyclic group of order n, then Aut (G) \cong U(n).

    In our case if we have \mathbb{Z}_{315}= \left\langle a \right\rangle

    Aut(\mathbb{Z}_{315}) = \{ a_i : \mathbb{Z}_{315} \to \mathbb{Z}_{315}, a \mapsto ia, i \in U(315) \} \cong U(315)

    315 = 5.7.3^2

    U(315) \cong \mathbb{Z}_5 \oplus \mathbb{Z}_7 \oplus \mathbb{Z}_9

    Does this mean the order of the largest cyclic subgroup of G is 63? Because 5 \times7 < 5 \times 9 < 7 \times 9 = 63.

    Am I right?
    Largely yes, this is all correct. Apart from your answer. There is a cyclic group of order 5.7.3 there...
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