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Thread: automorphism proof

  1. #1
    Senior Member Danneedshelp's Avatar
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    automorphism proof

    Theorem: for any positive integer n, $\displaystyle Aut(Z_{n})$ is isomorphic to $\displaystyle U(n)$.

    Q: Prove that every element $\displaystyle g\in\\Aut(Z_{n})$ is determined by $\displaystyle g(1)$; meaning, $\displaystyle g(1)$ results in a well-defined automorphism of $\displaystyle Z_{n}$.

    A: let $\displaystyle g\in\\Aut(Z_{n})$ be arbitrary. Then $\displaystyle f:Z_{n}\rightarrow\Z_{n}$ is an isomorphism. So, to show if something is well-defined I need to show if a=c, b=d, then a*b=c*d. I am not sure what to my a,b, and c's are. I know $\displaystyle f(k)=kf(1)$. so, do I let $\displaystyle f(m)=mf(1)$ and $\displaystyle f(p)=pf(1)$. Now, I want to show $\displaystyle f(p)*f(q)=mf(1)*pf(1)$?

    Thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    Theorem: for any positive integer n, $\displaystyle Aut(Z_{n})$ is isomorphic to $\displaystyle U(n)$.

    Q: Prove that every element $\displaystyle g\in\\Aut(Z_{n})$ is determined by $\displaystyle g(1)$; meaning, $\displaystyle g(1)$ results in a well-defined automorphism of $\displaystyle Z_{n}$.

    A: let $\displaystyle g\in\\Aut(Z_{n})$ be arbitrary. Then $\displaystyle f:Z_{n}\rightarrow\Z_{n}$ is an isomorphism. So, to show if something is well-defined I need to show if a=c, b=d, then a*b=c*d. I am not sure what to my a,b, and c's are. I know $\displaystyle f(k)=kf(1)$. so, do I let $\displaystyle f(m)=mf(1)$ and $\displaystyle f(p)=pf(1)$. Now, I want to show $\displaystyle f(p)*f(q)=mf(1)*pf(1)$?

    Thanks
    I'm sorry, what is $\displaystyle U(n)$? That can't be universal notation considering the very common Lie group called the Unitary Group is denoted $\displaystyle U(n)$.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    $\displaystyle U(n)$ represents the group of units modulo n. So, for example, $\displaystyle U(10)=\{1,3,7,9\}$.

    Sorry for not stating that.
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  4. #4
    Senior Member roninpro's Avatar
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    I think that your analysis of the problem is a little bit too complicated. For an easy example, let us look at the group $\displaystyle \mathbb{Z}_{10}$. If $\displaystyle f:\mathbb{Z}_{10}\to \mathbb{Z}_{10}$ is an automorphism, it must send a generator to a generator. To be explicit, a number $\displaystyle a$ is a generator of $\displaystyle \mathbb{Z}_n$ if and only if $\displaystyle \gcd(a,n)=1$. In our case, $\displaystyle a$ is a generator if and only if $\displaystyle \gcd(a,10)=1$. This means that $\displaystyle a=1, 3, 5, 7, 9$. So, we have the following possibilities: $\displaystyle f(1)=1$, $\displaystyle f(1)=3$, $\displaystyle f(1)=5$, $\displaystyle f(1)=7$, or $\displaystyle f(1)=9$. If you try composing these maps, you will find that you recover the multiplication table for your $\displaystyle U_{10}$.

    Give it a try. Good luck!
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Drexel28 View Post
    I'm sorry, what is $\displaystyle U(n)$? That can't be universal notation considering the very common Lie group called the Unitary Group is denoted $\displaystyle U(n)$.
    $\displaystyle U(R)$ is common for denoting the group of units of a ring. I suppose this is just a perversion of that.
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