Automorphisms of a Cyclic Group












0














To show that,




$operatorname{Aut}(mathbb{Z}/nmathbb{Z})cong (mathbb{Z}/nmathbb{Z})^times.$




What I feel is that if we show that "any automorphism $phi:mathbb{Z}/nmathbb{Z}longrightarrowmathbb{Z}/nmathbb{Z}$ is of the form $phi_a([k]_n):=[ak]_n$ for some $ainmathbb{Z}$ such that gcd$(a,n)=1.$ then we are done.



But I am not sure how to do this?
Thanks in advance.






abstract-algebra group-theory automorphism-group







share|cite|improve this question
























  • You have all the parts to prove your assertion.
    – Joel Pereira
    Nov 25 '18 at 16:31










  • Can you think of a map from $(mathbb{Z}/nmathbb{Z})^times$ to $mathrm{Aut}(mathbb{Z}/nmathbb{Z})$? The first one that you can think of is an isomorphism: you just need to prove it. That it's a homomorphism is easy, as is injectivity. Surjectivity is slightly harder, and is essentially what's in your thought process. As a hint for that, consider where $varphi$ sends $[1]_n$.
    – user3482749
    Nov 25 '18 at 16:50












  • As for LaTeX on here: see here.
    – user3482749
    Nov 25 '18 at 16:51
















0














To show that,




$operatorname{Aut}(mathbb{Z}/nmathbb{Z})cong (mathbb{Z}/nmathbb{Z})^times.$




What I feel is that if we show that "any automorphism $phi:mathbb{Z}/nmathbb{Z}longrightarrowmathbb{Z}/nmathbb{Z}$ is of the form $phi_a([k]_n):=[ak]_n$ for some $ainmathbb{Z}$ such that gcd$(a,n)=1.$ then we are done.



But I am not sure how to do this?
Thanks in advance.






abstract-algebra group-theory automorphism-group







share|cite|improve this question
























  • You have all the parts to prove your assertion.
    – Joel Pereira
    Nov 25 '18 at 16:31










  • Can you think of a map from $(mathbb{Z}/nmathbb{Z})^times$ to $mathrm{Aut}(mathbb{Z}/nmathbb{Z})$? The first one that you can think of is an isomorphism: you just need to prove it. That it's a homomorphism is easy, as is injectivity. Surjectivity is slightly harder, and is essentially what's in your thought process. As a hint for that, consider where $varphi$ sends $[1]_n$.
    – user3482749
    Nov 25 '18 at 16:50












  • As for LaTeX on here: see here.
    – user3482749
    Nov 25 '18 at 16:51














0












0








0


3





To show that,




$operatorname{Aut}(mathbb{Z}/nmathbb{Z})cong (mathbb{Z}/nmathbb{Z})^times.$




What I feel is that if we show that "any automorphism $phi:mathbb{Z}/nmathbb{Z}longrightarrowmathbb{Z}/nmathbb{Z}$ is of the form $phi_a([k]_n):=[ak]_n$ for some $ainmathbb{Z}$ such that gcd$(a,n)=1.$ then we are done.



But I am not sure how to do this?
Thanks in advance.






abstract-algebra group-theory automorphism-group







share|cite|improve this question















To show that,




$operatorname{Aut}(mathbb{Z}/nmathbb{Z})cong (mathbb{Z}/nmathbb{Z})^times.$




What I feel is that if we show that "any automorphism $phi:mathbb{Z}/nmathbb{Z}longrightarrowmathbb{Z}/nmathbb{Z}$ is of the form $phi_a([k]_n):=[ak]_n$ for some $ainmathbb{Z}$ such that gcd$(a,n)=1.$ then we are done.



But I am not sure how to do this?
Thanks in advance.






abstract-algebra group-theory automorphism-group




abstract-algebra group-theory automorphism-group






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 17:13









amWhy

191k28224439




191k28224439










asked Nov 25 '18 at 16:14









WhatsDUI

317112




317112












  • You have all the parts to prove your assertion.
    – Joel Pereira
    Nov 25 '18 at 16:31










  • Can you think of a map from $(mathbb{Z}/nmathbb{Z})^times$ to $mathrm{Aut}(mathbb{Z}/nmathbb{Z})$? The first one that you can think of is an isomorphism: you just need to prove it. That it's a homomorphism is easy, as is injectivity. Surjectivity is slightly harder, and is essentially what's in your thought process. As a hint for that, consider where $varphi$ sends $[1]_n$.
    – user3482749
    Nov 25 '18 at 16:50












  • As for LaTeX on here: see here.
    – user3482749
    Nov 25 '18 at 16:51


















  • You have all the parts to prove your assertion.
    – Joel Pereira
    Nov 25 '18 at 16:31










  • Can you think of a map from $(mathbb{Z}/nmathbb{Z})^times$ to $mathrm{Aut}(mathbb{Z}/nmathbb{Z})$? The first one that you can think of is an isomorphism: you just need to prove it. That it's a homomorphism is easy, as is injectivity. Surjectivity is slightly harder, and is essentially what's in your thought process. As a hint for that, consider where $varphi$ sends $[1]_n$.
    – user3482749
    Nov 25 '18 at 16:50












  • As for LaTeX on here: see here.
    – user3482749
    Nov 25 '18 at 16:51
















You have all the parts to prove your assertion.
– Joel Pereira
Nov 25 '18 at 16:31




You have all the parts to prove your assertion.
– Joel Pereira
Nov 25 '18 at 16:31












Can you think of a map from $(mathbb{Z}/nmathbb{Z})^times$ to $mathrm{Aut}(mathbb{Z}/nmathbb{Z})$? The first one that you can think of is an isomorphism: you just need to prove it. That it's a homomorphism is easy, as is injectivity. Surjectivity is slightly harder, and is essentially what's in your thought process. As a hint for that, consider where $varphi$ sends $[1]_n$.
– user3482749
Nov 25 '18 at 16:50






Can you think of a map from $(mathbb{Z}/nmathbb{Z})^times$ to $mathrm{Aut}(mathbb{Z}/nmathbb{Z})$? The first one that you can think of is an isomorphism: you just need to prove it. That it's a homomorphism is easy, as is injectivity. Surjectivity is slightly harder, and is essentially what's in your thought process. As a hint for that, consider where $varphi$ sends $[1]_n$.
– user3482749
Nov 25 '18 at 16:50














As for LaTeX on here: see here.
– user3482749
Nov 25 '18 at 16:51




As for LaTeX on here: see here.
– user3482749
Nov 25 '18 at 16:51










1 Answer
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$mathbb{Z}/nmathbb{Z}$ is cyclic generated by $1$. So an isomorphism $phi$ is completely determined by what 1 gets map to. Because suppose $phi(1)=k$ then $phi(m)=phi(1+1+1+...+1)=phi(1)+phi(1)+...+phi(1)= m*k$. Also, you need to make sure $0$ is the only element that gets map to $0$ as we have a isomorphism not a homomorphism. So you can't have $f(1)=k$ where $k$ is a zero divisor as then f(z)=zk=0 for some $z neq 0$. Which means $f(1)$ has to be invertible. Now you can check that $alpha :Aut(mathbb{Z}/nmathbb{Z}) rightarrow (mathbb{Z}/nmathbb{Z})^*$ by $alpha(f)=f(1)$ is an isomorphism.



Automorphism ring is a way to construct ring from groups. you can check that the reason why multiplication of two negative numbers is positive is because the ring of integers is actually automorphism of the abelian additive group $mathbb{Z}$. Because when you compose $f,g$ where $f(1)=-k$(sends all positive to negative),$g(1)=-n$(sends all positive to negative and vice-versa), you get that the composition of the two map sends all positive to all positive. Hope I got it right.






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    $mathbb{Z}/nmathbb{Z}$ is cyclic generated by $1$. So an isomorphism $phi$ is completely determined by what 1 gets map to. Because suppose $phi(1)=k$ then $phi(m)=phi(1+1+1+...+1)=phi(1)+phi(1)+...+phi(1)= m*k$. Also, you need to make sure $0$ is the only element that gets map to $0$ as we have a isomorphism not a homomorphism. So you can't have $f(1)=k$ where $k$ is a zero divisor as then f(z)=zk=0 for some $z neq 0$. Which means $f(1)$ has to be invertible. Now you can check that $alpha :Aut(mathbb{Z}/nmathbb{Z}) rightarrow (mathbb{Z}/nmathbb{Z})^*$ by $alpha(f)=f(1)$ is an isomorphism.



    Automorphism ring is a way to construct ring from groups. you can check that the reason why multiplication of two negative numbers is positive is because the ring of integers is actually automorphism of the abelian additive group $mathbb{Z}$. Because when you compose $f,g$ where $f(1)=-k$(sends all positive to negative),$g(1)=-n$(sends all positive to negative and vice-versa), you get that the composition of the two map sends all positive to all positive. Hope I got it right.






    share|cite|improve this answer


























      2














      $mathbb{Z}/nmathbb{Z}$ is cyclic generated by $1$. So an isomorphism $phi$ is completely determined by what 1 gets map to. Because suppose $phi(1)=k$ then $phi(m)=phi(1+1+1+...+1)=phi(1)+phi(1)+...+phi(1)= m*k$. Also, you need to make sure $0$ is the only element that gets map to $0$ as we have a isomorphism not a homomorphism. So you can't have $f(1)=k$ where $k$ is a zero divisor as then f(z)=zk=0 for some $z neq 0$. Which means $f(1)$ has to be invertible. Now you can check that $alpha :Aut(mathbb{Z}/nmathbb{Z}) rightarrow (mathbb{Z}/nmathbb{Z})^*$ by $alpha(f)=f(1)$ is an isomorphism.



      Automorphism ring is a way to construct ring from groups. you can check that the reason why multiplication of two negative numbers is positive is because the ring of integers is actually automorphism of the abelian additive group $mathbb{Z}$. Because when you compose $f,g$ where $f(1)=-k$(sends all positive to negative),$g(1)=-n$(sends all positive to negative and vice-versa), you get that the composition of the two map sends all positive to all positive. Hope I got it right.






      share|cite|improve this answer
























        2












        2








        2






        $mathbb{Z}/nmathbb{Z}$ is cyclic generated by $1$. So an isomorphism $phi$ is completely determined by what 1 gets map to. Because suppose $phi(1)=k$ then $phi(m)=phi(1+1+1+...+1)=phi(1)+phi(1)+...+phi(1)= m*k$. Also, you need to make sure $0$ is the only element that gets map to $0$ as we have a isomorphism not a homomorphism. So you can't have $f(1)=k$ where $k$ is a zero divisor as then f(z)=zk=0 for some $z neq 0$. Which means $f(1)$ has to be invertible. Now you can check that $alpha :Aut(mathbb{Z}/nmathbb{Z}) rightarrow (mathbb{Z}/nmathbb{Z})^*$ by $alpha(f)=f(1)$ is an isomorphism.



        Automorphism ring is a way to construct ring from groups. you can check that the reason why multiplication of two negative numbers is positive is because the ring of integers is actually automorphism of the abelian additive group $mathbb{Z}$. Because when you compose $f,g$ where $f(1)=-k$(sends all positive to negative),$g(1)=-n$(sends all positive to negative and vice-versa), you get that the composition of the two map sends all positive to all positive. Hope I got it right.






        share|cite|improve this answer












        $mathbb{Z}/nmathbb{Z}$ is cyclic generated by $1$. So an isomorphism $phi$ is completely determined by what 1 gets map to. Because suppose $phi(1)=k$ then $phi(m)=phi(1+1+1+...+1)=phi(1)+phi(1)+...+phi(1)= m*k$. Also, you need to make sure $0$ is the only element that gets map to $0$ as we have a isomorphism not a homomorphism. So you can't have $f(1)=k$ where $k$ is a zero divisor as then f(z)=zk=0 for some $z neq 0$. Which means $f(1)$ has to be invertible. Now you can check that $alpha :Aut(mathbb{Z}/nmathbb{Z}) rightarrow (mathbb{Z}/nmathbb{Z})^*$ by $alpha(f)=f(1)$ is an isomorphism.



        Automorphism ring is a way to construct ring from groups. you can check that the reason why multiplication of two negative numbers is positive is because the ring of integers is actually automorphism of the abelian additive group $mathbb{Z}$. Because when you compose $f,g$ where $f(1)=-k$(sends all positive to negative),$g(1)=-n$(sends all positive to negative and vice-versa), you get that the composition of the two map sends all positive to all positive. Hope I got it right.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 17:00









        mathnoob

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