Proof of U(n) being automorphism of $Z_n$.

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I want to prove that $rin U(n), alpha : Z_n to Z_n, alpha (s)=mod(rs, n)$ is an automorphism. I have proved that $alpha$ is one-one, maps identity to identity and $alpha(s+t) = alpha(s) + alpha(t)$.



I am not able to prove the onto part.










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    I want to prove that $rin U(n), alpha : Z_n to Z_n, alpha (s)=mod(rs, n)$ is an automorphism. I have proved that $alpha$ is one-one, maps identity to identity and $alpha(s+t) = alpha(s) + alpha(t)$.



    I am not able to prove the onto part.










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      up vote
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      down vote

      favorite









      up vote
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      down vote

      favorite











      I want to prove that $rin U(n), alpha : Z_n to Z_n, alpha (s)=mod(rs, n)$ is an automorphism. I have proved that $alpha$ is one-one, maps identity to identity and $alpha(s+t) = alpha(s) + alpha(t)$.



      I am not able to prove the onto part.










      share|cite|improve this question













      I want to prove that $rin U(n), alpha : Z_n to Z_n, alpha (s)=mod(rs, n)$ is an automorphism. I have proved that $alpha$ is one-one, maps identity to identity and $alpha(s+t) = alpha(s) + alpha(t)$.



      I am not able to prove the onto part.







      abstract-algebra automorphism-group






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      asked Sep 8 at 8:29









      Piyush Divyanakar

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          Recall that $r$ is a unit of $mathbbZ_n$ if and only if $r$ is coprime with $n$. Thus, by Bezout's identity, we have $s$ and $t$ so that $sn + tr = 1$. Modulo $n$, this gives $tr equiv 1 pmodn$. Therefore, for any $k in mathbbZ_n$,



          $$
          alpha(kt) = ktr equiv k pmodn .
          $$






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            1 Answer
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            1 Answer
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            active

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            up vote
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            accepted










            Recall that $r$ is a unit of $mathbbZ_n$ if and only if $r$ is coprime with $n$. Thus, by Bezout's identity, we have $s$ and $t$ so that $sn + tr = 1$. Modulo $n$, this gives $tr equiv 1 pmodn$. Therefore, for any $k in mathbbZ_n$,



            $$
            alpha(kt) = ktr equiv k pmodn .
            $$






            share|cite|improve this answer
























              up vote
              1
              down vote



              accepted










              Recall that $r$ is a unit of $mathbbZ_n$ if and only if $r$ is coprime with $n$. Thus, by Bezout's identity, we have $s$ and $t$ so that $sn + tr = 1$. Modulo $n$, this gives $tr equiv 1 pmodn$. Therefore, for any $k in mathbbZ_n$,



              $$
              alpha(kt) = ktr equiv k pmodn .
              $$






              share|cite|improve this answer






















                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Recall that $r$ is a unit of $mathbbZ_n$ if and only if $r$ is coprime with $n$. Thus, by Bezout's identity, we have $s$ and $t$ so that $sn + tr = 1$. Modulo $n$, this gives $tr equiv 1 pmodn$. Therefore, for any $k in mathbbZ_n$,



                $$
                alpha(kt) = ktr equiv k pmodn .
                $$






                share|cite|improve this answer












                Recall that $r$ is a unit of $mathbbZ_n$ if and only if $r$ is coprime with $n$. Thus, by Bezout's identity, we have $s$ and $t$ so that $sn + tr = 1$. Modulo $n$, this gives $tr equiv 1 pmodn$. Therefore, for any $k in mathbbZ_n$,



                $$
                alpha(kt) = ktr equiv k pmodn .
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 8 at 8:41









                Guido A.

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