Show that the map $alpha:mathbb{Z}_nto mathbb{Z}_n$ by $sto sr$ where $rin U(n)$ is an automorphism of...

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Show that the map $alpha:mathbb{Z}_nto mathbb{Z}_n$ by $sto sr$ where $rin U(n)$ is an automorphism of $mathbb{Z}_n$



The Next CEO of Stack OverflowThere is an automorphism of $mathbb Z_6$ which is not an inner automorphismThe notation $mathbb{Z}[alpha]$Showing $sigma_{a}(x) = x^{a}$ is an automorphism of $mathbb{Z}_{n}$ iff $gcd(a,n)=1$An isomorphic map from $mathbb R^{*} rightarrow mathbb R^{*}$Proof verification needed for showing a map is an automorphism iff gcd$(m,n)=1$Showing a group is surjective to show it's an automorphism.Is this map is an automorphism of $mathbb F_{p^n}$Show that the induced map $bar{alpha} : G rightarrow text{Im}(alpha)$ is an isomorphismLet $A$ be a group and fix $b ∈ A$. Show that the map $ϕ : A → A$ given by $ϕ(a) := b^{-1}ab$ is an automorphism of A.Every surjective homomorphism $alpha : mathbb{Z}^3 rightarrow mathbb{Z}^3$ is isomorphism












2












$begingroup$


$alpha :mathbb{Z}_ntomathbb{Z}_n$ by $sto sr$ where $rin U(n)$ is an automorphism.



Homomorphism:
$ alpha(s+t)=r(s+t)=rs+rt=alpha(s)+alpha(r)$



Injective:



Suppose $alpha(x)=alpha(y)$



Then $rx=ryiff x=y$



Im not sure how to show surjectivity however.



Also I'm not sure how this could be an isomorphism as $alpha(1)=r$ so its not mapping the identity to the identity.






group-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Consider $ r|n.$
    $endgroup$
    – Minz
    Mar 18 at 0:59












  • $begingroup$
    @Minz Notice that $rin U(n)$, which means $r$ is co-prime to $n$.
    $endgroup$
    – awllower
    Mar 18 at 1:01








  • 1




    $begingroup$
    @ awllower Thanks, I did not know the meaning of the notation $U(n).$ If $r$ is co-prime to $n$ then it has inverse modulo $n$.
    $endgroup$
    – Minz
    Mar 18 at 1:10


















2












$begingroup$


$alpha :mathbb{Z}_ntomathbb{Z}_n$ by $sto sr$ where $rin U(n)$ is an automorphism.



Homomorphism:
$ alpha(s+t)=r(s+t)=rs+rt=alpha(s)+alpha(r)$



Injective:



Suppose $alpha(x)=alpha(y)$



Then $rx=ryiff x=y$



Im not sure how to show surjectivity however.



Also I'm not sure how this could be an isomorphism as $alpha(1)=r$ so its not mapping the identity to the identity.






group-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Consider $ r|n.$
    $endgroup$
    – Minz
    Mar 18 at 0:59












  • $begingroup$
    @Minz Notice that $rin U(n)$, which means $r$ is co-prime to $n$.
    $endgroup$
    – awllower
    Mar 18 at 1:01








  • 1




    $begingroup$
    @ awllower Thanks, I did not know the meaning of the notation $U(n).$ If $r$ is co-prime to $n$ then it has inverse modulo $n$.
    $endgroup$
    – Minz
    Mar 18 at 1:10
















2












2








2





$begingroup$


$alpha :mathbb{Z}_ntomathbb{Z}_n$ by $sto sr$ where $rin U(n)$ is an automorphism.



Homomorphism:
$ alpha(s+t)=r(s+t)=rs+rt=alpha(s)+alpha(r)$



Injective:



Suppose $alpha(x)=alpha(y)$



Then $rx=ryiff x=y$



Im not sure how to show surjectivity however.



Also I'm not sure how this could be an isomorphism as $alpha(1)=r$ so its not mapping the identity to the identity.






group-theory







share|cite|improve this question









$endgroup$




$alpha :mathbb{Z}_ntomathbb{Z}_n$ by $sto sr$ where $rin U(n)$ is an automorphism.



Homomorphism:
$ alpha(s+t)=r(s+t)=rs+rt=alpha(s)+alpha(r)$



Injective:



Suppose $alpha(x)=alpha(y)$



Then $rx=ryiff x=y$



Im not sure how to show surjectivity however.



Also I'm not sure how this could be an isomorphism as $alpha(1)=r$ so its not mapping the identity to the identity.






group-theory




group-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 18 at 0:48









AColoredReptileAColoredReptile

401210




401210












  • $begingroup$
    Consider $ r|n.$
    $endgroup$
    – Minz
    Mar 18 at 0:59












  • $begingroup$
    @Minz Notice that $rin U(n)$, which means $r$ is co-prime to $n$.
    $endgroup$
    – awllower
    Mar 18 at 1:01








  • 1




    $begingroup$
    @ awllower Thanks, I did not know the meaning of the notation $U(n).$ If $r$ is co-prime to $n$ then it has inverse modulo $n$.
    $endgroup$
    – Minz
    Mar 18 at 1:10




















  • $begingroup$
    Consider $ r|n.$
    $endgroup$
    – Minz
    Mar 18 at 0:59












  • $begingroup$
    @Minz Notice that $rin U(n)$, which means $r$ is co-prime to $n$.
    $endgroup$
    – awllower
    Mar 18 at 1:01








  • 1




    $begingroup$
    @ awllower Thanks, I did not know the meaning of the notation $U(n).$ If $r$ is co-prime to $n$ then it has inverse modulo $n$.
    $endgroup$
    – Minz
    Mar 18 at 1:10


















$begingroup$
Consider $ r|n.$
$endgroup$
– Minz
Mar 18 at 0:59






$begingroup$
Consider $ r|n.$
$endgroup$
– Minz
Mar 18 at 0:59














$begingroup$
@Minz Notice that $rin U(n)$, which means $r$ is co-prime to $n$.
$endgroup$
– awllower
Mar 18 at 1:01






$begingroup$
@Minz Notice that $rin U(n)$, which means $r$ is co-prime to $n$.
$endgroup$
– awllower
Mar 18 at 1:01






1




1




$begingroup$
@ awllower Thanks, I did not know the meaning of the notation $U(n).$ If $r$ is co-prime to $n$ then it has inverse modulo $n$.
$endgroup$
– Minz
Mar 18 at 1:10






$begingroup$
@ awllower Thanks, I did not know the meaning of the notation $U(n).$ If $r$ is co-prime to $n$ then it has inverse modulo $n$.
$endgroup$
– Minz
Mar 18 at 1:10












2 Answers
2






active

oldest

votes


















2












$begingroup$

For a homomorphism to be an automorphism, it needs and it suffices for it to be bijective.



Since $rin U(n)$, there is some $r'$ such that $rcdot r'+ncdot n'=1$, for some integer $n'$. Hence the map $beta:mathbb Z_nrightarrowmathbb Z_n$ given by $smapsto sr'$ is the inverse of $alpha$. (Show this.) This shows that $alpha$ is bijective, and hence an automorphism.




As to your question regarding why it does not send $1$ to $1$, notice that the identity of $mathbb Z_n$ is $1_{mathbb Z_n}=0+nmathbb Z$, which is sent to the identity by $alpha$.




Hope this helps.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint:



    Directly find the inverse homomorphism. It is also the multiplication by a unit. Can you find which?






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      For a homomorphism to be an automorphism, it needs and it suffices for it to be bijective.



      Since $rin U(n)$, there is some $r'$ such that $rcdot r'+ncdot n'=1$, for some integer $n'$. Hence the map $beta:mathbb Z_nrightarrowmathbb Z_n$ given by $smapsto sr'$ is the inverse of $alpha$. (Show this.) This shows that $alpha$ is bijective, and hence an automorphism.




      As to your question regarding why it does not send $1$ to $1$, notice that the identity of $mathbb Z_n$ is $1_{mathbb Z_n}=0+nmathbb Z$, which is sent to the identity by $alpha$.




      Hope this helps.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        For a homomorphism to be an automorphism, it needs and it suffices for it to be bijective.



        Since $rin U(n)$, there is some $r'$ such that $rcdot r'+ncdot n'=1$, for some integer $n'$. Hence the map $beta:mathbb Z_nrightarrowmathbb Z_n$ given by $smapsto sr'$ is the inverse of $alpha$. (Show this.) This shows that $alpha$ is bijective, and hence an automorphism.




        As to your question regarding why it does not send $1$ to $1$, notice that the identity of $mathbb Z_n$ is $1_{mathbb Z_n}=0+nmathbb Z$, which is sent to the identity by $alpha$.




        Hope this helps.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          For a homomorphism to be an automorphism, it needs and it suffices for it to be bijective.



          Since $rin U(n)$, there is some $r'$ such that $rcdot r'+ncdot n'=1$, for some integer $n'$. Hence the map $beta:mathbb Z_nrightarrowmathbb Z_n$ given by $smapsto sr'$ is the inverse of $alpha$. (Show this.) This shows that $alpha$ is bijective, and hence an automorphism.




          As to your question regarding why it does not send $1$ to $1$, notice that the identity of $mathbb Z_n$ is $1_{mathbb Z_n}=0+nmathbb Z$, which is sent to the identity by $alpha$.




          Hope this helps.






          share|cite|improve this answer









          $endgroup$



          For a homomorphism to be an automorphism, it needs and it suffices for it to be bijective.



          Since $rin U(n)$, there is some $r'$ such that $rcdot r'+ncdot n'=1$, for some integer $n'$. Hence the map $beta:mathbb Z_nrightarrowmathbb Z_n$ given by $smapsto sr'$ is the inverse of $alpha$. (Show this.) This shows that $alpha$ is bijective, and hence an automorphism.




          As to your question regarding why it does not send $1$ to $1$, notice that the identity of $mathbb Z_n$ is $1_{mathbb Z_n}=0+nmathbb Z$, which is sent to the identity by $alpha$.




          Hope this helps.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 18 at 0:58









          awllowerawllower

          10.5k42672




          10.5k42672























              0












              $begingroup$

              Hint:



              Directly find the inverse homomorphism. It is also the multiplication by a unit. Can you find which?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint:



                Directly find the inverse homomorphism. It is also the multiplication by a unit. Can you find which?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint:



                  Directly find the inverse homomorphism. It is also the multiplication by a unit. Can you find which?






                  share|cite|improve this answer









                  $endgroup$



                  Hint:



                  Directly find the inverse homomorphism. It is also the multiplication by a unit. Can you find which?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 18 at 0:56









                  BernardBernard

                  124k741118




                  124k741118






























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