Is the following mappings automorphisms of their respective groups? (Herstein BOOK)

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Is the following map an automorphism of the cyclic group $G$ of order $12$ ?
$$T : x rightarrow x^3 $$




My attempt :



I thinks the answer is yes, because $G$ is cyclic,
$$T(xy) = (xy)^ 3= x^3y^3 =T(x)T(y)$$ So $T$ is a homomorphism



Also $T$ is one -one as $T(x) = T(y) $ implies $x=y$



Is my logic correct or not ?



Any hints/solution will be appreciated



Thanks




abstract-algebra group-theory group-homomorphism




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    up vote
    0
    down vote

    favorite













    Is the following map an automorphism of the cyclic group $G$ of order $12$ ?
    $$T : x rightarrow x^3 $$




    My attempt :



    I thinks the answer is yes, because $G$ is cyclic,
    $$T(xy) = (xy)^ 3= x^3y^3 =T(x)T(y)$$ So $T$ is a homomorphism



    Also $T$ is one -one as $T(x) = T(y) $ implies $x=y$



    Is my logic correct or not ?



    Any hints/solution will be appreciated



    Thanks




    abstract-algebra group-theory group-homomorphism




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      Is the following map an automorphism of the cyclic group $G$ of order $12$ ?
      $$T : x rightarrow x^3 $$




      My attempt :



      I thinks the answer is yes, because $G$ is cyclic,
      $$T(xy) = (xy)^ 3= x^3y^3 =T(x)T(y)$$ So $T$ is a homomorphism



      Also $T$ is one -one as $T(x) = T(y) $ implies $x=y$



      Is my logic correct or not ?



      Any hints/solution will be appreciated



      Thanks




      abstract-algebra group-theory group-homomorphism




      share|cite|improve this question














      Is the following map an automorphism of the cyclic group $G$ of order $12$ ?
      $$T : x rightarrow x^3 $$




      My attempt :



      I thinks the answer is yes, because $G$ is cyclic,
      $$T(xy) = (xy)^ 3= x^3y^3 =T(x)T(y)$$ So $T$ is a homomorphism



      Also $T$ is one -one as $T(x) = T(y) $ implies $x=y$



      Is my logic correct or not ?



      Any hints/solution will be appreciated



      Thanks





      abstract-algebra group-theory group-homomorphism





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      share|cite|improve this question




      share|cite|improve this question








      edited Jul 31 at 17:13









      Chinnapparaj R

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      asked Jul 31 at 16:34









      stupid

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      52218




















          2 Answers
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          Yes, $T$ is a homomorphism, but that's because $G$ is Abelian, which implies that $(xy)^3=x^3y^3$. And, no, it is not an automorphism. If $G=langle arangle$, then $(a^4)^3=(a^8)^3$, but $a^4neq a^8$.






          share|cite|improve this answer























          • what is $a$ here ?? im not getting ?? i mean what is the value of a ??
            – stupid
            Jul 31 at 17:00






          • 1




            The element $a$ is a generator of $G$. So, $a^12=e_G$ and the smallest natural $n$ such that $a^n=e_G$ is $n=12$.
            – José Carlos Santos
            Jul 31 at 17:13










          • oks i got its thanks u@ jose sir
            – stupid
            Jul 31 at 17:17

















          up vote
          2
          down vote













          It isn't an automorphism. $mathbbZ/12mathbbZ$ is a cyclic group of order 12. With your transformation you get $T(0)=0$ and $T(4)=0$ so it's not one to one.






          share|cite|improve this answer





















          • how T(4) = 0 ?? @ Mark can u exaplain more
            – stupid
            Jul 31 at 16:58






          • 1




            T(4)=4+4+4=12, and 12 equals to 0 in the group $mathbbZ/12mathbbZ$.
            – Mark
            Jul 31 at 17:01










          • oks thanks @Marks.
            – stupid
            Jul 31 at 17:05










          Your Answer




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          2 Answers
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          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Yes, $T$ is a homomorphism, but that's because $G$ is Abelian, which implies that $(xy)^3=x^3y^3$. And, no, it is not an automorphism. If $G=langle arangle$, then $(a^4)^3=(a^8)^3$, but $a^4neq a^8$.






          share|cite|improve this answer























          • what is $a$ here ?? im not getting ?? i mean what is the value of a ??
            – stupid
            Jul 31 at 17:00






          • 1




            The element $a$ is a generator of $G$. So, $a^12=e_G$ and the smallest natural $n$ such that $a^n=e_G$ is $n=12$.
            – José Carlos Santos
            Jul 31 at 17:13










          • oks i got its thanks u@ jose sir
            – stupid
            Jul 31 at 17:17














          up vote
          1
          down vote



          accepted










          Yes, $T$ is a homomorphism, but that's because $G$ is Abelian, which implies that $(xy)^3=x^3y^3$. And, no, it is not an automorphism. If $G=langle arangle$, then $(a^4)^3=(a^8)^3$, but $a^4neq a^8$.






          share|cite|improve this answer























          • what is $a$ here ?? im not getting ?? i mean what is the value of a ??
            – stupid
            Jul 31 at 17:00






          • 1




            The element $a$ is a generator of $G$. So, $a^12=e_G$ and the smallest natural $n$ such that $a^n=e_G$ is $n=12$.
            – José Carlos Santos
            Jul 31 at 17:13










          • oks i got its thanks u@ jose sir
            – stupid
            Jul 31 at 17:17












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Yes, $T$ is a homomorphism, but that's because $G$ is Abelian, which implies that $(xy)^3=x^3y^3$. And, no, it is not an automorphism. If $G=langle arangle$, then $(a^4)^3=(a^8)^3$, but $a^4neq a^8$.






          share|cite|improve this answer















          Yes, $T$ is a homomorphism, but that's because $G$ is Abelian, which implies that $(xy)^3=x^3y^3$. And, no, it is not an automorphism. If $G=langle arangle$, then $(a^4)^3=(a^8)^3$, but $a^4neq a^8$.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 31 at 16:43


























          answered Jul 31 at 16:38









          José Carlos Santos

          112k1696172




          112k1696172











          • what is $a$ here ?? im not getting ?? i mean what is the value of a ??
            – stupid
            Jul 31 at 17:00






          • 1




            The element $a$ is a generator of $G$. So, $a^12=e_G$ and the smallest natural $n$ such that $a^n=e_G$ is $n=12$.
            – José Carlos Santos
            Jul 31 at 17:13










          • oks i got its thanks u@ jose sir
            – stupid
            Jul 31 at 17:17
















          • what is $a$ here ?? im not getting ?? i mean what is the value of a ??
            – stupid
            Jul 31 at 17:00






          • 1




            The element $a$ is a generator of $G$. So, $a^12=e_G$ and the smallest natural $n$ such that $a^n=e_G$ is $n=12$.
            – José Carlos Santos
            Jul 31 at 17:13










          • oks i got its thanks u@ jose sir
            – stupid
            Jul 31 at 17:17















          what is $a$ here ?? im not getting ?? i mean what is the value of a ??
          – stupid
          Jul 31 at 17:00




          what is $a$ here ?? im not getting ?? i mean what is the value of a ??
          – stupid
          Jul 31 at 17:00




          1




          1




          The element $a$ is a generator of $G$. So, $a^12=e_G$ and the smallest natural $n$ such that $a^n=e_G$ is $n=12$.
          – José Carlos Santos
          Jul 31 at 17:13




          The element $a$ is a generator of $G$. So, $a^12=e_G$ and the smallest natural $n$ such that $a^n=e_G$ is $n=12$.
          – José Carlos Santos
          Jul 31 at 17:13












          oks i got its thanks u@ jose sir
          – stupid
          Jul 31 at 17:17




          oks i got its thanks u@ jose sir
          – stupid
          Jul 31 at 17:17










          up vote
          2
          down vote













          It isn't an automorphism. $mathbbZ/12mathbbZ$ is a cyclic group of order 12. With your transformation you get $T(0)=0$ and $T(4)=0$ so it's not one to one.






          share|cite|improve this answer





















          • how T(4) = 0 ?? @ Mark can u exaplain more
            – stupid
            Jul 31 at 16:58






          • 1




            T(4)=4+4+4=12, and 12 equals to 0 in the group $mathbbZ/12mathbbZ$.
            – Mark
            Jul 31 at 17:01










          • oks thanks @Marks.
            – stupid
            Jul 31 at 17:05














          up vote
          2
          down vote













          It isn't an automorphism. $mathbbZ/12mathbbZ$ is a cyclic group of order 12. With your transformation you get $T(0)=0$ and $T(4)=0$ so it's not one to one.






          share|cite|improve this answer





















          • how T(4) = 0 ?? @ Mark can u exaplain more
            – stupid
            Jul 31 at 16:58






          • 1




            T(4)=4+4+4=12, and 12 equals to 0 in the group $mathbbZ/12mathbbZ$.
            – Mark
            Jul 31 at 17:01










          • oks thanks @Marks.
            – stupid
            Jul 31 at 17:05












          up vote
          2
          down vote










          up vote
          2
          down vote









          It isn't an automorphism. $mathbbZ/12mathbbZ$ is a cyclic group of order 12. With your transformation you get $T(0)=0$ and $T(4)=0$ so it's not one to one.






          share|cite|improve this answer













          It isn't an automorphism. $mathbbZ/12mathbbZ$ is a cyclic group of order 12. With your transformation you get $T(0)=0$ and $T(4)=0$ so it's not one to one.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 31 at 16:38









          Mark

          5949




          5949











          • how T(4) = 0 ?? @ Mark can u exaplain more
            – stupid
            Jul 31 at 16:58






          • 1




            T(4)=4+4+4=12, and 12 equals to 0 in the group $mathbbZ/12mathbbZ$.
            – Mark
            Jul 31 at 17:01










          • oks thanks @Marks.
            – stupid
            Jul 31 at 17:05
















          • how T(4) = 0 ?? @ Mark can u exaplain more
            – stupid
            Jul 31 at 16:58






          • 1




            T(4)=4+4+4=12, and 12 equals to 0 in the group $mathbbZ/12mathbbZ$.
            – Mark
            Jul 31 at 17:01










          • oks thanks @Marks.
            – stupid
            Jul 31 at 17:05















          how T(4) = 0 ?? @ Mark can u exaplain more
          – stupid
          Jul 31 at 16:58




          how T(4) = 0 ?? @ Mark can u exaplain more
          – stupid
          Jul 31 at 16:58




          1




          1




          T(4)=4+4+4=12, and 12 equals to 0 in the group $mathbbZ/12mathbbZ$.
          – Mark
          Jul 31 at 17:01




          T(4)=4+4+4=12, and 12 equals to 0 in the group $mathbbZ/12mathbbZ$.
          – Mark
          Jul 31 at 17:01












          oks thanks @Marks.
          – stupid
          Jul 31 at 17:05




          oks thanks @Marks.
          – stupid
          Jul 31 at 17:05












           

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