$operatorname{Aut} (G)$ is isomorphic to $operatorname{Aut} (H)$ then is it necessary that $G$ is isomorphic...












7












$begingroup$


If $operatorname{Aut} (G)$ is isomorphic to $operatorname{Aut} (H)$ then is it necessary that $G$ is isomorphic to $H$?



My answer is no.
$operatorname{Aut} (mathbb{Z)}$ is isomorphic to $Z_2$ and $operatorname{Aut} (Z_3)$ is also isomorphic to $U(3)$, which is isomorphic to $Z_2$. But $mathbb Z$ is not isomorphic to $Z_3.$ Correct? Thanks






abstract-algebra group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sorry I meant$Z_3$
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:22










  • $begingroup$
    @Derek Holt I edited it to$Z_3$. Am I correct now?
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:23










  • $begingroup$
    Yes, you are correct. Also there's no need for introducing $U(3)$, $Aut(mathbb{Z}_3)$ has exactly two elements, thus it has to be $mathbb{Z}_2$.
    $endgroup$
    – freakish
    Dec 17 '18 at 11:24












  • $begingroup$
    @freakish thanks
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:25
















7












$begingroup$


If $operatorname{Aut} (G)$ is isomorphic to $operatorname{Aut} (H)$ then is it necessary that $G$ is isomorphic to $H$?



My answer is no.
$operatorname{Aut} (mathbb{Z)}$ is isomorphic to $Z_2$ and $operatorname{Aut} (Z_3)$ is also isomorphic to $U(3)$, which is isomorphic to $Z_2$. But $mathbb Z$ is not isomorphic to $Z_3.$ Correct? Thanks






abstract-algebra group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sorry I meant$Z_3$
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:22










  • $begingroup$
    @Derek Holt I edited it to$Z_3$. Am I correct now?
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:23










  • $begingroup$
    Yes, you are correct. Also there's no need for introducing $U(3)$, $Aut(mathbb{Z}_3)$ has exactly two elements, thus it has to be $mathbb{Z}_2$.
    $endgroup$
    – freakish
    Dec 17 '18 at 11:24












  • $begingroup$
    @freakish thanks
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:25














7












7








7


4



$begingroup$


If $operatorname{Aut} (G)$ is isomorphic to $operatorname{Aut} (H)$ then is it necessary that $G$ is isomorphic to $H$?



My answer is no.
$operatorname{Aut} (mathbb{Z)}$ is isomorphic to $Z_2$ and $operatorname{Aut} (Z_3)$ is also isomorphic to $U(3)$, which is isomorphic to $Z_2$. But $mathbb Z$ is not isomorphic to $Z_3.$ Correct? Thanks






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




If $operatorname{Aut} (G)$ is isomorphic to $operatorname{Aut} (H)$ then is it necessary that $G$ is isomorphic to $H$?



My answer is no.
$operatorname{Aut} (mathbb{Z)}$ is isomorphic to $Z_2$ and $operatorname{Aut} (Z_3)$ is also isomorphic to $U(3)$, which is isomorphic to $Z_2$. But $mathbb Z$ is not isomorphic to $Z_3.$ Correct? Thanks






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 12:26









user1729

17.4k64193




17.4k64193










asked Dec 17 '18 at 11:17









ramanujanramanujan

719713




719713












  • $begingroup$
    Sorry I meant$Z_3$
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:22










  • $begingroup$
    @Derek Holt I edited it to$Z_3$. Am I correct now?
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:23










  • $begingroup$
    Yes, you are correct. Also there's no need for introducing $U(3)$, $Aut(mathbb{Z}_3)$ has exactly two elements, thus it has to be $mathbb{Z}_2$.
    $endgroup$
    – freakish
    Dec 17 '18 at 11:24












  • $begingroup$
    @freakish thanks
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:25


















  • $begingroup$
    Sorry I meant$Z_3$
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:22










  • $begingroup$
    @Derek Holt I edited it to$Z_3$. Am I correct now?
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:23










  • $begingroup$
    Yes, you are correct. Also there's no need for introducing $U(3)$, $Aut(mathbb{Z}_3)$ has exactly two elements, thus it has to be $mathbb{Z}_2$.
    $endgroup$
    – freakish
    Dec 17 '18 at 11:24












  • $begingroup$
    @freakish thanks
    $endgroup$
    – ramanujan
    Dec 17 '18 at 11:25
















$begingroup$
Sorry I meant$Z_3$
$endgroup$
– ramanujan
Dec 17 '18 at 11:22




$begingroup$
Sorry I meant$Z_3$
$endgroup$
– ramanujan
Dec 17 '18 at 11:22












$begingroup$
@Derek Holt I edited it to$Z_3$. Am I correct now?
$endgroup$
– ramanujan
Dec 17 '18 at 11:23




$begingroup$
@Derek Holt I edited it to$Z_3$. Am I correct now?
$endgroup$
– ramanujan
Dec 17 '18 at 11:23












$begingroup$
Yes, you are correct. Also there's no need for introducing $U(3)$, $Aut(mathbb{Z}_3)$ has exactly two elements, thus it has to be $mathbb{Z}_2$.
$endgroup$
– freakish
Dec 17 '18 at 11:24






$begingroup$
Yes, you are correct. Also there's no need for introducing $U(3)$, $Aut(mathbb{Z}_3)$ has exactly two elements, thus it has to be $mathbb{Z}_2$.
$endgroup$
– freakish
Dec 17 '18 at 11:24














$begingroup$
@freakish thanks
$endgroup$
– ramanujan
Dec 17 '18 at 11:25




$begingroup$
@freakish thanks
$endgroup$
– ramanujan
Dec 17 '18 at 11:25










3 Answers
3






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4












$begingroup$

Besides your example, there is even an example with finite groups, as
$$
{rm Aut}(S_3)cong {rm Aut}(C_2times C_2)cong S_3,
$$
but $S_3$ is of course not isomorphic to $C_2times C_2$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    As another example, both the trivial group $Id$ and the cyclic group of order two $C_2$ have trivial automorphism group:
    $$operatorname{Aut}(Id)cong Idcongoperatorname{Aut(C_2)}$$



    This is the smallest possible example...



    (These are the only two groups $G$ with $operatorname{Aut}(G)cong Id$. See here for a proof.)






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      You can even use non-isomorphic finite groups of the same order. The smallest example is $Aut(C_4times C_2)cong Aut(D_8)cong D_8$.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

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        4












        $begingroup$

        Besides your example, there is even an example with finite groups, as
        $$
        {rm Aut}(S_3)cong {rm Aut}(C_2times C_2)cong S_3,
        $$
        but $S_3$ is of course not isomorphic to $C_2times C_2$.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          Besides your example, there is even an example with finite groups, as
          $$
          {rm Aut}(S_3)cong {rm Aut}(C_2times C_2)cong S_3,
          $$
          but $S_3$ is of course not isomorphic to $C_2times C_2$.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            Besides your example, there is even an example with finite groups, as
            $$
            {rm Aut}(S_3)cong {rm Aut}(C_2times C_2)cong S_3,
            $$
            but $S_3$ is of course not isomorphic to $C_2times C_2$.






            share|cite|improve this answer









            $endgroup$



            Besides your example, there is even an example with finite groups, as
            $$
            {rm Aut}(S_3)cong {rm Aut}(C_2times C_2)cong S_3,
            $$
            but $S_3$ is of course not isomorphic to $C_2times C_2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 17 '18 at 11:40









            Dietrich BurdeDietrich Burde

            81k648106




            81k648106























                2












                $begingroup$

                As another example, both the trivial group $Id$ and the cyclic group of order two $C_2$ have trivial automorphism group:
                $$operatorname{Aut}(Id)cong Idcongoperatorname{Aut(C_2)}$$



                This is the smallest possible example...



                (These are the only two groups $G$ with $operatorname{Aut}(G)cong Id$. See here for a proof.)






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  As another example, both the trivial group $Id$ and the cyclic group of order two $C_2$ have trivial automorphism group:
                  $$operatorname{Aut}(Id)cong Idcongoperatorname{Aut(C_2)}$$



                  This is the smallest possible example...



                  (These are the only two groups $G$ with $operatorname{Aut}(G)cong Id$. See here for a proof.)






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    As another example, both the trivial group $Id$ and the cyclic group of order two $C_2$ have trivial automorphism group:
                    $$operatorname{Aut}(Id)cong Idcongoperatorname{Aut(C_2)}$$



                    This is the smallest possible example...



                    (These are the only two groups $G$ with $operatorname{Aut}(G)cong Id$. See here for a proof.)






                    share|cite|improve this answer









                    $endgroup$



                    As another example, both the trivial group $Id$ and the cyclic group of order two $C_2$ have trivial automorphism group:
                    $$operatorname{Aut}(Id)cong Idcongoperatorname{Aut(C_2)}$$



                    This is the smallest possible example...



                    (These are the only two groups $G$ with $operatorname{Aut}(G)cong Id$. See here for a proof.)







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 17 '18 at 12:20









                    user1729user1729

                    17.4k64193




                    17.4k64193























                        1












                        $begingroup$

                        You can even use non-isomorphic finite groups of the same order. The smallest example is $Aut(C_4times C_2)cong Aut(D_8)cong D_8$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          You can even use non-isomorphic finite groups of the same order. The smallest example is $Aut(C_4times C_2)cong Aut(D_8)cong D_8$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            You can even use non-isomorphic finite groups of the same order. The smallest example is $Aut(C_4times C_2)cong Aut(D_8)cong D_8$.






                            share|cite|improve this answer









                            $endgroup$



                            You can even use non-isomorphic finite groups of the same order. The smallest example is $Aut(C_4times C_2)cong Aut(D_8)cong D_8$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 18 '18 at 1:19









                            C MonsourC Monsour

                            6,3041326




                            6,3041326






























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