Is $A_4 times mathbb{Z}_3$ isomorphic to $S_3 times S_3$?












2












$begingroup$



Ιs $A_4 times mathbb{Z}_3$ isomorphic to $S_3 times S_3$?




I am trying to find an element of $S_3 times S_3$ which has an order, let's say $a$, and $A_4 times mathbb{Z}_3$ has no element of such order.






abstract-algebra group-theory finite-groups symmetric-groups







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$endgroup$












  • $begingroup$
    You could count the elements of order $3$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 7:47










  • $begingroup$
    Both groups have elements of orders $1$, $2$, $3$ and $6$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 7:47










  • $begingroup$
    Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
    $endgroup$
    – Hagen von Eitzen
    Jan 18 at 7:50
















2












$begingroup$



Ιs $A_4 times mathbb{Z}_3$ isomorphic to $S_3 times S_3$?




I am trying to find an element of $S_3 times S_3$ which has an order, let's say $a$, and $A_4 times mathbb{Z}_3$ has no element of such order.






abstract-algebra group-theory finite-groups symmetric-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    You could count the elements of order $3$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 7:47










  • $begingroup$
    Both groups have elements of orders $1$, $2$, $3$ and $6$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 7:47










  • $begingroup$
    Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
    $endgroup$
    – Hagen von Eitzen
    Jan 18 at 7:50














2












2








2


1



$begingroup$



Ιs $A_4 times mathbb{Z}_3$ isomorphic to $S_3 times S_3$?




I am trying to find an element of $S_3 times S_3$ which has an order, let's say $a$, and $A_4 times mathbb{Z}_3$ has no element of such order.






abstract-algebra group-theory finite-groups symmetric-groups







share|cite|improve this question











$endgroup$





Ιs $A_4 times mathbb{Z}_3$ isomorphic to $S_3 times S_3$?




I am trying to find an element of $S_3 times S_3$ which has an order, let's say $a$, and $A_4 times mathbb{Z}_3$ has no element of such order.






abstract-algebra group-theory finite-groups symmetric-groups




abstract-algebra group-theory finite-groups symmetric-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 20:45









user26857

39.6k124284




39.6k124284










asked Jan 18 at 7:45









kotsos sgouraskotsos sgouras

157




157












  • $begingroup$
    You could count the elements of order $3$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 7:47










  • $begingroup$
    Both groups have elements of orders $1$, $2$, $3$ and $6$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 7:47










  • $begingroup$
    Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
    $endgroup$
    – Hagen von Eitzen
    Jan 18 at 7:50


















  • $begingroup$
    You could count the elements of order $3$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 7:47










  • $begingroup$
    Both groups have elements of orders $1$, $2$, $3$ and $6$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 7:47










  • $begingroup$
    Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
    $endgroup$
    – Hagen von Eitzen
    Jan 18 at 7:50
















$begingroup$
You could count the elements of order $3$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47




$begingroup$
You could count the elements of order $3$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47












$begingroup$
Both groups have elements of orders $1$, $2$, $3$ and $6$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47




$begingroup$
Both groups have elements of orders $1$, $2$, $3$ and $6$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47












$begingroup$
Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
$endgroup$
– Hagen von Eitzen
Jan 18 at 7:50




$begingroup$
Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
$endgroup$
– Hagen von Eitzen
Jan 18 at 7:50










2 Answers
2






active

oldest

votes


















4












$begingroup$

In $S_3times S_3$ there are





  1. $1$ element of order $1$


  2. $color{red}{15}$ elements of order $2$, $(tau_1,e),(e,tau_2),(tau_1,tau_2)$


  3. $8$ elements of order $3$


  4. $12$ elements of order $6$


while in $A_4timesmathbb{Z}_3$ there are





  1. $1$ element of order $1$


  2. $color{red}{3}$ elements of order $2$, $(tau_1 tau_2,e)$


  3. $26$ elements of order $3$


  4. $6$ elements of order $6$


so the easiest way to state $S_3times S_3 notsimeq A_4timesmathbb{Z}_3$ is probably to compare the number of involutions (elements with order $2$) in both groups. Or wonder about the largest abelian subgroups, which in the former case are isomorphic to $mathbb{Z}_3timesmathbb{Z}_3$ and in the latter case are isomorphic to $mathbb{Z}_2timesmathbb{Z}_2timesmathbb{Z}_3$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
    $endgroup$
    – user26857
    Jan 22 at 9:29








  • 1




    $begingroup$
    @user26857: you're right, answer fixed.
    $endgroup$
    – Jack D'Aurizio
    Jan 22 at 16:23



















3












$begingroup$

You could use the following useful facts.




  • Given $A$ and $B$ groups, $Z(A times B) = Z(A) times Z(B)$.


  • $Z(S_n)=lbrace id rbrace$ for all $n geq 3$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    In $S_3times S_3$ there are





    1. $1$ element of order $1$


    2. $color{red}{15}$ elements of order $2$, $(tau_1,e),(e,tau_2),(tau_1,tau_2)$


    3. $8$ elements of order $3$


    4. $12$ elements of order $6$


    while in $A_4timesmathbb{Z}_3$ there are





    1. $1$ element of order $1$


    2. $color{red}{3}$ elements of order $2$, $(tau_1 tau_2,e)$


    3. $26$ elements of order $3$


    4. $6$ elements of order $6$


    so the easiest way to state $S_3times S_3 notsimeq A_4timesmathbb{Z}_3$ is probably to compare the number of involutions (elements with order $2$) in both groups. Or wonder about the largest abelian subgroups, which in the former case are isomorphic to $mathbb{Z}_3timesmathbb{Z}_3$ and in the latter case are isomorphic to $mathbb{Z}_2timesmathbb{Z}_2timesmathbb{Z}_3$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
      $endgroup$
      – user26857
      Jan 22 at 9:29








    • 1




      $begingroup$
      @user26857: you're right, answer fixed.
      $endgroup$
      – Jack D'Aurizio
      Jan 22 at 16:23
















    4












    $begingroup$

    In $S_3times S_3$ there are





    1. $1$ element of order $1$


    2. $color{red}{15}$ elements of order $2$, $(tau_1,e),(e,tau_2),(tau_1,tau_2)$


    3. $8$ elements of order $3$


    4. $12$ elements of order $6$


    while in $A_4timesmathbb{Z}_3$ there are





    1. $1$ element of order $1$


    2. $color{red}{3}$ elements of order $2$, $(tau_1 tau_2,e)$


    3. $26$ elements of order $3$


    4. $6$ elements of order $6$


    so the easiest way to state $S_3times S_3 notsimeq A_4timesmathbb{Z}_3$ is probably to compare the number of involutions (elements with order $2$) in both groups. Or wonder about the largest abelian subgroups, which in the former case are isomorphic to $mathbb{Z}_3timesmathbb{Z}_3$ and in the latter case are isomorphic to $mathbb{Z}_2timesmathbb{Z}_2timesmathbb{Z}_3$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
      $endgroup$
      – user26857
      Jan 22 at 9:29








    • 1




      $begingroup$
      @user26857: you're right, answer fixed.
      $endgroup$
      – Jack D'Aurizio
      Jan 22 at 16:23














    4












    4








    4





    $begingroup$

    In $S_3times S_3$ there are





    1. $1$ element of order $1$


    2. $color{red}{15}$ elements of order $2$, $(tau_1,e),(e,tau_2),(tau_1,tau_2)$


    3. $8$ elements of order $3$


    4. $12$ elements of order $6$


    while in $A_4timesmathbb{Z}_3$ there are





    1. $1$ element of order $1$


    2. $color{red}{3}$ elements of order $2$, $(tau_1 tau_2,e)$


    3. $26$ elements of order $3$


    4. $6$ elements of order $6$


    so the easiest way to state $S_3times S_3 notsimeq A_4timesmathbb{Z}_3$ is probably to compare the number of involutions (elements with order $2$) in both groups. Or wonder about the largest abelian subgroups, which in the former case are isomorphic to $mathbb{Z}_3timesmathbb{Z}_3$ and in the latter case are isomorphic to $mathbb{Z}_2timesmathbb{Z}_2timesmathbb{Z}_3$.






    share|cite|improve this answer











    $endgroup$



    In $S_3times S_3$ there are





    1. $1$ element of order $1$


    2. $color{red}{15}$ elements of order $2$, $(tau_1,e),(e,tau_2),(tau_1,tau_2)$


    3. $8$ elements of order $3$


    4. $12$ elements of order $6$


    while in $A_4timesmathbb{Z}_3$ there are





    1. $1$ element of order $1$


    2. $color{red}{3}$ elements of order $2$, $(tau_1 tau_2,e)$


    3. $26$ elements of order $3$


    4. $6$ elements of order $6$


    so the easiest way to state $S_3times S_3 notsimeq A_4timesmathbb{Z}_3$ is probably to compare the number of involutions (elements with order $2$) in both groups. Or wonder about the largest abelian subgroups, which in the former case are isomorphic to $mathbb{Z}_3timesmathbb{Z}_3$ and in the latter case are isomorphic to $mathbb{Z}_2timesmathbb{Z}_2timesmathbb{Z}_3$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 22 at 16:23

























    answered Jan 18 at 9:02









    Jack D'AurizioJack D'Aurizio

    292k33284674




    292k33284674












    • $begingroup$
      $A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
      $endgroup$
      – user26857
      Jan 22 at 9:29








    • 1




      $begingroup$
      @user26857: you're right, answer fixed.
      $endgroup$
      – Jack D'Aurizio
      Jan 22 at 16:23


















    • $begingroup$
      $A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
      $endgroup$
      – user26857
      Jan 22 at 9:29








    • 1




      $begingroup$
      @user26857: you're right, answer fixed.
      $endgroup$
      – Jack D'Aurizio
      Jan 22 at 16:23
















    $begingroup$
    $A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
    $endgroup$
    – user26857
    Jan 22 at 9:29






    $begingroup$
    $A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
    $endgroup$
    – user26857
    Jan 22 at 9:29






    1




    1




    $begingroup$
    @user26857: you're right, answer fixed.
    $endgroup$
    – Jack D'Aurizio
    Jan 22 at 16:23




    $begingroup$
    @user26857: you're right, answer fixed.
    $endgroup$
    – Jack D'Aurizio
    Jan 22 at 16:23











    3












    $begingroup$

    You could use the following useful facts.




    • Given $A$ and $B$ groups, $Z(A times B) = Z(A) times Z(B)$.


    • $Z(S_n)=lbrace id rbrace$ for all $n geq 3$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      You could use the following useful facts.




      • Given $A$ and $B$ groups, $Z(A times B) = Z(A) times Z(B)$.


      • $Z(S_n)=lbrace id rbrace$ for all $n geq 3$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        You could use the following useful facts.




        • Given $A$ and $B$ groups, $Z(A times B) = Z(A) times Z(B)$.


        • $Z(S_n)=lbrace id rbrace$ for all $n geq 3$.






        share|cite|improve this answer









        $endgroup$



        You could use the following useful facts.




        • Given $A$ and $B$ groups, $Z(A times B) = Z(A) times Z(B)$.


        • $Z(S_n)=lbrace id rbrace$ for all $n geq 3$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 14:30









        Pietro GheriPietro Gheri

        1662




        1662






























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