Find a subgroup of $D_{42}$ that is isomorphic to $S_3$.












0












$begingroup$


Note that $D_{42}$ is generated by $rho, r$ where $ord(rho) = 21$, $ord(r)= 2$. To locate a copy of $S_3$, there must be a subgroup of order $3$, which can be $e, rho^{7}, rho^{14}$. What about order $2$ elements? Inevitably, I have to introduce some reflections which will enlarge the group.






abstract-algebra







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$endgroup$












  • $begingroup$
    Doesn't $S_3$ have 6 elements ?
    $endgroup$
    – Digitalis
    Dec 14 '18 at 17:49










  • $begingroup$
    Yes, and?@Digitalis
    $endgroup$
    – koch
    Dec 14 '18 at 17:55
















0












$begingroup$


Note that $D_{42}$ is generated by $rho, r$ where $ord(rho) = 21$, $ord(r)= 2$. To locate a copy of $S_3$, there must be a subgroup of order $3$, which can be $e, rho^{7}, rho^{14}$. What about order $2$ elements? Inevitably, I have to introduce some reflections which will enlarge the group.






abstract-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    Doesn't $S_3$ have 6 elements ?
    $endgroup$
    – Digitalis
    Dec 14 '18 at 17:49










  • $begingroup$
    Yes, and?@Digitalis
    $endgroup$
    – koch
    Dec 14 '18 at 17:55














0












0








0





$begingroup$


Note that $D_{42}$ is generated by $rho, r$ where $ord(rho) = 21$, $ord(r)= 2$. To locate a copy of $S_3$, there must be a subgroup of order $3$, which can be $e, rho^{7}, rho^{14}$. What about order $2$ elements? Inevitably, I have to introduce some reflections which will enlarge the group.






abstract-algebra







share|cite|improve this question









$endgroup$




Note that $D_{42}$ is generated by $rho, r$ where $ord(rho) = 21$, $ord(r)= 2$. To locate a copy of $S_3$, there must be a subgroup of order $3$, which can be $e, rho^{7}, rho^{14}$. What about order $2$ elements? Inevitably, I have to introduce some reflections which will enlarge the group.






abstract-algebra




abstract-algebra






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Dec 14 '18 at 17:47









kochkoch

18418




18418












  • $begingroup$
    Doesn't $S_3$ have 6 elements ?
    $endgroup$
    – Digitalis
    Dec 14 '18 at 17:49










  • $begingroup$
    Yes, and?@Digitalis
    $endgroup$
    – koch
    Dec 14 '18 at 17:55


















  • $begingroup$
    Doesn't $S_3$ have 6 elements ?
    $endgroup$
    – Digitalis
    Dec 14 '18 at 17:49










  • $begingroup$
    Yes, and?@Digitalis
    $endgroup$
    – koch
    Dec 14 '18 at 17:55
















$begingroup$
Doesn't $S_3$ have 6 elements ?
$endgroup$
– Digitalis
Dec 14 '18 at 17:49




$begingroup$
Doesn't $S_3$ have 6 elements ?
$endgroup$
– Digitalis
Dec 14 '18 at 17:49












$begingroup$
Yes, and?@Digitalis
$endgroup$
– koch
Dec 14 '18 at 17:55




$begingroup$
Yes, and?@Digitalis
$endgroup$
– koch
Dec 14 '18 at 17:55










2 Answers
2






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1












$begingroup$

$S_3$ has $3!=6$ elements and you need three more.
$1,rho^7,rho^{14},r,rrho^7,rrho^{14}$ does the trick nicely.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    We have a complete classification of all possible subgroups of $D_n$:



    Theorem: Every subgroup of $D_n=langle r,s rangle$ is is either cyclic or dihedral, and a complete listing of the subgroups is as follows:



    (1) $langle r^drangle$, where $dmid n$, with index $2d$,



    (2) $langle r^d, r^is rangle$, where $dmid n$, $0le ile d-1$, with index $d$.



    Every subgroup of $D_n$ occurs exactly once in this listing.
    For a proof see Theorem 3.1 of Keith Conrad's notes.



    For the appropriate $d$ we obtain the subgroup $D_3=S_3$ with $6$ elements in $(2)$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

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      1












      $begingroup$

      $S_3$ has $3!=6$ elements and you need three more.
      $1,rho^7,rho^{14},r,rrho^7,rrho^{14}$ does the trick nicely.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $S_3$ has $3!=6$ elements and you need three more.
        $1,rho^7,rho^{14},r,rrho^7,rrho^{14}$ does the trick nicely.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $S_3$ has $3!=6$ elements and you need three more.
          $1,rho^7,rho^{14},r,rrho^7,rrho^{14}$ does the trick nicely.






          share|cite|improve this answer









          $endgroup$



          $S_3$ has $3!=6$ elements and you need three more.
          $1,rho^7,rho^{14},r,rrho^7,rrho^{14}$ does the trick nicely.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 17:55









          Ross MillikanRoss Millikan

          297k23198371




          297k23198371























              0












              $begingroup$

              We have a complete classification of all possible subgroups of $D_n$:



              Theorem: Every subgroup of $D_n=langle r,s rangle$ is is either cyclic or dihedral, and a complete listing of the subgroups is as follows:



              (1) $langle r^drangle$, where $dmid n$, with index $2d$,



              (2) $langle r^d, r^is rangle$, where $dmid n$, $0le ile d-1$, with index $d$.



              Every subgroup of $D_n$ occurs exactly once in this listing.
              For a proof see Theorem 3.1 of Keith Conrad's notes.



              For the appropriate $d$ we obtain the subgroup $D_3=S_3$ with $6$ elements in $(2)$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                We have a complete classification of all possible subgroups of $D_n$:



                Theorem: Every subgroup of $D_n=langle r,s rangle$ is is either cyclic or dihedral, and a complete listing of the subgroups is as follows:



                (1) $langle r^drangle$, where $dmid n$, with index $2d$,



                (2) $langle r^d, r^is rangle$, where $dmid n$, $0le ile d-1$, with index $d$.



                Every subgroup of $D_n$ occurs exactly once in this listing.
                For a proof see Theorem 3.1 of Keith Conrad's notes.



                For the appropriate $d$ we obtain the subgroup $D_3=S_3$ with $6$ elements in $(2)$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  We have a complete classification of all possible subgroups of $D_n$:



                  Theorem: Every subgroup of $D_n=langle r,s rangle$ is is either cyclic or dihedral, and a complete listing of the subgroups is as follows:



                  (1) $langle r^drangle$, where $dmid n$, with index $2d$,



                  (2) $langle r^d, r^is rangle$, where $dmid n$, $0le ile d-1$, with index $d$.



                  Every subgroup of $D_n$ occurs exactly once in this listing.
                  For a proof see Theorem 3.1 of Keith Conrad's notes.



                  For the appropriate $d$ we obtain the subgroup $D_3=S_3$ with $6$ elements in $(2)$.






                  share|cite|improve this answer









                  $endgroup$



                  We have a complete classification of all possible subgroups of $D_n$:



                  Theorem: Every subgroup of $D_n=langle r,s rangle$ is is either cyclic or dihedral, and a complete listing of the subgroups is as follows:



                  (1) $langle r^drangle$, where $dmid n$, with index $2d$,



                  (2) $langle r^d, r^is rangle$, where $dmid n$, $0le ile d-1$, with index $d$.



                  Every subgroup of $D_n$ occurs exactly once in this listing.
                  For a proof see Theorem 3.1 of Keith Conrad's notes.



                  For the appropriate $d$ we obtain the subgroup $D_3=S_3$ with $6$ elements in $(2)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 14 '18 at 17:56









                  Dietrich BurdeDietrich Burde

                  79.3k647103




                  79.3k647103






























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