How do generators of a group work?












0












$begingroup$


$G$ is a group, $H$ is a subgroup of $G$, and $[G:H]$ stands for the index of $H$ in $G$ in the following example:



Let $G=S_3$, $H=left<(1,2)right>$. Then $[G:H]=3$.



I know the definition of group generators:
A set of generators $(g_1,...,g_n)$ is a set of group elements such that possibly repeated application of the generators on themselves and each other is capable of producing all the elements in the group.



What does the individual elements of $H=left<(1,2)right>$ look like? Any help would be greatly aprciated.



P.S. I know how to find the index when the groups don’t involve a group generator, the thing I need help with is understanding the group generator.






abstract-algebra finite-groups normal-subgroups







share|cite|improve this question











$endgroup$












  • $begingroup$
    What's the actual question here?
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:18










  • $begingroup$
    My question is more or less this: What does the individual elements of $H$ look like?
    $endgroup$
    – AMN52
    Dec 10 '18 at 4:21










  • $begingroup$
    $H$ must have the identity element. It must also have the element $(1 2)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:22










  • $begingroup$
    $(1,2)$ has order 2 so it's the only non identity element in $H$.
    $endgroup$
    – Justin Stevenson
    Dec 10 '18 at 4:22










  • $begingroup$
    Why does $H$ only contain $(1,2)$ and the identity element?
    $endgroup$
    – AMN52
    Dec 10 '18 at 4:25
















0












$begingroup$


$G$ is a group, $H$ is a subgroup of $G$, and $[G:H]$ stands for the index of $H$ in $G$ in the following example:



Let $G=S_3$, $H=left<(1,2)right>$. Then $[G:H]=3$.



I know the definition of group generators:
A set of generators $(g_1,...,g_n)$ is a set of group elements such that possibly repeated application of the generators on themselves and each other is capable of producing all the elements in the group.



What does the individual elements of $H=left<(1,2)right>$ look like? Any help would be greatly aprciated.



P.S. I know how to find the index when the groups don’t involve a group generator, the thing I need help with is understanding the group generator.






abstract-algebra finite-groups normal-subgroups







share|cite|improve this question











$endgroup$












  • $begingroup$
    What's the actual question here?
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:18










  • $begingroup$
    My question is more or less this: What does the individual elements of $H$ look like?
    $endgroup$
    – AMN52
    Dec 10 '18 at 4:21










  • $begingroup$
    $H$ must have the identity element. It must also have the element $(1 2)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:22










  • $begingroup$
    $(1,2)$ has order 2 so it's the only non identity element in $H$.
    $endgroup$
    – Justin Stevenson
    Dec 10 '18 at 4:22










  • $begingroup$
    Why does $H$ only contain $(1,2)$ and the identity element?
    $endgroup$
    – AMN52
    Dec 10 '18 at 4:25














0












0








0





$begingroup$


$G$ is a group, $H$ is a subgroup of $G$, and $[G:H]$ stands for the index of $H$ in $G$ in the following example:



Let $G=S_3$, $H=left<(1,2)right>$. Then $[G:H]=3$.



I know the definition of group generators:
A set of generators $(g_1,...,g_n)$ is a set of group elements such that possibly repeated application of the generators on themselves and each other is capable of producing all the elements in the group.



What does the individual elements of $H=left<(1,2)right>$ look like? Any help would be greatly aprciated.



P.S. I know how to find the index when the groups don’t involve a group generator, the thing I need help with is understanding the group generator.






abstract-algebra finite-groups normal-subgroups







share|cite|improve this question











$endgroup$




$G$ is a group, $H$ is a subgroup of $G$, and $[G:H]$ stands for the index of $H$ in $G$ in the following example:



Let $G=S_3$, $H=left<(1,2)right>$. Then $[G:H]=3$.



I know the definition of group generators:
A set of generators $(g_1,...,g_n)$ is a set of group elements such that possibly repeated application of the generators on themselves and each other is capable of producing all the elements in the group.



What does the individual elements of $H=left<(1,2)right>$ look like? Any help would be greatly aprciated.



P.S. I know how to find the index when the groups don’t involve a group generator, the thing I need help with is understanding the group generator.






abstract-algebra finite-groups normal-subgroups




abstract-algebra finite-groups normal-subgroups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 5:10









Shaun

8,951113682




8,951113682










asked Dec 10 '18 at 4:16









AMN52AMN52

326




326












  • $begingroup$
    What's the actual question here?
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:18










  • $begingroup$
    My question is more or less this: What does the individual elements of $H$ look like?
    $endgroup$
    – AMN52
    Dec 10 '18 at 4:21










  • $begingroup$
    $H$ must have the identity element. It must also have the element $(1 2)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:22










  • $begingroup$
    $(1,2)$ has order 2 so it's the only non identity element in $H$.
    $endgroup$
    – Justin Stevenson
    Dec 10 '18 at 4:22










  • $begingroup$
    Why does $H$ only contain $(1,2)$ and the identity element?
    $endgroup$
    – AMN52
    Dec 10 '18 at 4:25


















  • $begingroup$
    What's the actual question here?
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:18










  • $begingroup$
    My question is more or less this: What does the individual elements of $H$ look like?
    $endgroup$
    – AMN52
    Dec 10 '18 at 4:21










  • $begingroup$
    $H$ must have the identity element. It must also have the element $(1 2)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:22










  • $begingroup$
    $(1,2)$ has order 2 so it's the only non identity element in $H$.
    $endgroup$
    – Justin Stevenson
    Dec 10 '18 at 4:22










  • $begingroup$
    Why does $H$ only contain $(1,2)$ and the identity element?
    $endgroup$
    – AMN52
    Dec 10 '18 at 4:25
















$begingroup$
What's the actual question here?
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 4:18




$begingroup$
What's the actual question here?
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 4:18












$begingroup$
My question is more or less this: What does the individual elements of $H$ look like?
$endgroup$
– AMN52
Dec 10 '18 at 4:21




$begingroup$
My question is more or less this: What does the individual elements of $H$ look like?
$endgroup$
– AMN52
Dec 10 '18 at 4:21












$begingroup$
$H$ must have the identity element. It must also have the element $(1 2)$.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 4:22




$begingroup$
$H$ must have the identity element. It must also have the element $(1 2)$.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 4:22












$begingroup$
$(1,2)$ has order 2 so it's the only non identity element in $H$.
$endgroup$
– Justin Stevenson
Dec 10 '18 at 4:22




$begingroup$
$(1,2)$ has order 2 so it's the only non identity element in $H$.
$endgroup$
– Justin Stevenson
Dec 10 '18 at 4:22












$begingroup$
Why does $H$ only contain $(1,2)$ and the identity element?
$endgroup$
– AMN52
Dec 10 '18 at 4:25




$begingroup$
Why does $H$ only contain $(1,2)$ and the identity element?
$endgroup$
– AMN52
Dec 10 '18 at 4:25










2 Answers
2






active

oldest

votes


















2












$begingroup$

$(12)$ is a transposition. $(12)^{-1}=(12)$, that is, it's its own inverse. All that can be gotten by taking powers of $(12)$ is $(12)$ and $e$, the identity. (Note: In general, $langle arangle ={a^n:ninmathbb Z}$). Thus $langle (12)rangle ={(12),e}$, a two element group.



Since $mid S_3mid=6$, we get $[S_3:langle (12)rangle] =3$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Because $H$ is the set of all elements of the form $(1,2)^n$ and $(1,2)^2=e$.



    By Justin Stevenson.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      technically this is plagiarism...lol :)
      $endgroup$
      – zoidberg
      Dec 10 '18 at 4:36










    • $begingroup$
      Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
      $endgroup$
      – AMN52
      Dec 10 '18 at 4:50













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    2 Answers
    2






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    2 Answers
    2






    active

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    active

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    active

    oldest

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    2












    $begingroup$

    $(12)$ is a transposition. $(12)^{-1}=(12)$, that is, it's its own inverse. All that can be gotten by taking powers of $(12)$ is $(12)$ and $e$, the identity. (Note: In general, $langle arangle ={a^n:ninmathbb Z}$). Thus $langle (12)rangle ={(12),e}$, a two element group.



    Since $mid S_3mid=6$, we get $[S_3:langle (12)rangle] =3$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      $(12)$ is a transposition. $(12)^{-1}=(12)$, that is, it's its own inverse. All that can be gotten by taking powers of $(12)$ is $(12)$ and $e$, the identity. (Note: In general, $langle arangle ={a^n:ninmathbb Z}$). Thus $langle (12)rangle ={(12),e}$, a two element group.



      Since $mid S_3mid=6$, we get $[S_3:langle (12)rangle] =3$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        $(12)$ is a transposition. $(12)^{-1}=(12)$, that is, it's its own inverse. All that can be gotten by taking powers of $(12)$ is $(12)$ and $e$, the identity. (Note: In general, $langle arangle ={a^n:ninmathbb Z}$). Thus $langle (12)rangle ={(12),e}$, a two element group.



        Since $mid S_3mid=6$, we get $[S_3:langle (12)rangle] =3$.






        share|cite|improve this answer











        $endgroup$



        $(12)$ is a transposition. $(12)^{-1}=(12)$, that is, it's its own inverse. All that can be gotten by taking powers of $(12)$ is $(12)$ and $e$, the identity. (Note: In general, $langle arangle ={a^n:ninmathbb Z}$). Thus $langle (12)rangle ={(12),e}$, a two element group.



        Since $mid S_3mid=6$, we get $[S_3:langle (12)rangle] =3$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 10 '18 at 4:41

























        answered Dec 10 '18 at 4:33









        Chris CusterChris Custer

        11.9k3825




        11.9k3825























            1












            $begingroup$

            Because $H$ is the set of all elements of the form $(1,2)^n$ and $(1,2)^2=e$.



            By Justin Stevenson.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              technically this is plagiarism...lol :)
              $endgroup$
              – zoidberg
              Dec 10 '18 at 4:36










            • $begingroup$
              Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
              $endgroup$
              – AMN52
              Dec 10 '18 at 4:50


















            1












            $begingroup$

            Because $H$ is the set of all elements of the form $(1,2)^n$ and $(1,2)^2=e$.



            By Justin Stevenson.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              technically this is plagiarism...lol :)
              $endgroup$
              – zoidberg
              Dec 10 '18 at 4:36










            • $begingroup$
              Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
              $endgroup$
              – AMN52
              Dec 10 '18 at 4:50
















            1












            1








            1





            $begingroup$

            Because $H$ is the set of all elements of the form $(1,2)^n$ and $(1,2)^2=e$.



            By Justin Stevenson.






            share|cite|improve this answer











            $endgroup$



            Because $H$ is the set of all elements of the form $(1,2)^n$ and $(1,2)^2=e$.



            By Justin Stevenson.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 10 '18 at 4:51

























            answered Dec 10 '18 at 4:33









            AMN52AMN52

            326




            326












            • $begingroup$
              technically this is plagiarism...lol :)
              $endgroup$
              – zoidberg
              Dec 10 '18 at 4:36










            • $begingroup$
              Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
              $endgroup$
              – AMN52
              Dec 10 '18 at 4:50




















            • $begingroup$
              technically this is plagiarism...lol :)
              $endgroup$
              – zoidberg
              Dec 10 '18 at 4:36










            • $begingroup$
              Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
              $endgroup$
              – AMN52
              Dec 10 '18 at 4:50


















            $begingroup$
            technically this is plagiarism...lol :)
            $endgroup$
            – zoidberg
            Dec 10 '18 at 4:36




            $begingroup$
            technically this is plagiarism...lol :)
            $endgroup$
            – zoidberg
            Dec 10 '18 at 4:36












            $begingroup$
            Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
            $endgroup$
            – AMN52
            Dec 10 '18 at 4:50






            $begingroup$
            Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
            $endgroup$
            – AMN52
            Dec 10 '18 at 4:50




















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