Existence of subgroup of order power of prime in a finite abelian group?












5












$begingroup$


Say I have a finite abelian group $G$ such that $left | G right |=p_1^{n_1}...p_m^{n_m}$ where the $p_i$'s are distinct. Can I say that $G$ must have a subgroup of order $p_1^{n_1}$? I'm thinking that I can use the structure theorem to write $G$ as $mathbb{Z}/p_1^{n_{1_{1}}}times ...times mathbb{Z}/p_1^{n_{1_{s_{1}}}}times...timesmathbb{Z}/p_m^{n_{m_{1}}}times ...times mathbb{Z}/p_m^{n_{m_{s_{m}}}}$ then just take $mathbb{Z}/p_1^{n_{1_{1}}}times ...times mathbb{Z}/p_1^{n_{1_{s_{1}}}}timesleft { 1 right }...timesleft { 1 right }$ where $sum_i n_{1_{i}}=n_1$.



Is my reasoning correct? I would like to avoid using Sylow theorems.



Thanks in advance.






group-theory finite-groups abelian-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, you're on the right track (though those $;n_{1_r};$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
    $endgroup$
    – DonAntonio
    Dec 20 '18 at 21:49












  • $begingroup$
    @DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
    $endgroup$
    – John11
    Dec 20 '18 at 21:56
















5












$begingroup$


Say I have a finite abelian group $G$ such that $left | G right |=p_1^{n_1}...p_m^{n_m}$ where the $p_i$'s are distinct. Can I say that $G$ must have a subgroup of order $p_1^{n_1}$? I'm thinking that I can use the structure theorem to write $G$ as $mathbb{Z}/p_1^{n_{1_{1}}}times ...times mathbb{Z}/p_1^{n_{1_{s_{1}}}}times...timesmathbb{Z}/p_m^{n_{m_{1}}}times ...times mathbb{Z}/p_m^{n_{m_{s_{m}}}}$ then just take $mathbb{Z}/p_1^{n_{1_{1}}}times ...times mathbb{Z}/p_1^{n_{1_{s_{1}}}}timesleft { 1 right }...timesleft { 1 right }$ where $sum_i n_{1_{i}}=n_1$.



Is my reasoning correct? I would like to avoid using Sylow theorems.



Thanks in advance.






group-theory finite-groups abelian-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, you're on the right track (though those $;n_{1_r};$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
    $endgroup$
    – DonAntonio
    Dec 20 '18 at 21:49












  • $begingroup$
    @DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
    $endgroup$
    – John11
    Dec 20 '18 at 21:56














5












5








5


1



$begingroup$


Say I have a finite abelian group $G$ such that $left | G right |=p_1^{n_1}...p_m^{n_m}$ where the $p_i$'s are distinct. Can I say that $G$ must have a subgroup of order $p_1^{n_1}$? I'm thinking that I can use the structure theorem to write $G$ as $mathbb{Z}/p_1^{n_{1_{1}}}times ...times mathbb{Z}/p_1^{n_{1_{s_{1}}}}times...timesmathbb{Z}/p_m^{n_{m_{1}}}times ...times mathbb{Z}/p_m^{n_{m_{s_{m}}}}$ then just take $mathbb{Z}/p_1^{n_{1_{1}}}times ...times mathbb{Z}/p_1^{n_{1_{s_{1}}}}timesleft { 1 right }...timesleft { 1 right }$ where $sum_i n_{1_{i}}=n_1$.



Is my reasoning correct? I would like to avoid using Sylow theorems.



Thanks in advance.






group-theory finite-groups abelian-groups







share|cite|improve this question











$endgroup$




Say I have a finite abelian group $G$ such that $left | G right |=p_1^{n_1}...p_m^{n_m}$ where the $p_i$'s are distinct. Can I say that $G$ must have a subgroup of order $p_1^{n_1}$? I'm thinking that I can use the structure theorem to write $G$ as $mathbb{Z}/p_1^{n_{1_{1}}}times ...times mathbb{Z}/p_1^{n_{1_{s_{1}}}}times...timesmathbb{Z}/p_m^{n_{m_{1}}}times ...times mathbb{Z}/p_m^{n_{m_{s_{m}}}}$ then just take $mathbb{Z}/p_1^{n_{1_{1}}}times ...times mathbb{Z}/p_1^{n_{1_{s_{1}}}}timesleft { 1 right }...timesleft { 1 right }$ where $sum_i n_{1_{i}}=n_1$.



Is my reasoning correct? I would like to avoid using Sylow theorems.



Thanks in advance.






group-theory finite-groups abelian-groups




group-theory finite-groups abelian-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 22:05







John11

















asked Dec 20 '18 at 21:38









John11John11

1,0321821




1,0321821








  • 1




    $begingroup$
    Yes, you're on the right track (though those $;n_{1_r};$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
    $endgroup$
    – DonAntonio
    Dec 20 '18 at 21:49












  • $begingroup$
    @DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
    $endgroup$
    – John11
    Dec 20 '18 at 21:56














  • 1




    $begingroup$
    Yes, you're on the right track (though those $;n_{1_r};$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
    $endgroup$
    – DonAntonio
    Dec 20 '18 at 21:49












  • $begingroup$
    @DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
    $endgroup$
    – John11
    Dec 20 '18 at 21:56








1




1




$begingroup$
Yes, you're on the right track (though those $;n_{1_r};$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
$endgroup$
– DonAntonio
Dec 20 '18 at 21:49






$begingroup$
Yes, you're on the right track (though those $;n_{1_r};$ are weird...) . And fixing that a little, you can easily prove that a finite abelian group has a subgroup of any order dividing hte group's order.
$endgroup$
– DonAntonio
Dec 20 '18 at 21:49














$begingroup$
@DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
$endgroup$
– John11
Dec 20 '18 at 21:56




$begingroup$
@DonAntonio Yes that's exactly what I'm considering actually. What would be a better way to indicate the decomposition of each maximal power $n_i$ in the prime decomposition of $G$ as a sum?
$endgroup$
– John11
Dec 20 '18 at 21:56










2 Answers
2






active

oldest

votes


















3












$begingroup$

Some highlights:



First, you can take $;K:=Bbb Z/p_1^{n_1}Bbb Ztimes{1}timesldotstimes{1|];$ to get a subgroup of order $;p_1^{n_1};$ , and now you can generalize this to each prime $;p_1,..,p_m;$ and their powers.



Next, use the basic lemma that says that a finite $;p,-$ group of order $;p^n;$ has a (normal if you will, even when we're not in the abelian case!) subgroup of order $;p^k;$ , for any $;0le kle n;$ .



Finally, just use the direct product decomposition to get a subgroup of any order dividing the group's.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Here is a roadmap.



    Let $p$ be a prime dividing the order of $G$.




    • Let $P = { g in G : ord(g) text{ is a power of $p$} }$.


    • Then $P$ is a subgroup of $G$ (because $G$ is abelian).


    • The order of $P$ is a power of $p$ (by Cauchy's theorem).


    • The order of $P$ is the largest power of $p$ that divides the order of $G$ (by Cauchy's theorem applied to $G/P$).







    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Some highlights:



      First, you can take $;K:=Bbb Z/p_1^{n_1}Bbb Ztimes{1}timesldotstimes{1|];$ to get a subgroup of order $;p_1^{n_1};$ , and now you can generalize this to each prime $;p_1,..,p_m;$ and their powers.



      Next, use the basic lemma that says that a finite $;p,-$ group of order $;p^n;$ has a (normal if you will, even when we're not in the abelian case!) subgroup of order $;p^k;$ , for any $;0le kle n;$ .



      Finally, just use the direct product decomposition to get a subgroup of any order dividing the group's.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Some highlights:



        First, you can take $;K:=Bbb Z/p_1^{n_1}Bbb Ztimes{1}timesldotstimes{1|];$ to get a subgroup of order $;p_1^{n_1};$ , and now you can generalize this to each prime $;p_1,..,p_m;$ and their powers.



        Next, use the basic lemma that says that a finite $;p,-$ group of order $;p^n;$ has a (normal if you will, even when we're not in the abelian case!) subgroup of order $;p^k;$ , for any $;0le kle n;$ .



        Finally, just use the direct product decomposition to get a subgroup of any order dividing the group's.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Some highlights:



          First, you can take $;K:=Bbb Z/p_1^{n_1}Bbb Ztimes{1}timesldotstimes{1|];$ to get a subgroup of order $;p_1^{n_1};$ , and now you can generalize this to each prime $;p_1,..,p_m;$ and their powers.



          Next, use the basic lemma that says that a finite $;p,-$ group of order $;p^n;$ has a (normal if you will, even when we're not in the abelian case!) subgroup of order $;p^k;$ , for any $;0le kle n;$ .



          Finally, just use the direct product decomposition to get a subgroup of any order dividing the group's.






          share|cite|improve this answer









          $endgroup$



          Some highlights:



          First, you can take $;K:=Bbb Z/p_1^{n_1}Bbb Ztimes{1}timesldotstimes{1|];$ to get a subgroup of order $;p_1^{n_1};$ , and now you can generalize this to each prime $;p_1,..,p_m;$ and their powers.



          Next, use the basic lemma that says that a finite $;p,-$ group of order $;p^n;$ has a (normal if you will, even when we're not in the abelian case!) subgroup of order $;p^k;$ , for any $;0le kle n;$ .



          Finally, just use the direct product decomposition to get a subgroup of any order dividing the group's.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 23:00









          DonAntonioDonAntonio

          177k1492226




          177k1492226























              1












              $begingroup$

              Here is a roadmap.



              Let $p$ be a prime dividing the order of $G$.




              • Let $P = { g in G : ord(g) text{ is a power of $p$} }$.


              • Then $P$ is a subgroup of $G$ (because $G$ is abelian).


              • The order of $P$ is a power of $p$ (by Cauchy's theorem).


              • The order of $P$ is the largest power of $p$ that divides the order of $G$ (by Cauchy's theorem applied to $G/P$).







              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Here is a roadmap.



                Let $p$ be a prime dividing the order of $G$.




                • Let $P = { g in G : ord(g) text{ is a power of $p$} }$.


                • Then $P$ is a subgroup of $G$ (because $G$ is abelian).


                • The order of $P$ is a power of $p$ (by Cauchy's theorem).


                • The order of $P$ is the largest power of $p$ that divides the order of $G$ (by Cauchy's theorem applied to $G/P$).







                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Here is a roadmap.



                  Let $p$ be a prime dividing the order of $G$.




                  • Let $P = { g in G : ord(g) text{ is a power of $p$} }$.


                  • Then $P$ is a subgroup of $G$ (because $G$ is abelian).


                  • The order of $P$ is a power of $p$ (by Cauchy's theorem).


                  • The order of $P$ is the largest power of $p$ that divides the order of $G$ (by Cauchy's theorem applied to $G/P$).







                  share|cite|improve this answer









                  $endgroup$



                  Here is a roadmap.



                  Let $p$ be a prime dividing the order of $G$.




                  • Let $P = { g in G : ord(g) text{ is a power of $p$} }$.


                  • Then $P$ is a subgroup of $G$ (because $G$ is abelian).


                  • The order of $P$ is a power of $p$ (by Cauchy's theorem).


                  • The order of $P$ is the largest power of $p$ that divides the order of $G$ (by Cauchy's theorem applied to $G/P$).








                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '18 at 0:33









                  lhflhf

                  163k10168392




                  163k10168392






























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