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












4














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




















  • 1




    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.
    – DonAntonio
    3 hours ago












  • @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?
    – John11
    3 hours ago
















4














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




















  • 1




    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.
    – DonAntonio
    3 hours ago












  • @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?
    – John11
    3 hours ago














4












4








4


1





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















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 3 hours ago

























asked 3 hours ago









John11

9411721




9411721








  • 1




    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.
    – DonAntonio
    3 hours ago












  • @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?
    – John11
    3 hours ago














  • 1




    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.
    – DonAntonio
    3 hours ago












  • @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?
    – John11
    3 hours ago








1




1




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.
– DonAntonio
3 hours ago






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.
– DonAntonio
3 hours ago














@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?
– John11
3 hours ago




@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?
– John11
3 hours ago










3 Answers
3






active

oldest

votes


















2














By the Sylow Theorems, this must exist.






share|cite|improve this answer

















  • 1




    Yes, but I'm trying to avoid Sylow. Is it correct using the structure theorem like I did?
    – John11
    3 hours ago












  • I suggest you edit the question, @John11, to include that avoidance.
    – Shaun
    3 hours ago



















2














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





























    0














    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





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      By the Sylow Theorems, this must exist.






      share|cite|improve this answer

















      • 1




        Yes, but I'm trying to avoid Sylow. Is it correct using the structure theorem like I did?
        – John11
        3 hours ago












      • I suggest you edit the question, @John11, to include that avoidance.
        – Shaun
        3 hours ago
















      2














      By the Sylow Theorems, this must exist.






      share|cite|improve this answer

















      • 1




        Yes, but I'm trying to avoid Sylow. Is it correct using the structure theorem like I did?
        – John11
        3 hours ago












      • I suggest you edit the question, @John11, to include that avoidance.
        – Shaun
        3 hours ago














      2












      2








      2






      By the Sylow Theorems, this must exist.






      share|cite|improve this answer












      By the Sylow Theorems, this must exist.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 3 hours ago









      kmini

      943




      943








      • 1




        Yes, but I'm trying to avoid Sylow. Is it correct using the structure theorem like I did?
        – John11
        3 hours ago












      • I suggest you edit the question, @John11, to include that avoidance.
        – Shaun
        3 hours ago














      • 1




        Yes, but I'm trying to avoid Sylow. Is it correct using the structure theorem like I did?
        – John11
        3 hours ago












      • I suggest you edit the question, @John11, to include that avoidance.
        – Shaun
        3 hours ago








      1




      1




      Yes, but I'm trying to avoid Sylow. Is it correct using the structure theorem like I did?
      – John11
      3 hours ago






      Yes, but I'm trying to avoid Sylow. Is it correct using the structure theorem like I did?
      – John11
      3 hours ago














      I suggest you edit the question, @John11, to include that avoidance.
      – Shaun
      3 hours ago




      I suggest you edit the question, @John11, to include that avoidance.
      – Shaun
      3 hours ago











      2














      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


























        2














        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
























          2












          2








          2






          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












          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 2 hours ago









          DonAntonio

          176k1491225




          176k1491225























              0














              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


























                0














                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
























                  0












                  0








                  0






                  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












                  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 55 mins ago









                  lhf

                  162k9166385




                  162k9166385






























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