Find a subgroup of $mathbb{Z}_{12}oplus mathbb{Z}_{4}oplus mathbb{Z}_{15}$ of order 9












1












$begingroup$



Question:



Find a subgroup of $mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}$ that is of order 9.




By a certain theorem, $mathbb{Z}_{4}oplusmathbb{Z}_{15} cong mathbb{Z}_{60}$ from the fact that 4 and 15 are relatively prime.



This gives



$mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}cong mathbb{Z}_{12}oplus mathbb{Z}_{60}$



It now suffices to find all elements of order 9 in the group $mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}$; knowing that the order of any element in a group is the order of the cyclic subgroup generated by that element.



But, aside from listing the elements manually-which is tedious- what is a more efficient way to go about?






abstract-algebra number-theory elementary-number-theory finite-groups direct-product







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$endgroup$












  • $begingroup$
    Note that this isn't asking for a cyclic subgroup of order $9$: so there could be a subgroup of order $9$, but no element of order $9$. You should also keep an eye out for subgroups of the form $Bbb Z_3oplusBbb Z_3$.
    $endgroup$
    – Stahl
    Apr 9 '17 at 4:52












  • $begingroup$
    yes, you are correct. Apparently, the browser 'saves' the entire draft of the thread upon reopening and I conveniently forgotten this fact, assuming that I had already input a fitting title.
    $endgroup$
    – Mathematicing
    Apr 9 '17 at 4:54
















1












$begingroup$



Question:



Find a subgroup of $mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}$ that is of order 9.




By a certain theorem, $mathbb{Z}_{4}oplusmathbb{Z}_{15} cong mathbb{Z}_{60}$ from the fact that 4 and 15 are relatively prime.



This gives



$mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}cong mathbb{Z}_{12}oplus mathbb{Z}_{60}$



It now suffices to find all elements of order 9 in the group $mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}$; knowing that the order of any element in a group is the order of the cyclic subgroup generated by that element.



But, aside from listing the elements manually-which is tedious- what is a more efficient way to go about?






abstract-algebra number-theory elementary-number-theory finite-groups direct-product







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note that this isn't asking for a cyclic subgroup of order $9$: so there could be a subgroup of order $9$, but no element of order $9$. You should also keep an eye out for subgroups of the form $Bbb Z_3oplusBbb Z_3$.
    $endgroup$
    – Stahl
    Apr 9 '17 at 4:52












  • $begingroup$
    yes, you are correct. Apparently, the browser 'saves' the entire draft of the thread upon reopening and I conveniently forgotten this fact, assuming that I had already input a fitting title.
    $endgroup$
    – Mathematicing
    Apr 9 '17 at 4:54














1












1








1





$begingroup$



Question:



Find a subgroup of $mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}$ that is of order 9.




By a certain theorem, $mathbb{Z}_{4}oplusmathbb{Z}_{15} cong mathbb{Z}_{60}$ from the fact that 4 and 15 are relatively prime.



This gives



$mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}cong mathbb{Z}_{12}oplus mathbb{Z}_{60}$



It now suffices to find all elements of order 9 in the group $mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}$; knowing that the order of any element in a group is the order of the cyclic subgroup generated by that element.



But, aside from listing the elements manually-which is tedious- what is a more efficient way to go about?






abstract-algebra number-theory elementary-number-theory finite-groups direct-product







share|cite|improve this question











$endgroup$





Question:



Find a subgroup of $mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}$ that is of order 9.




By a certain theorem, $mathbb{Z}_{4}oplusmathbb{Z}_{15} cong mathbb{Z}_{60}$ from the fact that 4 and 15 are relatively prime.



This gives



$mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}cong mathbb{Z}_{12}oplus mathbb{Z}_{60}$



It now suffices to find all elements of order 9 in the group $mathbb{Z}_{12} oplus mathbb{Z}_{4} oplus mathbb{Z}_{15}$; knowing that the order of any element in a group is the order of the cyclic subgroup generated by that element.



But, aside from listing the elements manually-which is tedious- what is a more efficient way to go about?






abstract-algebra number-theory elementary-number-theory finite-groups direct-product




abstract-algebra number-theory elementary-number-theory finite-groups direct-product






share|cite|improve this question















share|cite|improve this question













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edited Dec 30 '18 at 14:30









Henning Makholm

239k17303540




239k17303540










asked Apr 9 '17 at 4:50









MathematicingMathematicing

2,44321854




2,44321854












  • $begingroup$
    Note that this isn't asking for a cyclic subgroup of order $9$: so there could be a subgroup of order $9$, but no element of order $9$. You should also keep an eye out for subgroups of the form $Bbb Z_3oplusBbb Z_3$.
    $endgroup$
    – Stahl
    Apr 9 '17 at 4:52












  • $begingroup$
    yes, you are correct. Apparently, the browser 'saves' the entire draft of the thread upon reopening and I conveniently forgotten this fact, assuming that I had already input a fitting title.
    $endgroup$
    – Mathematicing
    Apr 9 '17 at 4:54


















  • $begingroup$
    Note that this isn't asking for a cyclic subgroup of order $9$: so there could be a subgroup of order $9$, but no element of order $9$. You should also keep an eye out for subgroups of the form $Bbb Z_3oplusBbb Z_3$.
    $endgroup$
    – Stahl
    Apr 9 '17 at 4:52












  • $begingroup$
    yes, you are correct. Apparently, the browser 'saves' the entire draft of the thread upon reopening and I conveniently forgotten this fact, assuming that I had already input a fitting title.
    $endgroup$
    – Mathematicing
    Apr 9 '17 at 4:54
















$begingroup$
Note that this isn't asking for a cyclic subgroup of order $9$: so there could be a subgroup of order $9$, but no element of order $9$. You should also keep an eye out for subgroups of the form $Bbb Z_3oplusBbb Z_3$.
$endgroup$
– Stahl
Apr 9 '17 at 4:52






$begingroup$
Note that this isn't asking for a cyclic subgroup of order $9$: so there could be a subgroup of order $9$, but no element of order $9$. You should also keep an eye out for subgroups of the form $Bbb Z_3oplusBbb Z_3$.
$endgroup$
– Stahl
Apr 9 '17 at 4:52














$begingroup$
yes, you are correct. Apparently, the browser 'saves' the entire draft of the thread upon reopening and I conveniently forgotten this fact, assuming that I had already input a fitting title.
$endgroup$
– Mathematicing
Apr 9 '17 at 4:54




$begingroup$
yes, you are correct. Apparently, the browser 'saves' the entire draft of the thread upon reopening and I conveniently forgotten this fact, assuming that I had already input a fitting title.
$endgroup$
– Mathematicing
Apr 9 '17 at 4:54










2 Answers
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4












$begingroup$

The structure theorem for abelian groups is the right way to go but you are using it in an unhelpful direction. It implies that you can split up that sum further: $$mathbb{Z}_{12} oplus mathbb{Z}_4 oplus mathbb{Z}_{15} cong mathbb{Z}_4 oplus mathbb{Z}_3 oplus mathbb{Z}_4 oplus mathbb{Z}_3 oplus mathbb{Z}_5.$$ You can read off the subgroup of order $9$. To be explicit, the subgroup $mathbb{Z}_3 subseteq mathbb{Z}_{12}$ is cyclic generated by $4$ and the subgroup $mathbb{Z}_3 subseteq mathbb{Z}_{15}$ is cyclic generated by $5$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    While the decomposition into minimal cyclic subgroups as in the answer by user399601 is probably the easiest way to go, here is a somewhat difference approach. Whatever the structure of your subgroup, its elements have orders that are powers of$~3$, so you can restrict attention to the subgroup of such elements. The property of having order a power of$~3$ is equivalent to the same property for the components in each of the given factors in the direct sum. In $Bbb Z_4$ only the identity qualifies, in the other two factors the subgroup with the property is of order$~3$. That gives $3times1times3=9$ elements in all, and you have to search no longer.






    share|cite|improve this answer











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      2 Answers
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      2 Answers
      2






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      4












      $begingroup$

      The structure theorem for abelian groups is the right way to go but you are using it in an unhelpful direction. It implies that you can split up that sum further: $$mathbb{Z}_{12} oplus mathbb{Z}_4 oplus mathbb{Z}_{15} cong mathbb{Z}_4 oplus mathbb{Z}_3 oplus mathbb{Z}_4 oplus mathbb{Z}_3 oplus mathbb{Z}_5.$$ You can read off the subgroup of order $9$. To be explicit, the subgroup $mathbb{Z}_3 subseteq mathbb{Z}_{12}$ is cyclic generated by $4$ and the subgroup $mathbb{Z}_3 subseteq mathbb{Z}_{15}$ is cyclic generated by $5$.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        The structure theorem for abelian groups is the right way to go but you are using it in an unhelpful direction. It implies that you can split up that sum further: $$mathbb{Z}_{12} oplus mathbb{Z}_4 oplus mathbb{Z}_{15} cong mathbb{Z}_4 oplus mathbb{Z}_3 oplus mathbb{Z}_4 oplus mathbb{Z}_3 oplus mathbb{Z}_5.$$ You can read off the subgroup of order $9$. To be explicit, the subgroup $mathbb{Z}_3 subseteq mathbb{Z}_{12}$ is cyclic generated by $4$ and the subgroup $mathbb{Z}_3 subseteq mathbb{Z}_{15}$ is cyclic generated by $5$.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          The structure theorem for abelian groups is the right way to go but you are using it in an unhelpful direction. It implies that you can split up that sum further: $$mathbb{Z}_{12} oplus mathbb{Z}_4 oplus mathbb{Z}_{15} cong mathbb{Z}_4 oplus mathbb{Z}_3 oplus mathbb{Z}_4 oplus mathbb{Z}_3 oplus mathbb{Z}_5.$$ You can read off the subgroup of order $9$. To be explicit, the subgroup $mathbb{Z}_3 subseteq mathbb{Z}_{12}$ is cyclic generated by $4$ and the subgroup $mathbb{Z}_3 subseteq mathbb{Z}_{15}$ is cyclic generated by $5$.






          share|cite|improve this answer









          $endgroup$



          The structure theorem for abelian groups is the right way to go but you are using it in an unhelpful direction. It implies that you can split up that sum further: $$mathbb{Z}_{12} oplus mathbb{Z}_4 oplus mathbb{Z}_{15} cong mathbb{Z}_4 oplus mathbb{Z}_3 oplus mathbb{Z}_4 oplus mathbb{Z}_3 oplus mathbb{Z}_5.$$ You can read off the subgroup of order $9$. To be explicit, the subgroup $mathbb{Z}_3 subseteq mathbb{Z}_{12}$ is cyclic generated by $4$ and the subgroup $mathbb{Z}_3 subseteq mathbb{Z}_{15}$ is cyclic generated by $5$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 9 '17 at 4:56









          user399601user399601

          1,598412




          1,598412























              0












              $begingroup$

              While the decomposition into minimal cyclic subgroups as in the answer by user399601 is probably the easiest way to go, here is a somewhat difference approach. Whatever the structure of your subgroup, its elements have orders that are powers of$~3$, so you can restrict attention to the subgroup of such elements. The property of having order a power of$~3$ is equivalent to the same property for the components in each of the given factors in the direct sum. In $Bbb Z_4$ only the identity qualifies, in the other two factors the subgroup with the property is of order$~3$. That gives $3times1times3=9$ elements in all, and you have to search no longer.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                While the decomposition into minimal cyclic subgroups as in the answer by user399601 is probably the easiest way to go, here is a somewhat difference approach. Whatever the structure of your subgroup, its elements have orders that are powers of$~3$, so you can restrict attention to the subgroup of such elements. The property of having order a power of$~3$ is equivalent to the same property for the components in each of the given factors in the direct sum. In $Bbb Z_4$ only the identity qualifies, in the other two factors the subgroup with the property is of order$~3$. That gives $3times1times3=9$ elements in all, and you have to search no longer.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  While the decomposition into minimal cyclic subgroups as in the answer by user399601 is probably the easiest way to go, here is a somewhat difference approach. Whatever the structure of your subgroup, its elements have orders that are powers of$~3$, so you can restrict attention to the subgroup of such elements. The property of having order a power of$~3$ is equivalent to the same property for the components in each of the given factors in the direct sum. In $Bbb Z_4$ only the identity qualifies, in the other two factors the subgroup with the property is of order$~3$. That gives $3times1times3=9$ elements in all, and you have to search no longer.






                  share|cite|improve this answer











                  $endgroup$



                  While the decomposition into minimal cyclic subgroups as in the answer by user399601 is probably the easiest way to go, here is a somewhat difference approach. Whatever the structure of your subgroup, its elements have orders that are powers of$~3$, so you can restrict attention to the subgroup of such elements. The property of having order a power of$~3$ is equivalent to the same property for the components in each of the given factors in the direct sum. In $Bbb Z_4$ only the identity qualifies, in the other two factors the subgroup with the property is of order$~3$. That gives $3times1times3=9$ elements in all, and you have to search no longer.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Apr 9 '17 at 8:27

























                  answered Apr 9 '17 at 5:15









                  Marc van LeeuwenMarc van Leeuwen

                  86.6k5106220




                  86.6k5106220






























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