Problem on cyclic subgroup [closed]












1














Show that if $G$ is a group that contains cyclic subgroups of order $4,5$ then there must exist an element of order $20$ in $ G$.




I don't know how to prove it. I'm just beginner.Thanks for help .






abstract-algebra group-theory







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closed as off-topic by Derek Holt, Gibbs, Christopher, amWhy, Rebellos Nov 22 at 17:18


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Gibbs, Christopher, amWhy, Rebellos

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  • 1




    Why are you asking us to show something that is false? Where did this problem come from?
    – Derek Holt
    Nov 22 at 7:53
















1














Show that if $G$ is a group that contains cyclic subgroups of order $4,5$ then there must exist an element of order $20$ in $ G$.




I don't know how to prove it. I'm just beginner.Thanks for help .






abstract-algebra group-theory







share|cite|improve this question















closed as off-topic by Derek Holt, Gibbs, Christopher, amWhy, Rebellos Nov 22 at 17:18


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Gibbs, Christopher, amWhy, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    Why are you asking us to show something that is false? Where did this problem come from?
    – Derek Holt
    Nov 22 at 7:53














1












1








1







Show that if $G$ is a group that contains cyclic subgroups of order $4,5$ then there must exist an element of order $20$ in $ G$.




I don't know how to prove it. I'm just beginner.Thanks for help .






abstract-algebra group-theory







share|cite|improve this question















Show that if $G$ is a group that contains cyclic subgroups of order $4,5$ then there must exist an element of order $20$ in $ G$.




I don't know how to prove it. I'm just beginner.Thanks for help .






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 7:14









Chinnapparaj R

5,1601826




5,1601826










asked Nov 22 at 6:21









G C R

265




265




closed as off-topic by Derek Holt, Gibbs, Christopher, amWhy, Rebellos Nov 22 at 17:18


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Gibbs, Christopher, amWhy, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Derek Holt, Gibbs, Christopher, amWhy, Rebellos Nov 22 at 17:18


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Gibbs, Christopher, amWhy, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    Why are you asking us to show something that is false? Where did this problem come from?
    – Derek Holt
    Nov 22 at 7:53














  • 1




    Why are you asking us to show something that is false? Where did this problem come from?
    – Derek Holt
    Nov 22 at 7:53








1




1




Why are you asking us to show something that is false? Where did this problem come from?
– Derek Holt
Nov 22 at 7:53




Why are you asking us to show something that is false? Where did this problem come from?
– Derek Holt
Nov 22 at 7:53










1 Answer
1






active

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6














It is false, in general! For example, if we take $G=S_5$, then $G$ has cyclic subgroups of order $4$ and $5$,but $G$ has no element of order $20$.




If $G$ is Abelian then this result is true! Call $H$ the cyclic subgroup of order $4$ and $K$,the cyclic subgroups of order $5$ in $G$. So $exists a in H wedge b in K$ so that $vert a vert=4$ and $vert b vert=5$.



Now, all we know is about only $a$ and $b$, so use these two to produce the another element of order $20$.



Here $ab=ba$ implies $ab$ is the required element of order $20$.






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














    It is false, in general! For example, if we take $G=S_5$, then $G$ has cyclic subgroups of order $4$ and $5$,but $G$ has no element of order $20$.




    If $G$ is Abelian then this result is true! Call $H$ the cyclic subgroup of order $4$ and $K$,the cyclic subgroups of order $5$ in $G$. So $exists a in H wedge b in K$ so that $vert a vert=4$ and $vert b vert=5$.



    Now, all we know is about only $a$ and $b$, so use these two to produce the another element of order $20$.



    Here $ab=ba$ implies $ab$ is the required element of order $20$.






    share|cite|improve this answer


























      6














      It is false, in general! For example, if we take $G=S_5$, then $G$ has cyclic subgroups of order $4$ and $5$,but $G$ has no element of order $20$.




      If $G$ is Abelian then this result is true! Call $H$ the cyclic subgroup of order $4$ and $K$,the cyclic subgroups of order $5$ in $G$. So $exists a in H wedge b in K$ so that $vert a vert=4$ and $vert b vert=5$.



      Now, all we know is about only $a$ and $b$, so use these two to produce the another element of order $20$.



      Here $ab=ba$ implies $ab$ is the required element of order $20$.






      share|cite|improve this answer
























        6












        6








        6






        It is false, in general! For example, if we take $G=S_5$, then $G$ has cyclic subgroups of order $4$ and $5$,but $G$ has no element of order $20$.




        If $G$ is Abelian then this result is true! Call $H$ the cyclic subgroup of order $4$ and $K$,the cyclic subgroups of order $5$ in $G$. So $exists a in H wedge b in K$ so that $vert a vert=4$ and $vert b vert=5$.



        Now, all we know is about only $a$ and $b$, so use these two to produce the another element of order $20$.



        Here $ab=ba$ implies $ab$ is the required element of order $20$.






        share|cite|improve this answer












        It is false, in general! For example, if we take $G=S_5$, then $G$ has cyclic subgroups of order $4$ and $5$,but $G$ has no element of order $20$.




        If $G$ is Abelian then this result is true! Call $H$ the cyclic subgroup of order $4$ and $K$,the cyclic subgroups of order $5$ in $G$. So $exists a in H wedge b in K$ so that $vert a vert=4$ and $vert b vert=5$.



        Now, all we know is about only $a$ and $b$, so use these two to produce the another element of order $20$.



        Here $ab=ba$ implies $ab$ is the required element of order $20$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 7:05









        Chinnapparaj R

        5,1601826




        5,1601826















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