Problem on cyclic subgroup [closed]











up vote
1
down vote

favorite












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















up vote
1
down vote

favorite












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













up vote
1
down vote

favorite









up vote
1
down vote

favorite











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

4,7771825




4,7771825










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

oldest

votes

















up vote
6
down vote



accepted










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








    up vote
    6
    down vote



    accepted










    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

























      up vote
      6
      down vote



      accepted










      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























        up vote
        6
        down vote



        accepted







        up vote
        6
        down vote



        accepted






        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

        4,7771825




        4,7771825















            -

            Popular posts from this blog

            Contact image not getting when fetch all contact list from iPhone by CNContact

            Image
            up vote 0 down vote favorite I know this question already asked but not getting solution. From this code I will get all the information from the contact but image not found when open vcf files on mac os, also not getting when share this file. I use this stackoverflow link here but It's not help full. var contacts = [CNContact]() let keys = [CNContactVCardSerialization.descriptorForRequiredKeys() ] as [Any] let request = CNContactFetchRequest(keysToFetch: keys as! [CNKeyDescriptor]) do { try self.contactStore.enumerateContacts(with: request) { (contact, stop) in // Array containing all unified contacts from everywhere contacts.append(contact) } } catch { print("unable to fetch contacts") } do { let data = try
            Read more

            count number of partitions of a set with n elements into k subsets

            Image
            up vote 6 down vote favorite 1 This program is for count number of partitions of a set with n elements into k subsets I am confusing here return k*countP(n-1, k) + countP(n-1, k-1); can some one explain what is happening here? why we are multiplying with k? NOTE->I know this is not the best way to calculate number of partitions that would be DP // A C++ program to count number of partitions // of a set with n elements into k subsets #include<iostream> using namespace std; // Returns count of different partitions of n // elements in k subsets int countP(int n, int k) { // Base cases if (n == 0 || k == 0 || k > n) return 0; if (k == 1 || k == n) return 1; // S(n+1, k) = k*S(n, k) + S(n, k-1) return k*countP(n-1, k) + countP(n-1, k-1); } // Driver program int ma
            Read more

            A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

            Image
            0 I found a lot of questions abount appendices and ToC. Many users want appendices to be grouped in an Appendix part, however some problems arise with ToC, hyperref, PDF viewer bookmarks, and so on. There are different solutions which require extra packages, command patching and other extra code, however none of them satisfies me. I almost found an easy way to accomplish a good result, where appendices are added to bookmarks in the right way and hyperref links point to the right page. However, the number of the "Appendix" part page is wrong (it's the number of appendix A). Is there any EASY way to fix that? This is a MWE: documentclass{book} usepackage[nottoc,notlot,notlof]{tocbibind} usepackage{hyperref} begin{document} frontmatter tableofcontents mainmatter part{First} chapter{
            Read more