Give an example of a nonabelian group in which a product of elements of finite order can have infinite order....












4















This question already has an answer here:




  • Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite [duplicate]

    6 answers




So, I let a,b be elements in such a group. So |a|=n and |b|=m, n and m are finite. But |ab| needs to be infinite, but since |ab|=lcm(n,m), how can that be possible?






abstract-algebra group-theory examples-counterexamples abelian-groups







share|cite|improve this question















marked as duplicate by Daniel Fischer♦ Oct 26 '15 at 17:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 5




    The result you quote is not correct. In an Abelian group, the order of $ab$ is $le$ the lcm of the orders of $a$ and $b$.
    – André Nicolas
    Oct 26 '15 at 5:20










  • Some related posts: Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite and Examples and further results about the order of the product of two elements in a group
    – Martin Sleziak
    Oct 26 '15 at 10:51










  • Where does this problem come from? Is it from a book? From an assignment?
    – Martin Sleziak
    Oct 26 '15 at 10:52


















4















This question already has an answer here:




  • Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite [duplicate]

    6 answers




So, I let a,b be elements in such a group. So |a|=n and |b|=m, n and m are finite. But |ab| needs to be infinite, but since |ab|=lcm(n,m), how can that be possible?






abstract-algebra group-theory examples-counterexamples abelian-groups







share|cite|improve this question















marked as duplicate by Daniel Fischer♦ Oct 26 '15 at 17:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 5




    The result you quote is not correct. In an Abelian group, the order of $ab$ is $le$ the lcm of the orders of $a$ and $b$.
    – André Nicolas
    Oct 26 '15 at 5:20










  • Some related posts: Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite and Examples and further results about the order of the product of two elements in a group
    – Martin Sleziak
    Oct 26 '15 at 10:51










  • Where does this problem come from? Is it from a book? From an assignment?
    – Martin Sleziak
    Oct 26 '15 at 10:52
















4












4








4


0






This question already has an answer here:




  • Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite [duplicate]

    6 answers




So, I let a,b be elements in such a group. So |a|=n and |b|=m, n and m are finite. But |ab| needs to be infinite, but since |ab|=lcm(n,m), how can that be possible?






abstract-algebra group-theory examples-counterexamples abelian-groups







share|cite|improve this question
















This question already has an answer here:




  • Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite [duplicate]

    6 answers




So, I let a,b be elements in such a group. So |a|=n and |b|=m, n and m are finite. But |ab| needs to be infinite, but since |ab|=lcm(n,m), how can that be possible?





This question already has an answer here:




  • Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite [duplicate]

    6 answers







abstract-algebra group-theory examples-counterexamples abelian-groups




abstract-algebra group-theory examples-counterexamples abelian-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 26 '15 at 10:46









Martin Sleziak

44.6k7115270




44.6k7115270










asked Oct 26 '15 at 5:16









likelikelike

545




545




marked as duplicate by Daniel Fischer♦ Oct 26 '15 at 17:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Daniel Fischer♦ Oct 26 '15 at 17:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 5




    The result you quote is not correct. In an Abelian group, the order of $ab$ is $le$ the lcm of the orders of $a$ and $b$.
    – André Nicolas
    Oct 26 '15 at 5:20










  • Some related posts: Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite and Examples and further results about the order of the product of two elements in a group
    – Martin Sleziak
    Oct 26 '15 at 10:51










  • Where does this problem come from? Is it from a book? From an assignment?
    – Martin Sleziak
    Oct 26 '15 at 10:52
















  • 5




    The result you quote is not correct. In an Abelian group, the order of $ab$ is $le$ the lcm of the orders of $a$ and $b$.
    – André Nicolas
    Oct 26 '15 at 5:20










  • Some related posts: Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite and Examples and further results about the order of the product of two elements in a group
    – Martin Sleziak
    Oct 26 '15 at 10:51










  • Where does this problem come from? Is it from a book? From an assignment?
    – Martin Sleziak
    Oct 26 '15 at 10:52










5




5




The result you quote is not correct. In an Abelian group, the order of $ab$ is $le$ the lcm of the orders of $a$ and $b$.
– André Nicolas
Oct 26 '15 at 5:20




The result you quote is not correct. In an Abelian group, the order of $ab$ is $le$ the lcm of the orders of $a$ and $b$.
– André Nicolas
Oct 26 '15 at 5:20












Some related posts: Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite and Examples and further results about the order of the product of two elements in a group
– Martin Sleziak
Oct 26 '15 at 10:51




Some related posts: Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite and Examples and further results about the order of the product of two elements in a group
– Martin Sleziak
Oct 26 '15 at 10:51












Where does this problem come from? Is it from a book? From an assignment?
– Martin Sleziak
Oct 26 '15 at 10:52






Where does this problem come from? Is it from a book? From an assignment?
– Martin Sleziak
Oct 26 '15 at 10:52












4 Answers
4






active

oldest

votes


















8














Take the free group on letters $a,b$ with the sole relations that $a^2=b^2=1$. Now $ab$ has infinite order, since $$ababababcdots ab neq 1$$






share|cite|improve this answer

















  • 1




    Can you explain more on this? I don't quite understand.
    – likelikelike
    Oct 26 '15 at 5:28






  • 2




    @safaqfwrq Do you know how to define groups via generators and relations - that is, are you familiar with notation like "$langle a, bvert a^2, b^2rangle$"?
    – Noah Schweber
    Oct 26 '15 at 6:25






  • 6




    This is in some sense a restatement of the question: it's true that if $a^2 = 1$ and $b^2 = 1$ then $ab$ has infinite order iff it's true in this universal example. So you have not yet answered the question. You're implicitly using the fact that you think you know how to solve the word problem in this group...
    – Qiaochu Yuan
    Oct 26 '15 at 6:58



















13














Consider group of all permutations of $mathbb{Z}$; permutation group on $geq 3$ letters is always non-abelian.



Consider $sigma(x)=-x+1$ and $tau(x)=-x+2$, the two permutations of $mathbb{Z}$.




  • Show that $taucirctau=sigmacircsigma=I$, hence order of $sigma,tau$ is $2$.


  • Find $sigmacirc tau$.


  • Is it of finite order?







share|cite|improve this answer

















  • 3




    This simple answer (+1) has a nice relation to mine: Two reflections on parallel lines give a translation. This is then just the one-dimensional discrete case.
    – filipos
    Oct 26 '15 at 13:18



















9














Compose two reflections on the plane and you get a rotation. Reflections have order 2. The rotation can have arbitrary order, including infinite.






share|cite|improve this answer























  • I'm trying to imagine a rotation of infinite order. I think rotation through an irrational number of degrees, or a rational number of radians would work, right?
    – Todd Wilcox
    Oct 26 '15 at 13:43






  • 1




    Right. By the way, another interesting case is when the reflections are on parallel lines, which gives a translation (see Groups' answer).
    – filipos
    Oct 26 '15 at 15:01










  • An irrational number of degrees will work. An irrational number of radians might not. A rational number of radians will work though!
    – MJD
    Oct 26 '15 at 16:00



















3














Matrices, invertible ones are a group under multiplication and here is a good example






share|cite|improve this answer






























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8














    Take the free group on letters $a,b$ with the sole relations that $a^2=b^2=1$. Now $ab$ has infinite order, since $$ababababcdots ab neq 1$$






    share|cite|improve this answer

















    • 1




      Can you explain more on this? I don't quite understand.
      – likelikelike
      Oct 26 '15 at 5:28






    • 2




      @safaqfwrq Do you know how to define groups via generators and relations - that is, are you familiar with notation like "$langle a, bvert a^2, b^2rangle$"?
      – Noah Schweber
      Oct 26 '15 at 6:25






    • 6




      This is in some sense a restatement of the question: it's true that if $a^2 = 1$ and $b^2 = 1$ then $ab$ has infinite order iff it's true in this universal example. So you have not yet answered the question. You're implicitly using the fact that you think you know how to solve the word problem in this group...
      – Qiaochu Yuan
      Oct 26 '15 at 6:58
















    8














    Take the free group on letters $a,b$ with the sole relations that $a^2=b^2=1$. Now $ab$ has infinite order, since $$ababababcdots ab neq 1$$






    share|cite|improve this answer

















    • 1




      Can you explain more on this? I don't quite understand.
      – likelikelike
      Oct 26 '15 at 5:28






    • 2




      @safaqfwrq Do you know how to define groups via generators and relations - that is, are you familiar with notation like "$langle a, bvert a^2, b^2rangle$"?
      – Noah Schweber
      Oct 26 '15 at 6:25






    • 6




      This is in some sense a restatement of the question: it's true that if $a^2 = 1$ and $b^2 = 1$ then $ab$ has infinite order iff it's true in this universal example. So you have not yet answered the question. You're implicitly using the fact that you think you know how to solve the word problem in this group...
      – Qiaochu Yuan
      Oct 26 '15 at 6:58














    8












    8








    8






    Take the free group on letters $a,b$ with the sole relations that $a^2=b^2=1$. Now $ab$ has infinite order, since $$ababababcdots ab neq 1$$






    share|cite|improve this answer












    Take the free group on letters $a,b$ with the sole relations that $a^2=b^2=1$. Now $ab$ has infinite order, since $$ababababcdots ab neq 1$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 26 '15 at 5:20









    vadim123

    75.3k896187




    75.3k896187








    • 1




      Can you explain more on this? I don't quite understand.
      – likelikelike
      Oct 26 '15 at 5:28






    • 2




      @safaqfwrq Do you know how to define groups via generators and relations - that is, are you familiar with notation like "$langle a, bvert a^2, b^2rangle$"?
      – Noah Schweber
      Oct 26 '15 at 6:25






    • 6




      This is in some sense a restatement of the question: it's true that if $a^2 = 1$ and $b^2 = 1$ then $ab$ has infinite order iff it's true in this universal example. So you have not yet answered the question. You're implicitly using the fact that you think you know how to solve the word problem in this group...
      – Qiaochu Yuan
      Oct 26 '15 at 6:58














    • 1




      Can you explain more on this? I don't quite understand.
      – likelikelike
      Oct 26 '15 at 5:28






    • 2




      @safaqfwrq Do you know how to define groups via generators and relations - that is, are you familiar with notation like "$langle a, bvert a^2, b^2rangle$"?
      – Noah Schweber
      Oct 26 '15 at 6:25






    • 6




      This is in some sense a restatement of the question: it's true that if $a^2 = 1$ and $b^2 = 1$ then $ab$ has infinite order iff it's true in this universal example. So you have not yet answered the question. You're implicitly using the fact that you think you know how to solve the word problem in this group...
      – Qiaochu Yuan
      Oct 26 '15 at 6:58








    1




    1




    Can you explain more on this? I don't quite understand.
    – likelikelike
    Oct 26 '15 at 5:28




    Can you explain more on this? I don't quite understand.
    – likelikelike
    Oct 26 '15 at 5:28




    2




    2




    @safaqfwrq Do you know how to define groups via generators and relations - that is, are you familiar with notation like "$langle a, bvert a^2, b^2rangle$"?
    – Noah Schweber
    Oct 26 '15 at 6:25




    @safaqfwrq Do you know how to define groups via generators and relations - that is, are you familiar with notation like "$langle a, bvert a^2, b^2rangle$"?
    – Noah Schweber
    Oct 26 '15 at 6:25




    6




    6




    This is in some sense a restatement of the question: it's true that if $a^2 = 1$ and $b^2 = 1$ then $ab$ has infinite order iff it's true in this universal example. So you have not yet answered the question. You're implicitly using the fact that you think you know how to solve the word problem in this group...
    – Qiaochu Yuan
    Oct 26 '15 at 6:58




    This is in some sense a restatement of the question: it's true that if $a^2 = 1$ and $b^2 = 1$ then $ab$ has infinite order iff it's true in this universal example. So you have not yet answered the question. You're implicitly using the fact that you think you know how to solve the word problem in this group...
    – Qiaochu Yuan
    Oct 26 '15 at 6:58











    13














    Consider group of all permutations of $mathbb{Z}$; permutation group on $geq 3$ letters is always non-abelian.



    Consider $sigma(x)=-x+1$ and $tau(x)=-x+2$, the two permutations of $mathbb{Z}$.




    • Show that $taucirctau=sigmacircsigma=I$, hence order of $sigma,tau$ is $2$.


    • Find $sigmacirc tau$.


    • Is it of finite order?







    share|cite|improve this answer

















    • 3




      This simple answer (+1) has a nice relation to mine: Two reflections on parallel lines give a translation. This is then just the one-dimensional discrete case.
      – filipos
      Oct 26 '15 at 13:18
















    13














    Consider group of all permutations of $mathbb{Z}$; permutation group on $geq 3$ letters is always non-abelian.



    Consider $sigma(x)=-x+1$ and $tau(x)=-x+2$, the two permutations of $mathbb{Z}$.




    • Show that $taucirctau=sigmacircsigma=I$, hence order of $sigma,tau$ is $2$.


    • Find $sigmacirc tau$.


    • Is it of finite order?







    share|cite|improve this answer

















    • 3




      This simple answer (+1) has a nice relation to mine: Two reflections on parallel lines give a translation. This is then just the one-dimensional discrete case.
      – filipos
      Oct 26 '15 at 13:18














    13












    13








    13






    Consider group of all permutations of $mathbb{Z}$; permutation group on $geq 3$ letters is always non-abelian.



    Consider $sigma(x)=-x+1$ and $tau(x)=-x+2$, the two permutations of $mathbb{Z}$.




    • Show that $taucirctau=sigmacircsigma=I$, hence order of $sigma,tau$ is $2$.


    • Find $sigmacirc tau$.


    • Is it of finite order?







    share|cite|improve this answer












    Consider group of all permutations of $mathbb{Z}$; permutation group on $geq 3$ letters is always non-abelian.



    Consider $sigma(x)=-x+1$ and $tau(x)=-x+2$, the two permutations of $mathbb{Z}$.




    • Show that $taucirctau=sigmacircsigma=I$, hence order of $sigma,tau$ is $2$.


    • Find $sigmacirc tau$.


    • Is it of finite order?








    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 26 '15 at 5:43









    Groups

    5,54711038




    5,54711038








    • 3




      This simple answer (+1) has a nice relation to mine: Two reflections on parallel lines give a translation. This is then just the one-dimensional discrete case.
      – filipos
      Oct 26 '15 at 13:18














    • 3




      This simple answer (+1) has a nice relation to mine: Two reflections on parallel lines give a translation. This is then just the one-dimensional discrete case.
      – filipos
      Oct 26 '15 at 13:18








    3




    3




    This simple answer (+1) has a nice relation to mine: Two reflections on parallel lines give a translation. This is then just the one-dimensional discrete case.
    – filipos
    Oct 26 '15 at 13:18




    This simple answer (+1) has a nice relation to mine: Two reflections on parallel lines give a translation. This is then just the one-dimensional discrete case.
    – filipos
    Oct 26 '15 at 13:18











    9














    Compose two reflections on the plane and you get a rotation. Reflections have order 2. The rotation can have arbitrary order, including infinite.






    share|cite|improve this answer























    • I'm trying to imagine a rotation of infinite order. I think rotation through an irrational number of degrees, or a rational number of radians would work, right?
      – Todd Wilcox
      Oct 26 '15 at 13:43






    • 1




      Right. By the way, another interesting case is when the reflections are on parallel lines, which gives a translation (see Groups' answer).
      – filipos
      Oct 26 '15 at 15:01










    • An irrational number of degrees will work. An irrational number of radians might not. A rational number of radians will work though!
      – MJD
      Oct 26 '15 at 16:00
















    9














    Compose two reflections on the plane and you get a rotation. Reflections have order 2. The rotation can have arbitrary order, including infinite.






    share|cite|improve this answer























    • I'm trying to imagine a rotation of infinite order. I think rotation through an irrational number of degrees, or a rational number of radians would work, right?
      – Todd Wilcox
      Oct 26 '15 at 13:43






    • 1




      Right. By the way, another interesting case is when the reflections are on parallel lines, which gives a translation (see Groups' answer).
      – filipos
      Oct 26 '15 at 15:01










    • An irrational number of degrees will work. An irrational number of radians might not. A rational number of radians will work though!
      – MJD
      Oct 26 '15 at 16:00














    9












    9








    9






    Compose two reflections on the plane and you get a rotation. Reflections have order 2. The rotation can have arbitrary order, including infinite.






    share|cite|improve this answer














    Compose two reflections on the plane and you get a rotation. Reflections have order 2. The rotation can have arbitrary order, including infinite.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Oct 26 '15 at 12:53

























    answered Oct 26 '15 at 12:15









    filipos

    1,39546




    1,39546












    • I'm trying to imagine a rotation of infinite order. I think rotation through an irrational number of degrees, or a rational number of radians would work, right?
      – Todd Wilcox
      Oct 26 '15 at 13:43






    • 1




      Right. By the way, another interesting case is when the reflections are on parallel lines, which gives a translation (see Groups' answer).
      – filipos
      Oct 26 '15 at 15:01










    • An irrational number of degrees will work. An irrational number of radians might not. A rational number of radians will work though!
      – MJD
      Oct 26 '15 at 16:00


















    • I'm trying to imagine a rotation of infinite order. I think rotation through an irrational number of degrees, or a rational number of radians would work, right?
      – Todd Wilcox
      Oct 26 '15 at 13:43






    • 1




      Right. By the way, another interesting case is when the reflections are on parallel lines, which gives a translation (see Groups' answer).
      – filipos
      Oct 26 '15 at 15:01










    • An irrational number of degrees will work. An irrational number of radians might not. A rational number of radians will work though!
      – MJD
      Oct 26 '15 at 16:00
















    I'm trying to imagine a rotation of infinite order. I think rotation through an irrational number of degrees, or a rational number of radians would work, right?
    – Todd Wilcox
    Oct 26 '15 at 13:43




    I'm trying to imagine a rotation of infinite order. I think rotation through an irrational number of degrees, or a rational number of radians would work, right?
    – Todd Wilcox
    Oct 26 '15 at 13:43




    1




    1




    Right. By the way, another interesting case is when the reflections are on parallel lines, which gives a translation (see Groups' answer).
    – filipos
    Oct 26 '15 at 15:01




    Right. By the way, another interesting case is when the reflections are on parallel lines, which gives a translation (see Groups' answer).
    – filipos
    Oct 26 '15 at 15:01












    An irrational number of degrees will work. An irrational number of radians might not. A rational number of radians will work though!
    – MJD
    Oct 26 '15 at 16:00




    An irrational number of degrees will work. An irrational number of radians might not. A rational number of radians will work though!
    – MJD
    Oct 26 '15 at 16:00











    3














    Matrices, invertible ones are a group under multiplication and here is a good example






    share|cite|improve this answer




























      3














      Matrices, invertible ones are a group under multiplication and here is a good example






      share|cite|improve this answer


























        3












        3








        3






        Matrices, invertible ones are a group under multiplication and here is a good example






        share|cite|improve this answer














        Matrices, invertible ones are a group under multiplication and here is a good example







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 13 '17 at 12:20









        Community♦

        1




        1










        answered Oct 26 '15 at 5:21









        Zelos Malum

        5,4162822




        5,4162822















            -

            Popular posts from this blog

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
            Read more

            Sphinx de Gizeh

            Image
            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
            Read more

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
            Read more