Can an uncountable group have a countable number of subgroups?











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Can an uncountable group have only a countable number of subgroups?



Please give examples if any exist!




Edit: I want a group having uncountable cardinality but having a countable number of subgroups.



By countable number of subgroups, I mean the collection of all subgroups of a group is countable.










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  • I'm frankly a bit surprised at the negative reaction to this question.
    – Noah Schweber
    1 hour ago















up vote
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down vote

favorite
1













Can an uncountable group have only a countable number of subgroups?



Please give examples if any exist!




Edit: I want a group having uncountable cardinality but having a countable number of subgroups.



By countable number of subgroups, I mean the collection of all subgroups of a group is countable.










share|cite|improve this question
























  • I'm frankly a bit surprised at the negative reaction to this question.
    – Noah Schweber
    1 hour ago













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1






Can an uncountable group have only a countable number of subgroups?



Please give examples if any exist!




Edit: I want a group having uncountable cardinality but having a countable number of subgroups.



By countable number of subgroups, I mean the collection of all subgroups of a group is countable.










share|cite|improve this question
















Can an uncountable group have only a countable number of subgroups?



Please give examples if any exist!




Edit: I want a group having uncountable cardinality but having a countable number of subgroups.



By countable number of subgroups, I mean the collection of all subgroups of a group is countable.







group-theory examples-counterexamples infinite-groups






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edited 1 hour ago









Shaun

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8,075113577










asked 1 hour ago









Cloud JR

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  • I'm frankly a bit surprised at the negative reaction to this question.
    – Noah Schweber
    1 hour ago


















  • I'm frankly a bit surprised at the negative reaction to this question.
    – Noah Schweber
    1 hour ago
















I'm frankly a bit surprised at the negative reaction to this question.
– Noah Schweber
1 hour ago




I'm frankly a bit surprised at the negative reaction to this question.
– Noah Schweber
1 hour ago










2 Answers
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7
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No. Suppose $G$ is an uncountable group. Every element $g$ of $G$ belongs to a countable subgroup of $G$, namely the cyclic subgroup $langle grangle$. Thus $G$ is the union of all of its countable subgroups. Since a countable union of countable sets is countable, $G$ must have uncountably many countable subgroups.






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    up vote
    2
    down vote













    EDIT: bof's answer is the right one, but the construction below - while completely pointless overkill - is still an example of a useful technique, so I'm leaving this answer up.





    No, this cannot happen.



    Suppose $G$ is a group and $A$ is a countable subset of $G$. Then the closure of $A$ under the group operations ($*$ and $^{-1}$) of $G$, $langle Arangle$, is again countable - this is a good exercise (HINT: the set of finite strings from a countable set is countable).



    With this in hand, if $G$ is an uncountable group we can build an uncountable chain of subgroups of $G$, as follows:




    • We will define a countable subgroup $A_delta$ for every countable ordinal $delta$. There are uncountably many of these, so if we can do this we'll be done.


    • We let $A_0$ be the trivial subgroup.


    • Having defined $A_eta$ for every $eta<delta$, we let $a$ be some element of $G$ not in $bigcup_{eta<delta}A_eta$ - which exists, since this is a countable union of countable subgroups, and $G$ is uncountable - and let $A_delta=langle (bigcup_{eta<delta}A_eta)cup{a}rangle$.


    • It's easy to prove by transfinite induction that $(A_delta)_{delta<omega_1}$ is a strictly increasing chain of countable subgroups of $G$, so we're done.







    share|cite|improve this answer





















    • Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
      – bof
      44 mins ago











    Your Answer





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    2 Answers
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    2 Answers
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    up vote
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    down vote













    No. Suppose $G$ is an uncountable group. Every element $g$ of $G$ belongs to a countable subgroup of $G$, namely the cyclic subgroup $langle grangle$. Thus $G$ is the union of all of its countable subgroups. Since a countable union of countable sets is countable, $G$ must have uncountably many countable subgroups.






    share|cite|improve this answer

























      up vote
      7
      down vote













      No. Suppose $G$ is an uncountable group. Every element $g$ of $G$ belongs to a countable subgroup of $G$, namely the cyclic subgroup $langle grangle$. Thus $G$ is the union of all of its countable subgroups. Since a countable union of countable sets is countable, $G$ must have uncountably many countable subgroups.






      share|cite|improve this answer























        up vote
        7
        down vote










        up vote
        7
        down vote









        No. Suppose $G$ is an uncountable group. Every element $g$ of $G$ belongs to a countable subgroup of $G$, namely the cyclic subgroup $langle grangle$. Thus $G$ is the union of all of its countable subgroups. Since a countable union of countable sets is countable, $G$ must have uncountably many countable subgroups.






        share|cite|improve this answer












        No. Suppose $G$ is an uncountable group. Every element $g$ of $G$ belongs to a countable subgroup of $G$, namely the cyclic subgroup $langle grangle$. Thus $G$ is the union of all of its countable subgroups. Since a countable union of countable sets is countable, $G$ must have uncountably many countable subgroups.







        share|cite|improve this answer












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        share|cite|improve this answer










        answered 1 hour ago









        bof

        49.5k455118




        49.5k455118






















            up vote
            2
            down vote













            EDIT: bof's answer is the right one, but the construction below - while completely pointless overkill - is still an example of a useful technique, so I'm leaving this answer up.





            No, this cannot happen.



            Suppose $G$ is a group and $A$ is a countable subset of $G$. Then the closure of $A$ under the group operations ($*$ and $^{-1}$) of $G$, $langle Arangle$, is again countable - this is a good exercise (HINT: the set of finite strings from a countable set is countable).



            With this in hand, if $G$ is an uncountable group we can build an uncountable chain of subgroups of $G$, as follows:




            • We will define a countable subgroup $A_delta$ for every countable ordinal $delta$. There are uncountably many of these, so if we can do this we'll be done.


            • We let $A_0$ be the trivial subgroup.


            • Having defined $A_eta$ for every $eta<delta$, we let $a$ be some element of $G$ not in $bigcup_{eta<delta}A_eta$ - which exists, since this is a countable union of countable subgroups, and $G$ is uncountable - and let $A_delta=langle (bigcup_{eta<delta}A_eta)cup{a}rangle$.


            • It's easy to prove by transfinite induction that $(A_delta)_{delta<omega_1}$ is a strictly increasing chain of countable subgroups of $G$, so we're done.







            share|cite|improve this answer





















            • Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
              – bof
              44 mins ago















            up vote
            2
            down vote













            EDIT: bof's answer is the right one, but the construction below - while completely pointless overkill - is still an example of a useful technique, so I'm leaving this answer up.





            No, this cannot happen.



            Suppose $G$ is a group and $A$ is a countable subset of $G$. Then the closure of $A$ under the group operations ($*$ and $^{-1}$) of $G$, $langle Arangle$, is again countable - this is a good exercise (HINT: the set of finite strings from a countable set is countable).



            With this in hand, if $G$ is an uncountable group we can build an uncountable chain of subgroups of $G$, as follows:




            • We will define a countable subgroup $A_delta$ for every countable ordinal $delta$. There are uncountably many of these, so if we can do this we'll be done.


            • We let $A_0$ be the trivial subgroup.


            • Having defined $A_eta$ for every $eta<delta$, we let $a$ be some element of $G$ not in $bigcup_{eta<delta}A_eta$ - which exists, since this is a countable union of countable subgroups, and $G$ is uncountable - and let $A_delta=langle (bigcup_{eta<delta}A_eta)cup{a}rangle$.


            • It's easy to prove by transfinite induction that $(A_delta)_{delta<omega_1}$ is a strictly increasing chain of countable subgroups of $G$, so we're done.







            share|cite|improve this answer





















            • Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
              – bof
              44 mins ago













            up vote
            2
            down vote










            up vote
            2
            down vote









            EDIT: bof's answer is the right one, but the construction below - while completely pointless overkill - is still an example of a useful technique, so I'm leaving this answer up.





            No, this cannot happen.



            Suppose $G$ is a group and $A$ is a countable subset of $G$. Then the closure of $A$ under the group operations ($*$ and $^{-1}$) of $G$, $langle Arangle$, is again countable - this is a good exercise (HINT: the set of finite strings from a countable set is countable).



            With this in hand, if $G$ is an uncountable group we can build an uncountable chain of subgroups of $G$, as follows:




            • We will define a countable subgroup $A_delta$ for every countable ordinal $delta$. There are uncountably many of these, so if we can do this we'll be done.


            • We let $A_0$ be the trivial subgroup.


            • Having defined $A_eta$ for every $eta<delta$, we let $a$ be some element of $G$ not in $bigcup_{eta<delta}A_eta$ - which exists, since this is a countable union of countable subgroups, and $G$ is uncountable - and let $A_delta=langle (bigcup_{eta<delta}A_eta)cup{a}rangle$.


            • It's easy to prove by transfinite induction that $(A_delta)_{delta<omega_1}$ is a strictly increasing chain of countable subgroups of $G$, so we're done.







            share|cite|improve this answer












            EDIT: bof's answer is the right one, but the construction below - while completely pointless overkill - is still an example of a useful technique, so I'm leaving this answer up.





            No, this cannot happen.



            Suppose $G$ is a group and $A$ is a countable subset of $G$. Then the closure of $A$ under the group operations ($*$ and $^{-1}$) of $G$, $langle Arangle$, is again countable - this is a good exercise (HINT: the set of finite strings from a countable set is countable).



            With this in hand, if $G$ is an uncountable group we can build an uncountable chain of subgroups of $G$, as follows:




            • We will define a countable subgroup $A_delta$ for every countable ordinal $delta$. There are uncountably many of these, so if we can do this we'll be done.


            • We let $A_0$ be the trivial subgroup.


            • Having defined $A_eta$ for every $eta<delta$, we let $a$ be some element of $G$ not in $bigcup_{eta<delta}A_eta$ - which exists, since this is a countable union of countable subgroups, and $G$ is uncountable - and let $A_delta=langle (bigcup_{eta<delta}A_eta)cup{a}rangle$.


            • It's easy to prove by transfinite induction that $(A_delta)_{delta<omega_1}$ is a strictly increasing chain of countable subgroups of $G$, so we're done.








            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Noah Schweber

            119k10146278




            119k10146278












            • Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
              – bof
              44 mins ago


















            • Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
              – bof
              44 mins ago
















            Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
            – bof
            44 mins ago




            Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
            – bof
            44 mins ago


















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