countable group, uncountably many distinct subgroup?












26












$begingroup$


I need to know whether the following statement is true or false?



Every countable group $G$ has only countably many distinct subgroups.



I have not gotten any counter example to disprove the statement but an vague idea to disprove like: if it has uncountably many distinct subgroup then It must have uncountable number of element?






group-theory elementary-set-theory infinite-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you considered a countable direct product of countable groups?
    $endgroup$
    – user108903
    Dec 12 '12 at 16:23






  • 1




    $begingroup$
    The examples given in the answers prove distinct but not non-isomorphic. This is still true, and is true for certain groups with a single defining relator, by a paper of G. Baumslag and Miller (groups with a single defining relator are very natural).
    $endgroup$
    – user1729
    Dec 14 '12 at 12:10








  • 2




    $begingroup$
    (On the other hand, it is not true that a countable group can have uncountably many finitely generated subgroups. You might be interested in this MO question.)
    $endgroup$
    – user1729
    Dec 14 '12 at 12:13






  • 1




    $begingroup$
    @Shaun The full citation is G. Baumslag and C. F. Miller, A remark on the subgroups of finitely generated groups with one defining relation, Illinois J. Math., Volume 30, Issue 2 (1986), 255-257 MR840124
    $endgroup$
    – user1729
    Dec 11 '18 at 16:17
















26












$begingroup$


I need to know whether the following statement is true or false?



Every countable group $G$ has only countably many distinct subgroups.



I have not gotten any counter example to disprove the statement but an vague idea to disprove like: if it has uncountably many distinct subgroup then It must have uncountable number of element?






group-theory elementary-set-theory infinite-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you considered a countable direct product of countable groups?
    $endgroup$
    – user108903
    Dec 12 '12 at 16:23






  • 1




    $begingroup$
    The examples given in the answers prove distinct but not non-isomorphic. This is still true, and is true for certain groups with a single defining relator, by a paper of G. Baumslag and Miller (groups with a single defining relator are very natural).
    $endgroup$
    – user1729
    Dec 14 '12 at 12:10








  • 2




    $begingroup$
    (On the other hand, it is not true that a countable group can have uncountably many finitely generated subgroups. You might be interested in this MO question.)
    $endgroup$
    – user1729
    Dec 14 '12 at 12:13






  • 1




    $begingroup$
    @Shaun The full citation is G. Baumslag and C. F. Miller, A remark on the subgroups of finitely generated groups with one defining relation, Illinois J. Math., Volume 30, Issue 2 (1986), 255-257 MR840124
    $endgroup$
    – user1729
    Dec 11 '18 at 16:17














26












26








26


13



$begingroup$


I need to know whether the following statement is true or false?



Every countable group $G$ has only countably many distinct subgroups.



I have not gotten any counter example to disprove the statement but an vague idea to disprove like: if it has uncountably many distinct subgroup then It must have uncountable number of element?






group-theory elementary-set-theory infinite-groups







share|cite|improve this question











$endgroup$




I need to know whether the following statement is true or false?



Every countable group $G$ has only countably many distinct subgroups.



I have not gotten any counter example to disprove the statement but an vague idea to disprove like: if it has uncountably many distinct subgroup then It must have uncountable number of element?






group-theory elementary-set-theory infinite-groups




group-theory elementary-set-theory infinite-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 6:13









Shaun

9,804113684




9,804113684










asked Dec 12 '12 at 16:18









Ding DongDing Dong

17.4k1059183




17.4k1059183












  • $begingroup$
    Have you considered a countable direct product of countable groups?
    $endgroup$
    – user108903
    Dec 12 '12 at 16:23






  • 1




    $begingroup$
    The examples given in the answers prove distinct but not non-isomorphic. This is still true, and is true for certain groups with a single defining relator, by a paper of G. Baumslag and Miller (groups with a single defining relator are very natural).
    $endgroup$
    – user1729
    Dec 14 '12 at 12:10








  • 2




    $begingroup$
    (On the other hand, it is not true that a countable group can have uncountably many finitely generated subgroups. You might be interested in this MO question.)
    $endgroup$
    – user1729
    Dec 14 '12 at 12:13






  • 1




    $begingroup$
    @Shaun The full citation is G. Baumslag and C. F. Miller, A remark on the subgroups of finitely generated groups with one defining relation, Illinois J. Math., Volume 30, Issue 2 (1986), 255-257 MR840124
    $endgroup$
    – user1729
    Dec 11 '18 at 16:17


















  • $begingroup$
    Have you considered a countable direct product of countable groups?
    $endgroup$
    – user108903
    Dec 12 '12 at 16:23






  • 1




    $begingroup$
    The examples given in the answers prove distinct but not non-isomorphic. This is still true, and is true for certain groups with a single defining relator, by a paper of G. Baumslag and Miller (groups with a single defining relator are very natural).
    $endgroup$
    – user1729
    Dec 14 '12 at 12:10








  • 2




    $begingroup$
    (On the other hand, it is not true that a countable group can have uncountably many finitely generated subgroups. You might be interested in this MO question.)
    $endgroup$
    – user1729
    Dec 14 '12 at 12:13






  • 1




    $begingroup$
    @Shaun The full citation is G. Baumslag and C. F. Miller, A remark on the subgroups of finitely generated groups with one defining relation, Illinois J. Math., Volume 30, Issue 2 (1986), 255-257 MR840124
    $endgroup$
    – user1729
    Dec 11 '18 at 16:17
















$begingroup$
Have you considered a countable direct product of countable groups?
$endgroup$
– user108903
Dec 12 '12 at 16:23




$begingroup$
Have you considered a countable direct product of countable groups?
$endgroup$
– user108903
Dec 12 '12 at 16:23




1




1




$begingroup$
The examples given in the answers prove distinct but not non-isomorphic. This is still true, and is true for certain groups with a single defining relator, by a paper of G. Baumslag and Miller (groups with a single defining relator are very natural).
$endgroup$
– user1729
Dec 14 '12 at 12:10






$begingroup$
The examples given in the answers prove distinct but not non-isomorphic. This is still true, and is true for certain groups with a single defining relator, by a paper of G. Baumslag and Miller (groups with a single defining relator are very natural).
$endgroup$
– user1729
Dec 14 '12 at 12:10






2




2




$begingroup$
(On the other hand, it is not true that a countable group can have uncountably many finitely generated subgroups. You might be interested in this MO question.)
$endgroup$
– user1729
Dec 14 '12 at 12:13




$begingroup$
(On the other hand, it is not true that a countable group can have uncountably many finitely generated subgroups. You might be interested in this MO question.)
$endgroup$
– user1729
Dec 14 '12 at 12:13




1




1




$begingroup$
@Shaun The full citation is G. Baumslag and C. F. Miller, A remark on the subgroups of finitely generated groups with one defining relation, Illinois J. Math., Volume 30, Issue 2 (1986), 255-257 MR840124
$endgroup$
– user1729
Dec 11 '18 at 16:17




$begingroup$
@Shaun The full citation is G. Baumslag and C. F. Miller, A remark on the subgroups of finitely generated groups with one defining relation, Illinois J. Math., Volume 30, Issue 2 (1986), 255-257 MR840124
$endgroup$
– user1729
Dec 11 '18 at 16:17










2 Answers
2






active

oldest

votes


















31












$begingroup$

One example is the group consisting of all finite subsets of $mathbb N$, with the group operation being symmetric difference. The group is countably infinite, but for each finite or infinite $Asubseteq mathbb N$ there's a subgroup consisting of the finite subsets of $A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow.
    $endgroup$
    – Hagen von Eitzen
    Dec 12 '12 at 16:26










  • $begingroup$
    not getting in head, could it be elaborated a little?
    $endgroup$
    – Ding Dong
    Dec 12 '12 at 16:26






  • 1




    $begingroup$
    @Kuttus Which part gets stuck? That the main group is countable? Or that each subset produces a different subgroup? Or that there are uncountably many subsets of $mathbb N$?
    $endgroup$
    – Hagen von Eitzen
    Dec 12 '12 at 16:27






  • 1




    $begingroup$
    You are right, and I need a refresher course in proper reading.
    $endgroup$
    – gnometorule
    Dec 12 '12 at 17:10






  • 1




    $begingroup$
    @ramanujan: How would they produce identical subgroups? If $Ane B$ then there is an $x$ that is either in $Asetminus B$ or in $Bsetminus A$, so ${x}$ would only be in one of the subgroups (and therefore in particular they are not the same subgroup).
    $endgroup$
    – Henning Makholm
    Nov 13 '18 at 21:24





















35












$begingroup$

Let $(mathbb{Q},+)$ be the group of the rational numbers under addition. For any set $A$ of primes, let $G_A$ be the set of all rationals $a/b$ (in lowest terms) such that every prime factor of the denominator $b$ is in $A$. It is clear that $G_A$ is a subgroup of $mathbb{Q}$, and that $G_A = G_{A'}$ iff $A = A'$. Since there are uncountably many sets of primes, this produces uncountably many distinct subgroups of the countable group $mathbb{Q}$.






share|cite|improve this answer









$endgroup$









  • 4




    $begingroup$
    I like this example. In addition it shoulws that a countable ring can have uncountably many subrings.
    $endgroup$
    – Marc van Leeuwen
    Dec 14 '12 at 13:18






  • 4




    $begingroup$
    I like that this also shows that a countable group can have uncountably many distinct and non-isomorphic subgroups. Interestingly this extends to show that we can put continuum-many non-isomorphic group operations on a countable set.
    $endgroup$
    – Robert Wolfe
    Aug 17 '14 at 20:10






  • 2




    $begingroup$
    Given two distinct collections of primes $A$ and $A'$, let's say $A$ contains $p$ while $A'$ does not. Then $G_A$ is $p$-divisible while $G_{A'}$ is not.
    $endgroup$
    – Robert Wolfe
    Aug 17 '14 at 20:26






  • 1




    $begingroup$
    @Robert I'm not sure I follow. What does $p$-divisible mean here?
    $endgroup$
    – Anu
    Feb 5 '18 at 12:21






  • 2




    $begingroup$
    @Anu for every element $x$ of $G_A$ there is an element $y$ of $G_A$ such that $y^p=x$, or, in this case, $py=x$.
    $endgroup$
    – Robert Wolfe
    Feb 5 '18 at 14:04












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









31












$begingroup$

One example is the group consisting of all finite subsets of $mathbb N$, with the group operation being symmetric difference. The group is countably infinite, but for each finite or infinite $Asubseteq mathbb N$ there's a subgroup consisting of the finite subsets of $A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow.
    $endgroup$
    – Hagen von Eitzen
    Dec 12 '12 at 16:26










  • $begingroup$
    not getting in head, could it be elaborated a little?
    $endgroup$
    – Ding Dong
    Dec 12 '12 at 16:26






  • 1




    $begingroup$
    @Kuttus Which part gets stuck? That the main group is countable? Or that each subset produces a different subgroup? Or that there are uncountably many subsets of $mathbb N$?
    $endgroup$
    – Hagen von Eitzen
    Dec 12 '12 at 16:27






  • 1




    $begingroup$
    You are right, and I need a refresher course in proper reading.
    $endgroup$
    – gnometorule
    Dec 12 '12 at 17:10






  • 1




    $begingroup$
    @ramanujan: How would they produce identical subgroups? If $Ane B$ then there is an $x$ that is either in $Asetminus B$ or in $Bsetminus A$, so ${x}$ would only be in one of the subgroups (and therefore in particular they are not the same subgroup).
    $endgroup$
    – Henning Makholm
    Nov 13 '18 at 21:24


















31












$begingroup$

One example is the group consisting of all finite subsets of $mathbb N$, with the group operation being symmetric difference. The group is countably infinite, but for each finite or infinite $Asubseteq mathbb N$ there's a subgroup consisting of the finite subsets of $A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow.
    $endgroup$
    – Hagen von Eitzen
    Dec 12 '12 at 16:26










  • $begingroup$
    not getting in head, could it be elaborated a little?
    $endgroup$
    – Ding Dong
    Dec 12 '12 at 16:26






  • 1




    $begingroup$
    @Kuttus Which part gets stuck? That the main group is countable? Or that each subset produces a different subgroup? Or that there are uncountably many subsets of $mathbb N$?
    $endgroup$
    – Hagen von Eitzen
    Dec 12 '12 at 16:27






  • 1




    $begingroup$
    You are right, and I need a refresher course in proper reading.
    $endgroup$
    – gnometorule
    Dec 12 '12 at 17:10






  • 1




    $begingroup$
    @ramanujan: How would they produce identical subgroups? If $Ane B$ then there is an $x$ that is either in $Asetminus B$ or in $Bsetminus A$, so ${x}$ would only be in one of the subgroups (and therefore in particular they are not the same subgroup).
    $endgroup$
    – Henning Makholm
    Nov 13 '18 at 21:24
















31












31








31





$begingroup$

One example is the group consisting of all finite subsets of $mathbb N$, with the group operation being symmetric difference. The group is countably infinite, but for each finite or infinite $Asubseteq mathbb N$ there's a subgroup consisting of the finite subsets of $A$.






share|cite|improve this answer









$endgroup$



One example is the group consisting of all finite subsets of $mathbb N$, with the group operation being symmetric difference. The group is countably infinite, but for each finite or infinite $Asubseteq mathbb N$ there's a subgroup consisting of the finite subsets of $A$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '12 at 16:23









Henning MakholmHenning Makholm

242k17308552




242k17308552












  • $begingroup$
    +1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow.
    $endgroup$
    – Hagen von Eitzen
    Dec 12 '12 at 16:26










  • $begingroup$
    not getting in head, could it be elaborated a little?
    $endgroup$
    – Ding Dong
    Dec 12 '12 at 16:26






  • 1




    $begingroup$
    @Kuttus Which part gets stuck? That the main group is countable? Or that each subset produces a different subgroup? Or that there are uncountably many subsets of $mathbb N$?
    $endgroup$
    – Hagen von Eitzen
    Dec 12 '12 at 16:27






  • 1




    $begingroup$
    You are right, and I need a refresher course in proper reading.
    $endgroup$
    – gnometorule
    Dec 12 '12 at 17:10






  • 1




    $begingroup$
    @ramanujan: How would they produce identical subgroups? If $Ane B$ then there is an $x$ that is either in $Asetminus B$ or in $Bsetminus A$, so ${x}$ would only be in one of the subgroups (and therefore in particular they are not the same subgroup).
    $endgroup$
    – Henning Makholm
    Nov 13 '18 at 21:24




















  • $begingroup$
    +1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow.
    $endgroup$
    – Hagen von Eitzen
    Dec 12 '12 at 16:26










  • $begingroup$
    not getting in head, could it be elaborated a little?
    $endgroup$
    – Ding Dong
    Dec 12 '12 at 16:26






  • 1




    $begingroup$
    @Kuttus Which part gets stuck? That the main group is countable? Or that each subset produces a different subgroup? Or that there are uncountably many subsets of $mathbb N$?
    $endgroup$
    – Hagen von Eitzen
    Dec 12 '12 at 16:27






  • 1




    $begingroup$
    You are right, and I need a refresher course in proper reading.
    $endgroup$
    – gnometorule
    Dec 12 '12 at 17:10






  • 1




    $begingroup$
    @ramanujan: How would they produce identical subgroups? If $Ane B$ then there is an $x$ that is either in $Asetminus B$ or in $Bsetminus A$, so ${x}$ would only be in one of the subgroups (and therefore in particular they are not the same subgroup).
    $endgroup$
    – Henning Makholm
    Nov 13 '18 at 21:24


















$begingroup$
+1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow.
$endgroup$
– Hagen von Eitzen
Dec 12 '12 at 16:26




$begingroup$
+1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow.
$endgroup$
– Hagen von Eitzen
Dec 12 '12 at 16:26












$begingroup$
not getting in head, could it be elaborated a little?
$endgroup$
– Ding Dong
Dec 12 '12 at 16:26




$begingroup$
not getting in head, could it be elaborated a little?
$endgroup$
– Ding Dong
Dec 12 '12 at 16:26




1




1




$begingroup$
@Kuttus Which part gets stuck? That the main group is countable? Or that each subset produces a different subgroup? Or that there are uncountably many subsets of $mathbb N$?
$endgroup$
– Hagen von Eitzen
Dec 12 '12 at 16:27




$begingroup$
@Kuttus Which part gets stuck? That the main group is countable? Or that each subset produces a different subgroup? Or that there are uncountably many subsets of $mathbb N$?
$endgroup$
– Hagen von Eitzen
Dec 12 '12 at 16:27




1




1




$begingroup$
You are right, and I need a refresher course in proper reading.
$endgroup$
– gnometorule
Dec 12 '12 at 17:10




$begingroup$
You are right, and I need a refresher course in proper reading.
$endgroup$
– gnometorule
Dec 12 '12 at 17:10




1




1




$begingroup$
@ramanujan: How would they produce identical subgroups? If $Ane B$ then there is an $x$ that is either in $Asetminus B$ or in $Bsetminus A$, so ${x}$ would only be in one of the subgroups (and therefore in particular they are not the same subgroup).
$endgroup$
– Henning Makholm
Nov 13 '18 at 21:24






$begingroup$
@ramanujan: How would they produce identical subgroups? If $Ane B$ then there is an $x$ that is either in $Asetminus B$ or in $Bsetminus A$, so ${x}$ would only be in one of the subgroups (and therefore in particular they are not the same subgroup).
$endgroup$
– Henning Makholm
Nov 13 '18 at 21:24













35












$begingroup$

Let $(mathbb{Q},+)$ be the group of the rational numbers under addition. For any set $A$ of primes, let $G_A$ be the set of all rationals $a/b$ (in lowest terms) such that every prime factor of the denominator $b$ is in $A$. It is clear that $G_A$ is a subgroup of $mathbb{Q}$, and that $G_A = G_{A'}$ iff $A = A'$. Since there are uncountably many sets of primes, this produces uncountably many distinct subgroups of the countable group $mathbb{Q}$.






share|cite|improve this answer









$endgroup$









  • 4




    $begingroup$
    I like this example. In addition it shoulws that a countable ring can have uncountably many subrings.
    $endgroup$
    – Marc van Leeuwen
    Dec 14 '12 at 13:18






  • 4




    $begingroup$
    I like that this also shows that a countable group can have uncountably many distinct and non-isomorphic subgroups. Interestingly this extends to show that we can put continuum-many non-isomorphic group operations on a countable set.
    $endgroup$
    – Robert Wolfe
    Aug 17 '14 at 20:10






  • 2




    $begingroup$
    Given two distinct collections of primes $A$ and $A'$, let's say $A$ contains $p$ while $A'$ does not. Then $G_A$ is $p$-divisible while $G_{A'}$ is not.
    $endgroup$
    – Robert Wolfe
    Aug 17 '14 at 20:26






  • 1




    $begingroup$
    @Robert I'm not sure I follow. What does $p$-divisible mean here?
    $endgroup$
    – Anu
    Feb 5 '18 at 12:21






  • 2




    $begingroup$
    @Anu for every element $x$ of $G_A$ there is an element $y$ of $G_A$ such that $y^p=x$, or, in this case, $py=x$.
    $endgroup$
    – Robert Wolfe
    Feb 5 '18 at 14:04
















35












$begingroup$

Let $(mathbb{Q},+)$ be the group of the rational numbers under addition. For any set $A$ of primes, let $G_A$ be the set of all rationals $a/b$ (in lowest terms) such that every prime factor of the denominator $b$ is in $A$. It is clear that $G_A$ is a subgroup of $mathbb{Q}$, and that $G_A = G_{A'}$ iff $A = A'$. Since there are uncountably many sets of primes, this produces uncountably many distinct subgroups of the countable group $mathbb{Q}$.






share|cite|improve this answer









$endgroup$









  • 4




    $begingroup$
    I like this example. In addition it shoulws that a countable ring can have uncountably many subrings.
    $endgroup$
    – Marc van Leeuwen
    Dec 14 '12 at 13:18






  • 4




    $begingroup$
    I like that this also shows that a countable group can have uncountably many distinct and non-isomorphic subgroups. Interestingly this extends to show that we can put continuum-many non-isomorphic group operations on a countable set.
    $endgroup$
    – Robert Wolfe
    Aug 17 '14 at 20:10






  • 2




    $begingroup$
    Given two distinct collections of primes $A$ and $A'$, let's say $A$ contains $p$ while $A'$ does not. Then $G_A$ is $p$-divisible while $G_{A'}$ is not.
    $endgroup$
    – Robert Wolfe
    Aug 17 '14 at 20:26






  • 1




    $begingroup$
    @Robert I'm not sure I follow. What does $p$-divisible mean here?
    $endgroup$
    – Anu
    Feb 5 '18 at 12:21






  • 2




    $begingroup$
    @Anu for every element $x$ of $G_A$ there is an element $y$ of $G_A$ such that $y^p=x$, or, in this case, $py=x$.
    $endgroup$
    – Robert Wolfe
    Feb 5 '18 at 14:04














35












35








35





$begingroup$

Let $(mathbb{Q},+)$ be the group of the rational numbers under addition. For any set $A$ of primes, let $G_A$ be the set of all rationals $a/b$ (in lowest terms) such that every prime factor of the denominator $b$ is in $A$. It is clear that $G_A$ is a subgroup of $mathbb{Q}$, and that $G_A = G_{A'}$ iff $A = A'$. Since there are uncountably many sets of primes, this produces uncountably many distinct subgroups of the countable group $mathbb{Q}$.






share|cite|improve this answer









$endgroup$



Let $(mathbb{Q},+)$ be the group of the rational numbers under addition. For any set $A$ of primes, let $G_A$ be the set of all rationals $a/b$ (in lowest terms) such that every prime factor of the denominator $b$ is in $A$. It is clear that $G_A$ is a subgroup of $mathbb{Q}$, and that $G_A = G_{A'}$ iff $A = A'$. Since there are uncountably many sets of primes, this produces uncountably many distinct subgroups of the countable group $mathbb{Q}$.







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answered Dec 12 '12 at 17:13









Nate EldredgeNate Eldredge

64.4k682174




64.4k682174








  • 4




    $begingroup$
    I like this example. In addition it shoulws that a countable ring can have uncountably many subrings.
    $endgroup$
    – Marc van Leeuwen
    Dec 14 '12 at 13:18






  • 4




    $begingroup$
    I like that this also shows that a countable group can have uncountably many distinct and non-isomorphic subgroups. Interestingly this extends to show that we can put continuum-many non-isomorphic group operations on a countable set.
    $endgroup$
    – Robert Wolfe
    Aug 17 '14 at 20:10






  • 2




    $begingroup$
    Given two distinct collections of primes $A$ and $A'$, let's say $A$ contains $p$ while $A'$ does not. Then $G_A$ is $p$-divisible while $G_{A'}$ is not.
    $endgroup$
    – Robert Wolfe
    Aug 17 '14 at 20:26






  • 1




    $begingroup$
    @Robert I'm not sure I follow. What does $p$-divisible mean here?
    $endgroup$
    – Anu
    Feb 5 '18 at 12:21






  • 2




    $begingroup$
    @Anu for every element $x$ of $G_A$ there is an element $y$ of $G_A$ such that $y^p=x$, or, in this case, $py=x$.
    $endgroup$
    – Robert Wolfe
    Feb 5 '18 at 14:04














  • 4




    $begingroup$
    I like this example. In addition it shoulws that a countable ring can have uncountably many subrings.
    $endgroup$
    – Marc van Leeuwen
    Dec 14 '12 at 13:18






  • 4




    $begingroup$
    I like that this also shows that a countable group can have uncountably many distinct and non-isomorphic subgroups. Interestingly this extends to show that we can put continuum-many non-isomorphic group operations on a countable set.
    $endgroup$
    – Robert Wolfe
    Aug 17 '14 at 20:10






  • 2




    $begingroup$
    Given two distinct collections of primes $A$ and $A'$, let's say $A$ contains $p$ while $A'$ does not. Then $G_A$ is $p$-divisible while $G_{A'}$ is not.
    $endgroup$
    – Robert Wolfe
    Aug 17 '14 at 20:26






  • 1




    $begingroup$
    @Robert I'm not sure I follow. What does $p$-divisible mean here?
    $endgroup$
    – Anu
    Feb 5 '18 at 12:21






  • 2




    $begingroup$
    @Anu for every element $x$ of $G_A$ there is an element $y$ of $G_A$ such that $y^p=x$, or, in this case, $py=x$.
    $endgroup$
    – Robert Wolfe
    Feb 5 '18 at 14:04








4




4




$begingroup$
I like this example. In addition it shoulws that a countable ring can have uncountably many subrings.
$endgroup$
– Marc van Leeuwen
Dec 14 '12 at 13:18




$begingroup$
I like this example. In addition it shoulws that a countable ring can have uncountably many subrings.
$endgroup$
– Marc van Leeuwen
Dec 14 '12 at 13:18




4




4




$begingroup$
I like that this also shows that a countable group can have uncountably many distinct and non-isomorphic subgroups. Interestingly this extends to show that we can put continuum-many non-isomorphic group operations on a countable set.
$endgroup$
– Robert Wolfe
Aug 17 '14 at 20:10




$begingroup$
I like that this also shows that a countable group can have uncountably many distinct and non-isomorphic subgroups. Interestingly this extends to show that we can put continuum-many non-isomorphic group operations on a countable set.
$endgroup$
– Robert Wolfe
Aug 17 '14 at 20:10




2




2




$begingroup$
Given two distinct collections of primes $A$ and $A'$, let's say $A$ contains $p$ while $A'$ does not. Then $G_A$ is $p$-divisible while $G_{A'}$ is not.
$endgroup$
– Robert Wolfe
Aug 17 '14 at 20:26




$begingroup$
Given two distinct collections of primes $A$ and $A'$, let's say $A$ contains $p$ while $A'$ does not. Then $G_A$ is $p$-divisible while $G_{A'}$ is not.
$endgroup$
– Robert Wolfe
Aug 17 '14 at 20:26




1




1




$begingroup$
@Robert I'm not sure I follow. What does $p$-divisible mean here?
$endgroup$
– Anu
Feb 5 '18 at 12:21




$begingroup$
@Robert I'm not sure I follow. What does $p$-divisible mean here?
$endgroup$
– Anu
Feb 5 '18 at 12:21




2




2




$begingroup$
@Anu for every element $x$ of $G_A$ there is an element $y$ of $G_A$ such that $y^p=x$, or, in this case, $py=x$.
$endgroup$
– Robert Wolfe
Feb 5 '18 at 14:04




$begingroup$
@Anu for every element $x$ of $G_A$ there is an element $y$ of $G_A$ such that $y^p=x$, or, in this case, $py=x$.
$endgroup$
– Robert Wolfe
Feb 5 '18 at 14:04


















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