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
edited Dec 11 '18 at 6:13
Shaun
9,804113684
asked Dec 12 '12 at 16:18
Ding DongDing Dong
17.4k1059183
$endgroup$
add a comment |
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
edited Dec 11 '18 at 6:13
Shaun
9,804113684
asked Dec 12 '12 at 16:18
Ding DongDing Dong
17.4k1059183
$endgroup$
$begingroup$
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
add a comment |
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
edited Dec 11 '18 at 6:13
Shaun
9,804113684
asked Dec 12 '12 at 16:18
Ding DongDing Dong
17.4k1059183
$endgroup$
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
edited Dec 11 '18 at 6:13
Shaun
9,804113684
asked Dec 12 '12 at 16:18
Ding DongDing Dong
17.4k1059183
edited Dec 11 '18 at 6:13
Shaun
9,804113684
asked Dec 12 '12 at 16:18
Ding DongDing Dong
17.4k1059183
edited Dec 11 '18 at 6:13
Shaun
9,804113684
edited Dec 11 '18 at 6:13
Shaun
9,804113684
edited Dec 11 '18 at 6:13
Shaun
9,804113684
9,804113684
asked Dec 12 '12 at 16:18
Ding DongDing Dong
17.4k1059183
asked Dec 12 '12 at 16:18
Ding DongDing Dong
17.4k1059183
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
add a comment |
$begingroup$
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
add a comment |
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 $A$.
answered Dec 12 '12 at 16:23
Henning MakholmHenning Makholm
242k17308552
$endgroup$
$begingroup$
+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
|
show 6 more comments
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 $mathbb{Q}$.
answered Dec 12 '12 at 17:13
Nate EldredgeNate Eldredge
64.4k682174
$endgroup$
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
|
show 6 more comments
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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 $A$.
answered Dec 12 '12 at 16:23
Henning MakholmHenning Makholm
242k17308552
$endgroup$
$begingroup$
+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
|
show 6 more comments
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 $A$.
answered Dec 12 '12 at 16:23
Henning MakholmHenning Makholm
242k17308552
$endgroup$
$begingroup$
+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
|
show 6 more comments
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 $A$.
answered Dec 12 '12 at 16:23
Henning MakholmHenning Makholm
242k17308552
$endgroup$
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 $A$.
answered Dec 12 '12 at 16:23
Henning MakholmHenning Makholm
242k17308552
answered Dec 12 '12 at 16:23
Henning MakholmHenning Makholm
242k17308552
answered Dec 12 '12 at 16:23
Henning MakholmHenning Makholm
242k17308552
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
|
show 6 more comments
$begingroup$
+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
|
show 6 more comments
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 $mathbb{Q}$.
answered Dec 12 '12 at 17:13
Nate EldredgeNate Eldredge
64.4k682174
$endgroup$
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
|
show 6 more comments
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 $mathbb{Q}$.
answered Dec 12 '12 at 17:13
Nate EldredgeNate Eldredge
64.4k682174
$endgroup$
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
|
show 6 more comments
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 $mathbb{Q}$.
answered Dec 12 '12 at 17:13
Nate EldredgeNate Eldredge
64.4k682174
$endgroup$
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 $mathbb{Q}$.
answered Dec 12 '12 at 17:13
Nate EldredgeNate Eldredge
64.4k682174
answered Dec 12 '12 at 17:13
Nate EldredgeNate Eldredge
64.4k682174
answered Dec 12 '12 at 17:13
Nate EldredgeNate Eldredge
64.4k682174
answered Dec 12 '12 at 17:13
Nate EldredgeNate Eldredge
64.4k682174
64.4k682174
4
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I like this example. In addition it shoulws that a countable ring can have uncountably many subrings.
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– Marc van Leeuwen
Dec 14 '12 at 13:18
4
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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.
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– Robert Wolfe
Aug 17 '14 at 20:10
2
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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.
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– Robert Wolfe
Aug 17 '14 at 20:26
1
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@Robert I'm not sure I follow. What does $p$-divisible mean here?
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– Anu
Feb 5 '18 at 12:21
2
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@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, $py=x$.
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– Robert Wolfe
Feb 5 '18 at 14:04
|
show 6 more comments
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
|
show 6 more comments
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$begingroup$
Have you considered a countable direct product of countable groups?
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– user108903
Dec 12 '12 at 16:23
1
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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).
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– user1729
Dec 14 '12 at 12:10
2
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(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.)
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– user1729
Dec 14 '12 at 12:13
1
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@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
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– user1729
Dec 11 '18 at 16:17