Can an uncountable group have a countable number of subgroups? [closed]
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14
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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.
group-theory examples-counterexamples infinite-groups
closed as off-topic by Shaun, Brian Borchers, user10354138, Cesareo, TheSimpliFire Dec 9 at 19:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Brian Borchers, user10354138, Cesareo, TheSimpliFire
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
14
down vote
favorite
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
closed as off-topic by Shaun, Brian Borchers, user10354138, Cesareo, TheSimpliFire Dec 9 at 19:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Brian Borchers, user10354138, Cesareo, TheSimpliFire
If this question can be reworded to fit the rules in the help center, please edit the question.
5
I'm frankly a bit surprised at the negative reaction to this question.
– Noah Schweber
Dec 9 at 3:00
2
Possible duplicate of Countable number of subgroups $implies $ countable group
– Carmeister
Dec 9 at 10:05
Then why is it marked as "off-topic" rather than "duplicate"?
– C Monsour
Dec 9 at 20:42
I'm voting to reopen. If it should be closed please give the correct reason.
– C Monsour
Dec 9 at 20:44
@CMonsour It was closed because there is no context. It's just a problem statement with no details. The closure reason is correct. While it might be better that it be listed as a duplicate, I don't see that as a reason to reopen.
– jgon
Dec 9 at 21:10
add a comment |
up vote
14
down vote
favorite
up vote
14
down vote
favorite
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
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
group-theory examples-counterexamples infinite-groups
edited Dec 9 at 3:01
Shaun
8,215113578
8,215113578
asked Dec 9 at 2:47
Cloud JR
806417
806417
closed as off-topic by Shaun, Brian Borchers, user10354138, Cesareo, TheSimpliFire Dec 9 at 19:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Brian Borchers, user10354138, Cesareo, TheSimpliFire
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shaun, Brian Borchers, user10354138, Cesareo, TheSimpliFire Dec 9 at 19:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Brian Borchers, user10354138, Cesareo, TheSimpliFire
If this question can be reworded to fit the rules in the help center, please edit the question.
5
I'm frankly a bit surprised at the negative reaction to this question.
– Noah Schweber
Dec 9 at 3:00
2
Possible duplicate of Countable number of subgroups $implies $ countable group
– Carmeister
Dec 9 at 10:05
Then why is it marked as "off-topic" rather than "duplicate"?
– C Monsour
Dec 9 at 20:42
I'm voting to reopen. If it should be closed please give the correct reason.
– C Monsour
Dec 9 at 20:44
@CMonsour It was closed because there is no context. It's just a problem statement with no details. The closure reason is correct. While it might be better that it be listed as a duplicate, I don't see that as a reason to reopen.
– jgon
Dec 9 at 21:10
add a comment |
5
I'm frankly a bit surprised at the negative reaction to this question.
– Noah Schweber
Dec 9 at 3:00
2
Possible duplicate of Countable number of subgroups $implies $ countable group
– Carmeister
Dec 9 at 10:05
Then why is it marked as "off-topic" rather than "duplicate"?
– C Monsour
Dec 9 at 20:42
I'm voting to reopen. If it should be closed please give the correct reason.
– C Monsour
Dec 9 at 20:44
@CMonsour It was closed because there is no context. It's just a problem statement with no details. The closure reason is correct. While it might be better that it be listed as a duplicate, I don't see that as a reason to reopen.
– jgon
Dec 9 at 21:10
5
5
I'm frankly a bit surprised at the negative reaction to this question.
– Noah Schweber
Dec 9 at 3:00
I'm frankly a bit surprised at the negative reaction to this question.
– Noah Schweber
Dec 9 at 3:00
2
2
Possible duplicate of Countable number of subgroups $implies $ countable group
– Carmeister
Dec 9 at 10:05
Possible duplicate of Countable number of subgroups $implies $ countable group
– Carmeister
Dec 9 at 10:05
Then why is it marked as "off-topic" rather than "duplicate"?
– C Monsour
Dec 9 at 20:42
Then why is it marked as "off-topic" rather than "duplicate"?
– C Monsour
Dec 9 at 20:42
I'm voting to reopen. If it should be closed please give the correct reason.
– C Monsour
Dec 9 at 20:44
I'm voting to reopen. If it should be closed please give the correct reason.
– C Monsour
Dec 9 at 20:44
@CMonsour It was closed because there is no context. It's just a problem statement with no details. The closure reason is correct. While it might be better that it be listed as a duplicate, I don't see that as a reason to reopen.
– jgon
Dec 9 at 21:10
@CMonsour It was closed because there is no context. It's just a problem statement with no details. The closure reason is correct. While it might be better that it be listed as a duplicate, I don't see that as a reason to reopen.
– jgon
Dec 9 at 21:10
add a comment |
2 Answers
2
active
oldest
votes
up vote
27
down vote
accepted
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.
1
How can we be sure that the subgroups generated by $g$ and $g'$ (for $gnot= g'$) are not the same?
– BenjaminH
Dec 9 at 8:54
With a bit of refinement, this argument can be turned into the group being a countable union of finite sets. I forgot if this makes it countable without choice.
– Tobias Kildetoft
Dec 9 at 9:00
@TobiasKildetoft The axiom of choice is needed to prove that a countable union of two-element sets is countable. (If the union were countable, you could prove the axiom of choice for two-element sets.)
– bof
Dec 9 at 10:17
4
@BenjaminH : You can't, for a specific pair $g,g'$. However, both $langle grangle$ and $langle g'rangle$ are countable, so there are indeed uncountably many such subgroups (otherwise a countable union of countable sets would exhaust $G$, which is not possible).
– MPW
Dec 9 at 13:23
add a comment |
up vote
9
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.
Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
– bof
Dec 9 at 3:43
2
@bof This question that I asked and Noah answered a while ago should answer your question
– Paul Plummer
Dec 9 at 4:51
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
27
down vote
accepted
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.
1
How can we be sure that the subgroups generated by $g$ and $g'$ (for $gnot= g'$) are not the same?
– BenjaminH
Dec 9 at 8:54
With a bit of refinement, this argument can be turned into the group being a countable union of finite sets. I forgot if this makes it countable without choice.
– Tobias Kildetoft
Dec 9 at 9:00
@TobiasKildetoft The axiom of choice is needed to prove that a countable union of two-element sets is countable. (If the union were countable, you could prove the axiom of choice for two-element sets.)
– bof
Dec 9 at 10:17
4
@BenjaminH : You can't, for a specific pair $g,g'$. However, both $langle grangle$ and $langle g'rangle$ are countable, so there are indeed uncountably many such subgroups (otherwise a countable union of countable sets would exhaust $G$, which is not possible).
– MPW
Dec 9 at 13:23
add a comment |
up vote
27
down vote
accepted
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.
1
How can we be sure that the subgroups generated by $g$ and $g'$ (for $gnot= g'$) are not the same?
– BenjaminH
Dec 9 at 8:54
With a bit of refinement, this argument can be turned into the group being a countable union of finite sets. I forgot if this makes it countable without choice.
– Tobias Kildetoft
Dec 9 at 9:00
@TobiasKildetoft The axiom of choice is needed to prove that a countable union of two-element sets is countable. (If the union were countable, you could prove the axiom of choice for two-element sets.)
– bof
Dec 9 at 10:17
4
@BenjaminH : You can't, for a specific pair $g,g'$. However, both $langle grangle$ and $langle g'rangle$ are countable, so there are indeed uncountably many such subgroups (otherwise a countable union of countable sets would exhaust $G$, which is not possible).
– MPW
Dec 9 at 13:23
add a comment |
up vote
27
down vote
accepted
up vote
27
down vote
accepted
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.
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.
answered Dec 9 at 2:56
bof
49.9k457119
49.9k457119
1
How can we be sure that the subgroups generated by $g$ and $g'$ (for $gnot= g'$) are not the same?
– BenjaminH
Dec 9 at 8:54
With a bit of refinement, this argument can be turned into the group being a countable union of finite sets. I forgot if this makes it countable without choice.
– Tobias Kildetoft
Dec 9 at 9:00
@TobiasKildetoft The axiom of choice is needed to prove that a countable union of two-element sets is countable. (If the union were countable, you could prove the axiom of choice for two-element sets.)
– bof
Dec 9 at 10:17
4
@BenjaminH : You can't, for a specific pair $g,g'$. However, both $langle grangle$ and $langle g'rangle$ are countable, so there are indeed uncountably many such subgroups (otherwise a countable union of countable sets would exhaust $G$, which is not possible).
– MPW
Dec 9 at 13:23
add a comment |
1
How can we be sure that the subgroups generated by $g$ and $g'$ (for $gnot= g'$) are not the same?
– BenjaminH
Dec 9 at 8:54
With a bit of refinement, this argument can be turned into the group being a countable union of finite sets. I forgot if this makes it countable without choice.
– Tobias Kildetoft
Dec 9 at 9:00
@TobiasKildetoft The axiom of choice is needed to prove that a countable union of two-element sets is countable. (If the union were countable, you could prove the axiom of choice for two-element sets.)
– bof
Dec 9 at 10:17
4
@BenjaminH : You can't, for a specific pair $g,g'$. However, both $langle grangle$ and $langle g'rangle$ are countable, so there are indeed uncountably many such subgroups (otherwise a countable union of countable sets would exhaust $G$, which is not possible).
– MPW
Dec 9 at 13:23
1
1
How can we be sure that the subgroups generated by $g$ and $g'$ (for $gnot= g'$) are not the same?
– BenjaminH
Dec 9 at 8:54
How can we be sure that the subgroups generated by $g$ and $g'$ (for $gnot= g'$) are not the same?
– BenjaminH
Dec 9 at 8:54
With a bit of refinement, this argument can be turned into the group being a countable union of finite sets. I forgot if this makes it countable without choice.
– Tobias Kildetoft
Dec 9 at 9:00
With a bit of refinement, this argument can be turned into the group being a countable union of finite sets. I forgot if this makes it countable without choice.
– Tobias Kildetoft
Dec 9 at 9:00
@TobiasKildetoft The axiom of choice is needed to prove that a countable union of two-element sets is countable. (If the union were countable, you could prove the axiom of choice for two-element sets.)
– bof
Dec 9 at 10:17
@TobiasKildetoft The axiom of choice is needed to prove that a countable union of two-element sets is countable. (If the union were countable, you could prove the axiom of choice for two-element sets.)
– bof
Dec 9 at 10:17
4
4
@BenjaminH : You can't, for a specific pair $g,g'$. However, both $langle grangle$ and $langle g'rangle$ are countable, so there are indeed uncountably many such subgroups (otherwise a countable union of countable sets would exhaust $G$, which is not possible).
– MPW
Dec 9 at 13:23
@BenjaminH : You can't, for a specific pair $g,g'$. However, both $langle grangle$ and $langle g'rangle$ are countable, so there are indeed uncountably many such subgroups (otherwise a countable union of countable sets would exhaust $G$, which is not possible).
– MPW
Dec 9 at 13:23
add a comment |
up vote
9
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.
Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
– bof
Dec 9 at 3:43
2
@bof This question that I asked and Noah answered a while ago should answer your question
– Paul Plummer
Dec 9 at 4:51
add a comment |
up vote
9
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.
Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
– bof
Dec 9 at 3:43
2
@bof This question that I asked and Noah answered a while ago should answer your question
– Paul Plummer
Dec 9 at 4:51
add a comment |
up vote
9
down vote
up vote
9
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.
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.
answered Dec 9 at 2:58
Noah Schweber
120k10146278
120k10146278
Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
– bof
Dec 9 at 3:43
2
@bof This question that I asked and Noah answered a while ago should answer your question
– Paul Plummer
Dec 9 at 4:51
add a comment |
Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
– bof
Dec 9 at 3:43
2
@bof This question that I asked and Noah answered a while ago should answer your question
– Paul Plummer
Dec 9 at 4:51
Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
– bof
Dec 9 at 3:43
Both answers use the axiom of choice. In ZF, can there be an uncountable group with only a countable number of subgroups?
– bof
Dec 9 at 3:43
2
2
@bof This question that I asked and Noah answered a while ago should answer your question
– Paul Plummer
Dec 9 at 4:51
@bof This question that I asked and Noah answered a while ago should answer your question
– Paul Plummer
Dec 9 at 4:51
add a comment |
5
I'm frankly a bit surprised at the negative reaction to this question.
– Noah Schweber
Dec 9 at 3:00
2
Possible duplicate of Countable number of subgroups $implies $ countable group
– Carmeister
Dec 9 at 10:05
Then why is it marked as "off-topic" rather than "duplicate"?
– C Monsour
Dec 9 at 20:42
I'm voting to reopen. If it should be closed please give the correct reason.
– C Monsour
Dec 9 at 20:44
@CMonsour It was closed because there is no context. It's just a problem statement with no details. The closure reason is correct. While it might be better that it be listed as a duplicate, I don't see that as a reason to reopen.
– jgon
Dec 9 at 21:10