Mixing
Hereditarily indecomposable groups
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Question. Is it true that each uncountable group $G$ contains an uncountable subgroup $A$ and an infinite subgroup $B$ such that $Acap B=1$? What will be the answer if we additionally require that $ab=ba$ for all $ain A$ and $bin B$?
gr.group-theory set-theory
asked 58 mins ago
Taras Banakh
14k12882
add a comment |Â
up vote
2
down vote
favorite
Question. Is it true that each uncountable group $G$ contains an uncountable subgroup $A$ and an infinite subgroup $B$ such that $Acap B=1$? What will be the answer if we additionally require that $ab=ba$ for all $ain A$ and $bin B$?
gr.group-theory set-theory
asked 58 mins ago
Taras Banakh
14k12882
If I recall correctly, for the abelian case the answer is positive. However it is easy to arrange a model without choice where there is an uncountable group which is indecomposable. In fact, you can even have it as a vector space over your favorite field if you want to.
– Asaf Karagila
31 mins ago
Yes, the positive answer in the abelian case (in ZFC) is an exercise (any uncountable abelian group of cardinal $alpha$ admits a subgroup that is a free $Z/pZ$-module of rank $alpha$ for some $p$ either prime or zero). The nilpotent case seems to be not too hard to deduce.
– YCor
25 mins ago
An alternative approach to Mohammad's answer would be to have uncountable groups in which the intersection of all nontrivial subgroups is nonzero (which forces being torsion), or just such that the intersection any two nontrivial subgroups is nonzero (which forces being either torsion or torsion-free). However, I have not found anything in a quick Google search.
– YCor
21 mins ago
PS beware that the use of "indecomposable" and "hereditarily" are quite far from the standard uses.
– YCor
19 mins ago
@YCor I just wanted to call the problem somehow in an attractive way. But it relates anyway, if "decomposable" means "representable as the direct sum of two non-trivial factors".
– Taras Banakh
14 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Question. Is it true that each uncountable group $G$ contains an uncountable subgroup $A$ and an infinite subgroup $B$ such that $Acap B=1$? What will be the answer if we additionally require that $ab=ba$ for all $ain A$ and $bin B$?
gr.group-theory set-theory
asked 58 mins ago
Taras Banakh
14k12882
Question. Is it true that each uncountable group $G$ contains an uncountable subgroup $A$ and an infinite subgroup $B$ such that $Acap B=1$? What will be the answer if we additionally require that $ab=ba$ for all $ain A$ and $bin B$?
gr.group-theory set-theory
gr.group-theory set-theory
asked 58 mins ago
Taras Banakh
14k12882
asked 58 mins ago
Taras Banakh
14k12882
asked 58 mins ago
Taras Banakh
14k12882
asked 58 mins ago
Taras Banakh
14k12882
asked 58 mins ago
Taras Banakh
14k12882
14k12882
If I recall correctly, for the abelian case the answer is positive. However it is easy to arrange a model without choice where there is an uncountable group which is indecomposable. In fact, you can even have it as a vector space over your favorite field if you want to.
– Asaf Karagila
31 mins ago
Yes, the positive answer in the abelian case (in ZFC) is an exercise (any uncountable abelian group of cardinal $alpha$ admits a subgroup that is a free $Z/pZ$-module of rank $alpha$ for some $p$ either prime or zero). The nilpotent case seems to be not too hard to deduce.
– YCor
25 mins ago
An alternative approach to Mohammad's answer would be to have uncountable groups in which the intersection of all nontrivial subgroups is nonzero (which forces being torsion), or just such that the intersection any two nontrivial subgroups is nonzero (which forces being either torsion or torsion-free). However, I have not found anything in a quick Google search.
– YCor
21 mins ago
PS beware that the use of "indecomposable" and "hereditarily" are quite far from the standard uses.
– YCor
19 mins ago
@YCor I just wanted to call the problem somehow in an attractive way. But it relates anyway, if "decomposable" means "representable as the direct sum of two non-trivial factors".
– Taras Banakh
14 mins ago
add a comment |Â
If I recall correctly, for the abelian case the answer is positive. However it is easy to arrange a model without choice where there is an uncountable group which is indecomposable. In fact, you can even have it as a vector space over your favorite field if you want to.
– Asaf Karagila
31 mins ago
Yes, the positive answer in the abelian case (in ZFC) is an exercise (any uncountable abelian group of cardinal $alpha$ admits a subgroup that is a free $Z/pZ$-module of rank $alpha$ for some $p$ either prime or zero). The nilpotent case seems to be not too hard to deduce.
– YCor
25 mins ago
An alternative approach to Mohammad's answer would be to have uncountable groups in which the intersection of all nontrivial subgroups is nonzero (which forces being torsion), or just such that the intersection any two nontrivial subgroups is nonzero (which forces being either torsion or torsion-free). However, I have not found anything in a quick Google search.
– YCor
21 mins ago
PS beware that the use of "indecomposable" and "hereditarily" are quite far from the standard uses.
– YCor
19 mins ago
@YCor I just wanted to call the problem somehow in an attractive way. But it relates anyway, if "decomposable" means "representable as the direct sum of two non-trivial factors".
– Taras Banakh
14 mins ago
If I recall correctly, for the abelian case the answer is positive. However it is easy to arrange a model without choice where there is an uncountable group which is indecomposable. In fact, you can even have it as a vector space over your favorite field if you want to.
– Asaf Karagila
31 mins ago
If I recall correctly, for the abelian case the answer is positive. However it is easy to arrange a model without choice where there is an uncountable group which is indecomposable. In fact, you can even have it as a vector space over your favorite field if you want to.
– Asaf Karagila
31 mins ago
Yes, the positive answer in the abelian case (in ZFC) is an exercise (any uncountable abelian group of cardinal $alpha$ admits a subgroup that is a free $Z/pZ$-module of rank $alpha$ for some $p$ either prime or zero). The nilpotent case seems to be not too hard to deduce.
– YCor
25 mins ago
Yes, the positive answer in the abelian case (in ZFC) is an exercise (any uncountable abelian group of cardinal $alpha$ admits a subgroup that is a free $Z/pZ$-module of rank $alpha$ for some $p$ either prime or zero). The nilpotent case seems to be not too hard to deduce.
– YCor
25 mins ago
An alternative approach to Mohammad's answer would be to have uncountable groups in which the intersection of all nontrivial subgroups is nonzero (which forces being torsion), or just such that the intersection any two nontrivial subgroups is nonzero (which forces being either torsion or torsion-free). However, I have not found anything in a quick Google search.
– YCor
21 mins ago
An alternative approach to Mohammad's answer would be to have uncountable groups in which the intersection of all nontrivial subgroups is nonzero (which forces being torsion), or just such that the intersection any two nontrivial subgroups is nonzero (which forces being either torsion or torsion-free). However, I have not found anything in a quick Google search.
– YCor
21 mins ago
PS beware that the use of "indecomposable" and "hereditarily" are quite far from the standard uses.
– YCor
19 mins ago
PS beware that the use of "indecomposable" and "hereditarily" are quite far from the standard uses.
– YCor
19 mins ago
@YCor I just wanted to call the problem somehow in an attractive way. But it relates anyway, if "decomposable" means "representable as the direct sum of two non-trivial factors".
– Taras Banakh
14 mins ago
@YCor I just wanted to call the problem somehow in an attractive way. But it relates anyway, if "decomposable" means "representable as the direct sum of two non-trivial factors".
– Taras Banakh
14 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
This is not rue, by Shelah's construction of a Jonsson group
of cardinality $aleph_1$ (a group of size $aleph_1$ for which every proper subgroup is countable). See On a problem of Kurosh, Jónsson groups, and applications
answered 42 mins ago
Mohammad Golshani
18.4k264144
I cannot open the paper of Shelah :( Does his construction hold in ZFC or require some axiom (like diamond)?
– Taras Banakh
15 mins ago
1
@TarasBanakh it's a ZFC construction.
– YCor
11 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This is not rue, by Shelah's construction of a Jonsson group
of cardinality $aleph_1$ (a group of size $aleph_1$ for which every proper subgroup is countable). See On a problem of Kurosh, Jónsson groups, and applications
answered 42 mins ago
Mohammad Golshani
18.4k264144
I cannot open the paper of Shelah :( Does his construction hold in ZFC or require some axiom (like diamond)?
– Taras Banakh
15 mins ago
1
@TarasBanakh it's a ZFC construction.
– YCor
11 mins ago
add a comment |Â
up vote
3
down vote
accepted
This is not rue, by Shelah's construction of a Jonsson group
of cardinality $aleph_1$ (a group of size $aleph_1$ for which every proper subgroup is countable). See On a problem of Kurosh, Jónsson groups, and applications
answered 42 mins ago
Mohammad Golshani
18.4k264144
I cannot open the paper of Shelah :( Does his construction hold in ZFC or require some axiom (like diamond)?
– Taras Banakh
15 mins ago
1
@TarasBanakh it's a ZFC construction.
– YCor
11 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This is not rue, by Shelah's construction of a Jonsson group
of cardinality $aleph_1$ (a group of size $aleph_1$ for which every proper subgroup is countable). See On a problem of Kurosh, Jónsson groups, and applications
answered 42 mins ago
Mohammad Golshani
18.4k264144
This is not rue, by Shelah's construction of a Jonsson group
of cardinality $aleph_1$ (a group of size $aleph_1$ for which every proper subgroup is countable). See On a problem of Kurosh, Jónsson groups, and applications
answered 42 mins ago
Mohammad Golshani
18.4k264144
answered 42 mins ago
Mohammad Golshani
18.4k264144
answered 42 mins ago
Mohammad Golshani
18.4k264144
answered 42 mins ago
Mohammad Golshani
18.4k264144
18.4k264144
I cannot open the paper of Shelah :( Does his construction hold in ZFC or require some axiom (like diamond)?
– Taras Banakh
15 mins ago
1
@TarasBanakh it's a ZFC construction.
– YCor
11 mins ago
add a comment |Â
I cannot open the paper of Shelah :( Does his construction hold in ZFC or require some axiom (like diamond)?
– Taras Banakh
15 mins ago
1
@TarasBanakh it's a ZFC construction.
– YCor
11 mins ago
I cannot open the paper of Shelah :( Does his construction hold in ZFC or require some axiom (like diamond)?
– Taras Banakh
15 mins ago
I cannot open the paper of Shelah :( Does his construction hold in ZFC or require some axiom (like diamond)?
– Taras Banakh
15 mins ago
1
1
@TarasBanakh it's a ZFC construction.
– YCor
11 mins ago
@TarasBanakh it's a ZFC construction.
– YCor
11 mins ago
add a comment |Â
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If I recall correctly, for the abelian case the answer is positive. However it is easy to arrange a model without choice where there is an uncountable group which is indecomposable. In fact, you can even have it as a vector space over your favorite field if you want to.
– Asaf Karagila
31 mins ago
Yes, the positive answer in the abelian case (in ZFC) is an exercise (any uncountable abelian group of cardinal $alpha$ admits a subgroup that is a free $Z/pZ$-module of rank $alpha$ for some $p$ either prime or zero). The nilpotent case seems to be not too hard to deduce.
– YCor
25 mins ago
An alternative approach to Mohammad's answer would be to have uncountable groups in which the intersection of all nontrivial subgroups is nonzero (which forces being torsion), or just such that the intersection any two nontrivial subgroups is nonzero (which forces being either torsion or torsion-free). However, I have not found anything in a quick Google search.
– YCor
21 mins ago
PS beware that the use of "indecomposable" and "hereditarily" are quite far from the standard uses.
– YCor
19 mins ago
@YCor I just wanted to call the problem somehow in an attractive way. But it relates anyway, if "decomposable" means "representable as the direct sum of two non-trivial factors".
– Taras Banakh
14 mins ago