Subgroup of $C^*$ (nonzero complex) with finite index.
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True or false:
Let $C^*$ be the set of all nonzero complex numbers and $H$ be a subgroup of $C^*$(with respect to multiplication) be such that $[C^*:H]$ is finite then $H=C^*$.
I'm guessing it true as I am thinking that if suppose there is such a proper subgroup $H$ for which the number of coset will be finite then I'm guessing that there is a gap between $C^*$ and $H$ and that gap cannot be filled up by finite union. But I am unable to give a concrete prove.
Thanks in advance.
abstract-algebra group-theory
asked Mar 20 '16 at 20:01
user322390
9915
add a comment |
up vote
4
down vote
favorite
True or false:
Let $C^*$ be the set of all nonzero complex numbers and $H$ be a subgroup of $C^*$(with respect to multiplication) be such that $[C^*:H]$ is finite then $H=C^*$.
I'm guessing it true as I am thinking that if suppose there is such a proper subgroup $H$ for which the number of coset will be finite then I'm guessing that there is a gap between $C^*$ and $H$ and that gap cannot be filled up by finite union. But I am unable to give a concrete prove.
Thanks in advance.
abstract-algebra group-theory
asked Mar 20 '16 at 20:01
user322390
9915
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
True or false:
Let $C^*$ be the set of all nonzero complex numbers and $H$ be a subgroup of $C^*$(with respect to multiplication) be such that $[C^*:H]$ is finite then $H=C^*$.
I'm guessing it true as I am thinking that if suppose there is such a proper subgroup $H$ for which the number of coset will be finite then I'm guessing that there is a gap between $C^*$ and $H$ and that gap cannot be filled up by finite union. But I am unable to give a concrete prove.
Thanks in advance.
abstract-algebra group-theory
asked Mar 20 '16 at 20:01
user322390
9915
True or false:
Let $C^*$ be the set of all nonzero complex numbers and $H$ be a subgroup of $C^*$(with respect to multiplication) be such that $[C^*:H]$ is finite then $H=C^*$.
I'm guessing it true as I am thinking that if suppose there is such a proper subgroup $H$ for which the number of coset will be finite then I'm guessing that there is a gap between $C^*$ and $H$ and that gap cannot be filled up by finite union. But I am unable to give a concrete prove.
Thanks in advance.
abstract-algebra group-theory
abstract-algebra group-theory
asked Mar 20 '16 at 20:01
user322390
9915
asked Mar 20 '16 at 20:01
user322390
9915
asked Mar 20 '16 at 20:01
user322390
9915
asked Mar 20 '16 at 20:01
user322390
9915
asked Mar 20 '16 at 20:01
user322390
9915
9915
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
Suppose $H$ has finite index in $Bbb C^times$, let $m=[Bbb C^times:H]$. Then for any nonzero complex number, $z^min H$. Now given $winBbb C$ we can always solve $z^m-w=0$, so $win H$.
Alternatively, a finite divisible abelian group is trivial. Now $Bbb C^times /H$ is finite, and it is divisible, so it must be trivial.
answered Mar 20 '16 at 20:23
Pedro Tamaroff♦
95.8k10150295
Why is $z^m in H$? I don't follow that one step in this proof.
– Mehdi2277
Mar 21 '16 at 6:04
$(zH)^m=H$ as the order of the group ${gH: gin C^*}$ is m. Now $(zH)^m = (z^m)H = H$, so $z^m in H$.
– user322390
Apr 1 '16 at 10:32
The claim that a torsion divisible abelian group is trivial is blatantly false. However it holds that a finite divisible abelian group is trivial.
– YCor
Jun 2 at 9:44
@YCor True, what I wanted to say is that finite (or torsion of finite index?) plus divisible implies trivial. An obvious counterexample is $mathbb Q/mathbb Z$.
– Pedro Tamaroff♦
Jun 2 at 9:48
add a comment |
up vote
4
down vote
More generally, an (abelian) divisible group has no nontrivial subgroups of finite index. The proof is essentially the same as for this question.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 20 '16 at 20:14
user324381
1112
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Suppose $H$ has finite index in $Bbb C^times$, let $m=[Bbb C^times:H]$. Then for any nonzero complex number, $z^min H$. Now given $winBbb C$ we can always solve $z^m-w=0$, so $win H$.
Alternatively, a finite divisible abelian group is trivial. Now $Bbb C^times /H$ is finite, and it is divisible, so it must be trivial.
answered Mar 20 '16 at 20:23
Pedro Tamaroff♦
95.8k10150295
Why is $z^m in H$? I don't follow that one step in this proof.
– Mehdi2277
Mar 21 '16 at 6:04
$(zH)^m=H$ as the order of the group ${gH: gin C^*}$ is m. Now $(zH)^m = (z^m)H = H$, so $z^m in H$.
– user322390
Apr 1 '16 at 10:32
The claim that a torsion divisible abelian group is trivial is blatantly false. However it holds that a finite divisible abelian group is trivial.
– YCor
Jun 2 at 9:44
@YCor True, what I wanted to say is that finite (or torsion of finite index?) plus divisible implies trivial. An obvious counterexample is $mathbb Q/mathbb Z$.
– Pedro Tamaroff♦
Jun 2 at 9:48
add a comment |
up vote
7
down vote
accepted
Suppose $H$ has finite index in $Bbb C^times$, let $m=[Bbb C^times:H]$. Then for any nonzero complex number, $z^min H$. Now given $winBbb C$ we can always solve $z^m-w=0$, so $win H$.
Alternatively, a finite divisible abelian group is trivial. Now $Bbb C^times /H$ is finite, and it is divisible, so it must be trivial.
answered Mar 20 '16 at 20:23
Pedro Tamaroff♦
95.8k10150295
Why is $z^m in H$? I don't follow that one step in this proof.
– Mehdi2277
Mar 21 '16 at 6:04
$(zH)^m=H$ as the order of the group ${gH: gin C^*}$ is m. Now $(zH)^m = (z^m)H = H$, so $z^m in H$.
– user322390
Apr 1 '16 at 10:32
The claim that a torsion divisible abelian group is trivial is blatantly false. However it holds that a finite divisible abelian group is trivial.
– YCor
Jun 2 at 9:44
@YCor True, what I wanted to say is that finite (or torsion of finite index?) plus divisible implies trivial. An obvious counterexample is $mathbb Q/mathbb Z$.
– Pedro Tamaroff♦
Jun 2 at 9:48
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Suppose $H$ has finite index in $Bbb C^times$, let $m=[Bbb C^times:H]$. Then for any nonzero complex number, $z^min H$. Now given $winBbb C$ we can always solve $z^m-w=0$, so $win H$.
Alternatively, a finite divisible abelian group is trivial. Now $Bbb C^times /H$ is finite, and it is divisible, so it must be trivial.
answered Mar 20 '16 at 20:23
Pedro Tamaroff♦
95.8k10150295
Suppose $H$ has finite index in $Bbb C^times$, let $m=[Bbb C^times:H]$. Then for any nonzero complex number, $z^min H$. Now given $winBbb C$ we can always solve $z^m-w=0$, so $win H$.
Alternatively, a finite divisible abelian group is trivial. Now $Bbb C^times /H$ is finite, and it is divisible, so it must be trivial.
answered Mar 20 '16 at 20:23
Pedro Tamaroff♦
95.8k10150295
edited Jun 2 at 9:49
answered Mar 20 '16 at 20:23
Pedro Tamaroff♦
95.8k10150295
answered Mar 20 '16 at 20:23
Pedro Tamaroff♦
95.8k10150295
answered Mar 20 '16 at 20:23
Pedro Tamaroff♦
95.8k10150295
95.8k10150295
Why is $z^m in H$? I don't follow that one step in this proof.
– Mehdi2277
Mar 21 '16 at 6:04
$(zH)^m=H$ as the order of the group ${gH: gin C^*}$ is m. Now $(zH)^m = (z^m)H = H$, so $z^m in H$.
– user322390
Apr 1 '16 at 10:32
The claim that a torsion divisible abelian group is trivial is blatantly false. However it holds that a finite divisible abelian group is trivial.
– YCor
Jun 2 at 9:44
@YCor True, what I wanted to say is that finite (or torsion of finite index?) plus divisible implies trivial. An obvious counterexample is $mathbb Q/mathbb Z$.
– Pedro Tamaroff♦
Jun 2 at 9:48
add a comment |
Why is $z^m in H$? I don't follow that one step in this proof.
– Mehdi2277
Mar 21 '16 at 6:04
$(zH)^m=H$ as the order of the group ${gH: gin C^*}$ is m. Now $(zH)^m = (z^m)H = H$, so $z^m in H$.
– user322390
Apr 1 '16 at 10:32
The claim that a torsion divisible abelian group is trivial is blatantly false. However it holds that a finite divisible abelian group is trivial.
– YCor
Jun 2 at 9:44
@YCor True, what I wanted to say is that finite (or torsion of finite index?) plus divisible implies trivial. An obvious counterexample is $mathbb Q/mathbb Z$.
– Pedro Tamaroff♦
Jun 2 at 9:48
Why is $z^m in H$? I don't follow that one step in this proof.
– Mehdi2277
Mar 21 '16 at 6:04
Why is $z^m in H$? I don't follow that one step in this proof.
– Mehdi2277
Mar 21 '16 at 6:04
$(zH)^m=H$ as the order of the group ${gH: gin C^*}$ is m. Now $(zH)^m = (z^m)H = H$, so $z^m in H$.
– user322390
Apr 1 '16 at 10:32
$(zH)^m=H$ as the order of the group ${gH: gin C^*}$ is m. Now $(zH)^m = (z^m)H = H$, so $z^m in H$.
– user322390
Apr 1 '16 at 10:32
The claim that a torsion divisible abelian group is trivial is blatantly false. However it holds that a finite divisible abelian group is trivial.
– YCor
Jun 2 at 9:44
The claim that a torsion divisible abelian group is trivial is blatantly false. However it holds that a finite divisible abelian group is trivial.
– YCor
Jun 2 at 9:44
@YCor True, what I wanted to say is that finite (or torsion of finite index?) plus divisible implies trivial. An obvious counterexample is $mathbb Q/mathbb Z$.
– Pedro Tamaroff♦
Jun 2 at 9:48
@YCor True, what I wanted to say is that finite (or torsion of finite index?) plus divisible implies trivial. An obvious counterexample is $mathbb Q/mathbb Z$.
– Pedro Tamaroff♦
Jun 2 at 9:48
add a comment |
up vote
4
down vote
More generally, an (abelian) divisible group has no nontrivial subgroups of finite index. The proof is essentially the same as for this question.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 20 '16 at 20:14
user324381
1112
add a comment |
up vote
4
down vote
More generally, an (abelian) divisible group has no nontrivial subgroups of finite index. The proof is essentially the same as for this question.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 20 '16 at 20:14
user324381
1112
add a comment |
up vote
4
down vote
up vote
4
down vote
More generally, an (abelian) divisible group has no nontrivial subgroups of finite index. The proof is essentially the same as for this question.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 20 '16 at 20:14
user324381
1112
More generally, an (abelian) divisible group has no nontrivial subgroups of finite index. The proof is essentially the same as for this question.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 20 '16 at 20:14
user324381
1112
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Mar 20 '16 at 20:14
user324381
1112
answered Mar 20 '16 at 20:14
user324381
1112
answered Mar 20 '16 at 20:14
user324381
1112
1112
add a comment |
add a comment |
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