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If a single integer power kills every element of abelian group $G$, then $G$ is finite (UPDATE after shown false)
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Is it true that for any abelian group $G$, if there is some integer $k$ satisfying $g^k = e$ for every $g in G$, then $G$ is finite?
This is true if $G$ is finitely generated.
I believe that it is true in the general case but am struggling to prove this. Any help is appreciated.
EDIT: Alan Wang points out a simple counterexample to the proposition.
My original idea for this question came from Keith Conrad's algebra notes, which has the following exercise.
Show a finitely generated $A$-module $M$ is a torsion module if and only if there is some $a ne 0$ in $A$ such that $aM = 0$. (This is false without a hypothesis of finite generatedness even for $A = mathbbZ$, since infinite torsion abelian groups exist.)
It's the parenthetical which now has me confused. How is an infinite torsion abelian group an automatic counterexample?
abstract-algebra group-theory
asked Jul 21 at 6:41
CuriousKid7
1,438517
add a comment |Â
up vote
1
down vote
favorite
Is it true that for any abelian group $G$, if there is some integer $k$ satisfying $g^k = e$ for every $g in G$, then $G$ is finite?
This is true if $G$ is finitely generated.
I believe that it is true in the general case but am struggling to prove this. Any help is appreciated.
EDIT: Alan Wang points out a simple counterexample to the proposition.
My original idea for this question came from Keith Conrad's algebra notes, which has the following exercise.
Show a finitely generated $A$-module $M$ is a torsion module if and only if there is some $a ne 0$ in $A$ such that $aM = 0$. (This is false without a hypothesis of finite generatedness even for $A = mathbbZ$, since infinite torsion abelian groups exist.)
It's the parenthetical which now has me confused. How is an infinite torsion abelian group an automatic counterexample?
abstract-algebra group-theory
asked Jul 21 at 6:41
CuriousKid7
1,438517
4
What about infinite direct product of $BbbZ_2$?
– Alan Wang
Jul 21 at 6:43
1
It is false, take $C_p^mathbbN$ for example.
– jgon
Jul 21 at 6:43
Why couldn't the group have an infinite number of elements all of finite order?
– fleablood
Jul 21 at 6:44
"This is true if G is finitely generated." And it'd be false if it where infinitely generated. So the question is: is it possible to have an infinitely generated group where every element has a finite order? And ... why the heck not? As Alan Wang suggests why not $mathbb Z_2^mathbb N$ or, more generally, $ times_iin mathbb N mathbb Z_k_i$
– fleablood
Jul 21 at 6:49
1
For nonabelian groups you might want to explore the Burnside Problem. It is now known that there are infinite groups which are finitely generated and in which all the elements have bounded order.
– Mark Bennet
Jul 21 at 7:03
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is it true that for any abelian group $G$, if there is some integer $k$ satisfying $g^k = e$ for every $g in G$, then $G$ is finite?
This is true if $G$ is finitely generated.
I believe that it is true in the general case but am struggling to prove this. Any help is appreciated.
EDIT: Alan Wang points out a simple counterexample to the proposition.
My original idea for this question came from Keith Conrad's algebra notes, which has the following exercise.
Show a finitely generated $A$-module $M$ is a torsion module if and only if there is some $a ne 0$ in $A$ such that $aM = 0$. (This is false without a hypothesis of finite generatedness even for $A = mathbbZ$, since infinite torsion abelian groups exist.)
It's the parenthetical which now has me confused. How is an infinite torsion abelian group an automatic counterexample?
abstract-algebra group-theory
asked Jul 21 at 6:41
CuriousKid7
1,438517
Is it true that for any abelian group $G$, if there is some integer $k$ satisfying $g^k = e$ for every $g in G$, then $G$ is finite?
This is true if $G$ is finitely generated.
I believe that it is true in the general case but am struggling to prove this. Any help is appreciated.
EDIT: Alan Wang points out a simple counterexample to the proposition.
My original idea for this question came from Keith Conrad's algebra notes, which has the following exercise.
Show a finitely generated $A$-module $M$ is a torsion module if and only if there is some $a ne 0$ in $A$ such that $aM = 0$. (This is false without a hypothesis of finite generatedness even for $A = mathbbZ$, since infinite torsion abelian groups exist.)
It's the parenthetical which now has me confused. How is an infinite torsion abelian group an automatic counterexample?
abstract-algebra group-theory
asked Jul 21 at 6:41
CuriousKid7
1,438517
edited Jul 21 at 7:02
asked Jul 21 at 6:41
CuriousKid7
1,438517
asked Jul 21 at 6:41
CuriousKid7
1,438517
asked Jul 21 at 6:41
CuriousKid7
1,438517
1,438517
4
What about infinite direct product of $BbbZ_2$?
– Alan Wang
Jul 21 at 6:43
1
It is false, take $C_p^mathbbN$ for example.
– jgon
Jul 21 at 6:43
Why couldn't the group have an infinite number of elements all of finite order?
– fleablood
Jul 21 at 6:44
"This is true if G is finitely generated." And it'd be false if it where infinitely generated. So the question is: is it possible to have an infinitely generated group where every element has a finite order? And ... why the heck not? As Alan Wang suggests why not $mathbb Z_2^mathbb N$ or, more generally, $ times_iin mathbb N mathbb Z_k_i$
– fleablood
Jul 21 at 6:49
1
For nonabelian groups you might want to explore the Burnside Problem. It is now known that there are infinite groups which are finitely generated and in which all the elements have bounded order.
– Mark Bennet
Jul 21 at 7:03
add a comment |Â
4
What about infinite direct product of $BbbZ_2$?
– Alan Wang
Jul 21 at 6:43
1
It is false, take $C_p^mathbbN$ for example.
– jgon
Jul 21 at 6:43
Why couldn't the group have an infinite number of elements all of finite order?
– fleablood
Jul 21 at 6:44
"This is true if G is finitely generated." And it'd be false if it where infinitely generated. So the question is: is it possible to have an infinitely generated group where every element has a finite order? And ... why the heck not? As Alan Wang suggests why not $mathbb Z_2^mathbb N$ or, more generally, $ times_iin mathbb N mathbb Z_k_i$
– fleablood
Jul 21 at 6:49
1
For nonabelian groups you might want to explore the Burnside Problem. It is now known that there are infinite groups which are finitely generated and in which all the elements have bounded order.
– Mark Bennet
Jul 21 at 7:03
4
4
What about infinite direct product of $BbbZ_2$?
– Alan Wang
Jul 21 at 6:43
What about infinite direct product of $BbbZ_2$?
– Alan Wang
Jul 21 at 6:43
1
1
It is false, take $C_p^mathbbN$ for example.
– jgon
Jul 21 at 6:43
It is false, take $C_p^mathbbN$ for example.
– jgon
Jul 21 at 6:43
Why couldn't the group have an infinite number of elements all of finite order?
– fleablood
Jul 21 at 6:44
Why couldn't the group have an infinite number of elements all of finite order?
– fleablood
Jul 21 at 6:44
"This is true if G is finitely generated." And it'd be false if it where infinitely generated. So the question is: is it possible to have an infinitely generated group where every element has a finite order? And ... why the heck not? As Alan Wang suggests why not $mathbb Z_2^mathbb N$ or, more generally, $ times_iin mathbb N mathbb Z_k_i$
– fleablood
Jul 21 at 6:49
"This is true if G is finitely generated." And it'd be false if it where infinitely generated. So the question is: is it possible to have an infinitely generated group where every element has a finite order? And ... why the heck not? As Alan Wang suggests why not $mathbb Z_2^mathbb N$ or, more generally, $ times_iin mathbb N mathbb Z_k_i$
– fleablood
Jul 21 at 6:49
1
1
For nonabelian groups you might want to explore the Burnside Problem. It is now known that there are infinite groups which are finitely generated and in which all the elements have bounded order.
– Mark Bennet
Jul 21 at 7:03
For nonabelian groups you might want to explore the Burnside Problem. It is now known that there are infinite groups which are finitely generated and in which all the elements have bounded order.
– Mark Bennet
Jul 21 at 7:03
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Existence of an infinite torsion group in itself does not necessarily provide a counterexample for the (parenthesised part of the) proposition.
However, consider e.g. $Bbb Q/Bbb Z$. Here each element $a/b$ is killed by an integer ($b$), but no integer kills the whole group.
answered Jul 21 at 8:19
Berci
56.4k23570
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Existence of an infinite torsion group in itself does not necessarily provide a counterexample for the (parenthesised part of the) proposition.
However, consider e.g. $Bbb Q/Bbb Z$. Here each element $a/b$ is killed by an integer ($b$), but no integer kills the whole group.
answered Jul 21 at 8:19
Berci
56.4k23570
add a comment |Â
up vote
2
down vote
accepted
Existence of an infinite torsion group in itself does not necessarily provide a counterexample for the (parenthesised part of the) proposition.
However, consider e.g. $Bbb Q/Bbb Z$. Here each element $a/b$ is killed by an integer ($b$), but no integer kills the whole group.
answered Jul 21 at 8:19
Berci
56.4k23570
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Existence of an infinite torsion group in itself does not necessarily provide a counterexample for the (parenthesised part of the) proposition.
However, consider e.g. $Bbb Q/Bbb Z$. Here each element $a/b$ is killed by an integer ($b$), but no integer kills the whole group.
answered Jul 21 at 8:19
Berci
56.4k23570
Existence of an infinite torsion group in itself does not necessarily provide a counterexample for the (parenthesised part of the) proposition.
However, consider e.g. $Bbb Q/Bbb Z$. Here each element $a/b$ is killed by an integer ($b$), but no integer kills the whole group.
answered Jul 21 at 8:19
Berci
56.4k23570
answered Jul 21 at 8:19
Berci
56.4k23570
answered Jul 21 at 8:19
Berci
56.4k23570
answered Jul 21 at 8:19
Berci
56.4k23570
56.4k23570
add a comment |Â
add a comment |Â
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4
What about infinite direct product of $BbbZ_2$?
– Alan Wang
Jul 21 at 6:43
1
It is false, take $C_p^mathbbN$ for example.
– jgon
Jul 21 at 6:43
Why couldn't the group have an infinite number of elements all of finite order?
– fleablood
Jul 21 at 6:44
"This is true if G is finitely generated." And it'd be false if it where infinitely generated. So the question is: is it possible to have an infinitely generated group where every element has a finite order? And ... why the heck not? As Alan Wang suggests why not $mathbb Z_2^mathbb N$ or, more generally, $ times_iin mathbb N mathbb Z_k_i$
– fleablood
Jul 21 at 6:49
1
For nonabelian groups you might want to explore the Burnside Problem. It is now known that there are infinite groups which are finitely generated and in which all the elements have bounded order.
– Mark Bennet
Jul 21 at 7:03