Can an infinite group have a finite number of elements with order k?Example of a group which is abelian and has finite (except the $e$) and infinite order elements.Infinite group must have infinite subgroups.Cyclic Groups of Infinite and Finite orderthe number of elements of order 2 in infinite groupInfinite group with finite order elementsGive an example of a nonabelian group in which a product of elements of finite order can have infinite order.Number of elements of order $n$ in an infinite group?Infinite non abelian group with finite order elementsExample of infinite abelian group having exactly $n$ elements of finite order.infinite group with exactly $n$ elements of finite order.
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Can an infinite group have a finite number of elements with order k?
Example of a group which is abelian and has finite (except the $e$) and infinite order elements.Infinite group must have infinite subgroups.Cyclic Groups of Infinite and Finite orderthe number of elements of order 2 in infinite groupInfinite group with finite order elementsGive an example of a nonabelian group in which a product of elements of finite order can have infinite order.Number of elements of order $n$ in an infinite group?Infinite non abelian group with finite order elementsExample of infinite abelian group having exactly $n$ elements of finite order.infinite group with exactly $n$ elements of finite order.
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$begingroup$
As stated in the title, is there an infinite group with finite number of elements with finite order k?
group-theory examples-counterexamples infinite-groups
edited 8 hours ago
Shaun
11.3k12 gold badges36 silver badges90 bronze badges
asked 9 hours ago
Utkarsh KumarUtkarsh Kumar
504 bronze badges
$endgroup$
add a comment |
$begingroup$
As stated in the title, is there an infinite group with finite number of elements with finite order k?
group-theory examples-counterexamples infinite-groups
edited 8 hours ago
Shaun
11.3k12 gold badges36 silver badges90 bronze badges
asked 9 hours ago
Utkarsh KumarUtkarsh Kumar
504 bronze badges
$endgroup$
$begingroup$
Got it. Thank you. I wish I could accept multiple answers.
$endgroup$
– Utkarsh Kumar
8 hours ago
$begingroup$
You could at least upvote them, @UtkarshKumar.
$endgroup$
– Shaun
8 hours ago
2
$begingroup$
Yes, I did that. Yours was the latest, and hence I hadn't seen it yet.
$endgroup$
– Utkarsh Kumar
8 hours ago
2
$begingroup$
Wouldn't it be sufficient to consider a group with no elements of finite order except the identity? (E.g. $mathbbZ$) I may have misunderstood the question.
$endgroup$
– Sambo
8 hours ago
add a comment |
$begingroup$
As stated in the title, is there an infinite group with finite number of elements with finite order k?
group-theory examples-counterexamples infinite-groups
edited 8 hours ago
Shaun
11.3k12 gold badges36 silver badges90 bronze badges
asked 9 hours ago
Utkarsh KumarUtkarsh Kumar
504 bronze badges
$endgroup$
As stated in the title, is there an infinite group with finite number of elements with finite order k?
group-theory examples-counterexamples infinite-groups
group-theory examples-counterexamples infinite-groups
edited 8 hours ago
Shaun
11.3k12 gold badges36 silver badges90 bronze badges
asked 9 hours ago
Utkarsh KumarUtkarsh Kumar
504 bronze badges
edited 8 hours ago
Shaun
11.3k12 gold badges36 silver badges90 bronze badges
asked 9 hours ago
Utkarsh KumarUtkarsh Kumar
504 bronze badges
edited 8 hours ago
Shaun
11.3k12 gold badges36 silver badges90 bronze badges
edited 8 hours ago
Shaun
11.3k12 gold badges36 silver badges90 bronze badges
edited 8 hours ago
Shaun
11.3k12 gold badges36 silver badges90 bronze badges
11.3k12 gold badges36 silver badges90 bronze badges
asked 9 hours ago
Utkarsh KumarUtkarsh Kumar
504 bronze badges
asked 9 hours ago
Utkarsh KumarUtkarsh Kumar
504 bronze badges
asked 9 hours ago
Utkarsh KumarUtkarsh Kumar
504 bronze badges
504 bronze badges
$begingroup$
Got it. Thank you. I wish I could accept multiple answers.
$endgroup$
– Utkarsh Kumar
8 hours ago
$begingroup$
You could at least upvote them, @UtkarshKumar.
$endgroup$
– Shaun
8 hours ago
2
$begingroup$
Yes, I did that. Yours was the latest, and hence I hadn't seen it yet.
$endgroup$
– Utkarsh Kumar
8 hours ago
2
$begingroup$
Wouldn't it be sufficient to consider a group with no elements of finite order except the identity? (E.g. $mathbbZ$) I may have misunderstood the question.
$endgroup$
– Sambo
8 hours ago
add a comment |
$begingroup$
Got it. Thank you. I wish I could accept multiple answers.
$endgroup$
– Utkarsh Kumar
8 hours ago
$begingroup$
You could at least upvote them, @UtkarshKumar.
$endgroup$
– Shaun
8 hours ago
2
$begingroup$
Yes, I did that. Yours was the latest, and hence I hadn't seen it yet.
$endgroup$
– Utkarsh Kumar
8 hours ago
2
$begingroup$
Wouldn't it be sufficient to consider a group with no elements of finite order except the identity? (E.g. $mathbbZ$) I may have misunderstood the question.
$endgroup$
– Sambo
8 hours ago
$begingroup$
Got it. Thank you. I wish I could accept multiple answers.
$endgroup$
– Utkarsh Kumar
8 hours ago
$begingroup$
Got it. Thank you. I wish I could accept multiple answers.
$endgroup$
– Utkarsh Kumar
8 hours ago
$begingroup$
You could at least upvote them, @UtkarshKumar.
$endgroup$
– Shaun
8 hours ago
$begingroup$
You could at least upvote them, @UtkarshKumar.
$endgroup$
– Shaun
8 hours ago
2
2
$begingroup$
Yes, I did that. Yours was the latest, and hence I hadn't seen it yet.
$endgroup$
– Utkarsh Kumar
8 hours ago
$begingroup$
Yes, I did that. Yours was the latest, and hence I hadn't seen it yet.
$endgroup$
– Utkarsh Kumar
8 hours ago
2
2
$begingroup$
Wouldn't it be sufficient to consider a group with no elements of finite order except the identity? (E.g. $mathbbZ$) I may have misunderstood the question.
$endgroup$
– Sambo
8 hours ago
$begingroup$
Wouldn't it be sufficient to consider a group with no elements of finite order except the identity? (E.g. $mathbbZ$) I may have misunderstood the question.
$endgroup$
– Sambo
8 hours ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Consider $mathbbZtimesmathbbZ_k$. Then, $(0,a)$ is the set, which is finite.
(To prove this, first assume $(b,a)$ has order $k$, then $kb=0Rightarrow b=0$. Then note that $a$ has order $k$ in $mathbbZ_k$ iff $gcd(a,k)=1$)
answered 9 hours ago
C.ParkC.Park
7182 silver badges11 bronze badges
$endgroup$
add a comment |
$begingroup$
Yes. Take $mathbbZ_k times mathbbZ$ with pointwise addition as the operation. Only the elements of the form $(n, 0)$ have finite order, of which there are at most $k$.
answered 8 hours ago
Duncan RamageDuncan Ramage
4,03411 silver badges26 bronze badges
$endgroup$
add a comment |
$begingroup$
The group of rotations of the circle has a finite number of elements of each integral order.
In the infinite group of sequences $(a_2, a_3, a_5, ldots)$ where each $a_p in mathbbZ_p$ (addition modulo the prime $p$) and all but finitely many are $0$ every element has finite order and there are only finitely many of each order.
answered 8 hours ago
Ethan BolkerEthan Bolker
52.9k5 gold badges61 silver badges131 bronze badges
$endgroup$
add a comment |
$begingroup$
This answer was made community wiki due to a technicality. Hopefully, others will learn from it.
Consider the free product given by the following (not necessarily unique) group presentation $$Bbb Z_k+1astBbb Zconglangle x, ymid x^k+1rangle.$$
$endgroup$
$begingroup$
There are infinitely many elements with finite order in this group; there are finitely many up to conjugation.
$endgroup$
– Paul Plummer
7 hours ago
$begingroup$
Ah, yes, of course! I overlooked the possibility of, say, $$underbrace(y^ixy^-1)cdots (y^ixy^-1)_k+1,texttimes$$ for each $i$ and all such products with fewer than $k+1$ terms. Thank you, @PaulPlummer; I should know better by now.
$endgroup$
– Shaun
7 hours ago
add a comment |
$begingroup$
$(mathbbQ,+)$ has finite (none) elements of order $k$ where $k>1$.
answered 5 hours ago
EpsilonDeltaEpsilonDelta
6521 gold badge1 silver badge9 bronze badges
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider $mathbbZtimesmathbbZ_k$. Then, $(0,a)$ is the set, which is finite.
(To prove this, first assume $(b,a)$ has order $k$, then $kb=0Rightarrow b=0$. Then note that $a$ has order $k$ in $mathbbZ_k$ iff $gcd(a,k)=1$)
answered 9 hours ago
C.ParkC.Park
7182 silver badges11 bronze badges
$endgroup$
add a comment |
$begingroup$
Consider $mathbbZtimesmathbbZ_k$. Then, $(0,a)$ is the set, which is finite.
(To prove this, first assume $(b,a)$ has order $k$, then $kb=0Rightarrow b=0$. Then note that $a$ has order $k$ in $mathbbZ_k$ iff $gcd(a,k)=1$)
answered 9 hours ago
C.ParkC.Park
7182 silver badges11 bronze badges
$endgroup$
add a comment |
$begingroup$
Consider $mathbbZtimesmathbbZ_k$. Then, $(0,a)$ is the set, which is finite.
(To prove this, first assume $(b,a)$ has order $k$, then $kb=0Rightarrow b=0$. Then note that $a$ has order $k$ in $mathbbZ_k$ iff $gcd(a,k)=1$)
answered 9 hours ago
C.ParkC.Park
7182 silver badges11 bronze badges
$endgroup$
Consider $mathbbZtimesmathbbZ_k$. Then, $(0,a)$ is the set, which is finite.
(To prove this, first assume $(b,a)$ has order $k$, then $kb=0Rightarrow b=0$. Then note that $a$ has order $k$ in $mathbbZ_k$ iff $gcd(a,k)=1$)
answered 9 hours ago
C.ParkC.Park
7182 silver badges11 bronze badges
answered 9 hours ago
C.ParkC.Park
7182 silver badges11 bronze badges
answered 9 hours ago
C.ParkC.Park
7182 silver badges11 bronze badges
answered 9 hours ago
C.ParkC.Park
7182 silver badges11 bronze badges
7182 silver badges11 bronze badges
add a comment |
add a comment |
$begingroup$
Yes. Take $mathbbZ_k times mathbbZ$ with pointwise addition as the operation. Only the elements of the form $(n, 0)$ have finite order, of which there are at most $k$.
answered 8 hours ago
Duncan RamageDuncan Ramage
4,03411 silver badges26 bronze badges
$endgroup$
add a comment |
$begingroup$
Yes. Take $mathbbZ_k times mathbbZ$ with pointwise addition as the operation. Only the elements of the form $(n, 0)$ have finite order, of which there are at most $k$.
answered 8 hours ago
Duncan RamageDuncan Ramage
4,03411 silver badges26 bronze badges
$endgroup$
add a comment |
$begingroup$
Yes. Take $mathbbZ_k times mathbbZ$ with pointwise addition as the operation. Only the elements of the form $(n, 0)$ have finite order, of which there are at most $k$.
answered 8 hours ago
Duncan RamageDuncan Ramage
4,03411 silver badges26 bronze badges
$endgroup$
Yes. Take $mathbbZ_k times mathbbZ$ with pointwise addition as the operation. Only the elements of the form $(n, 0)$ have finite order, of which there are at most $k$.
answered 8 hours ago
Duncan RamageDuncan Ramage
4,03411 silver badges26 bronze badges
answered 8 hours ago
Duncan RamageDuncan Ramage
4,03411 silver badges26 bronze badges
answered 8 hours ago
Duncan RamageDuncan Ramage
4,03411 silver badges26 bronze badges
answered 8 hours ago
Duncan RamageDuncan Ramage
4,03411 silver badges26 bronze badges
4,03411 silver badges26 bronze badges
add a comment |
add a comment |
$begingroup$
The group of rotations of the circle has a finite number of elements of each integral order.
In the infinite group of sequences $(a_2, a_3, a_5, ldots)$ where each $a_p in mathbbZ_p$ (addition modulo the prime $p$) and all but finitely many are $0$ every element has finite order and there are only finitely many of each order.
answered 8 hours ago
Ethan BolkerEthan Bolker
52.9k5 gold badges61 silver badges131 bronze badges
$endgroup$
add a comment |
$begingroup$
The group of rotations of the circle has a finite number of elements of each integral order.
In the infinite group of sequences $(a_2, a_3, a_5, ldots)$ where each $a_p in mathbbZ_p$ (addition modulo the prime $p$) and all but finitely many are $0$ every element has finite order and there are only finitely many of each order.
answered 8 hours ago
Ethan BolkerEthan Bolker
52.9k5 gold badges61 silver badges131 bronze badges
$endgroup$
add a comment |
$begingroup$
The group of rotations of the circle has a finite number of elements of each integral order.
In the infinite group of sequences $(a_2, a_3, a_5, ldots)$ where each $a_p in mathbbZ_p$ (addition modulo the prime $p$) and all but finitely many are $0$ every element has finite order and there are only finitely many of each order.
answered 8 hours ago
Ethan BolkerEthan Bolker
52.9k5 gold badges61 silver badges131 bronze badges
$endgroup$
The group of rotations of the circle has a finite number of elements of each integral order.
In the infinite group of sequences $(a_2, a_3, a_5, ldots)$ where each $a_p in mathbbZ_p$ (addition modulo the prime $p$) and all but finitely many are $0$ every element has finite order and there are only finitely many of each order.
answered 8 hours ago
Ethan BolkerEthan Bolker
52.9k5 gold badges61 silver badges131 bronze badges
edited 8 hours ago
answered 8 hours ago
Ethan BolkerEthan Bolker
52.9k5 gold badges61 silver badges131 bronze badges
answered 8 hours ago
Ethan BolkerEthan Bolker
52.9k5 gold badges61 silver badges131 bronze badges
answered 8 hours ago
Ethan BolkerEthan Bolker
52.9k5 gold badges61 silver badges131 bronze badges
52.9k5 gold badges61 silver badges131 bronze badges
add a comment |
add a comment |
$begingroup$
This answer was made community wiki due to a technicality. Hopefully, others will learn from it.
Consider the free product given by the following (not necessarily unique) group presentation $$Bbb Z_k+1astBbb Zconglangle x, ymid x^k+1rangle.$$
$endgroup$
$begingroup$
There are infinitely many elements with finite order in this group; there are finitely many up to conjugation.
$endgroup$
– Paul Plummer
7 hours ago
$begingroup$
Ah, yes, of course! I overlooked the possibility of, say, $$underbrace(y^ixy^-1)cdots (y^ixy^-1)_k+1,texttimes$$ for each $i$ and all such products with fewer than $k+1$ terms. Thank you, @PaulPlummer; I should know better by now.
$endgroup$
– Shaun
7 hours ago
add a comment |
$begingroup$
This answer was made community wiki due to a technicality. Hopefully, others will learn from it.
Consider the free product given by the following (not necessarily unique) group presentation $$Bbb Z_k+1astBbb Zconglangle x, ymid x^k+1rangle.$$
$endgroup$
$begingroup$
There are infinitely many elements with finite order in this group; there are finitely many up to conjugation.
$endgroup$
– Paul Plummer
7 hours ago
$begingroup$
Ah, yes, of course! I overlooked the possibility of, say, $$underbrace(y^ixy^-1)cdots (y^ixy^-1)_k+1,texttimes$$ for each $i$ and all such products with fewer than $k+1$ terms. Thank you, @PaulPlummer; I should know better by now.
$endgroup$
– Shaun
7 hours ago
add a comment |
$begingroup$
This answer was made community wiki due to a technicality. Hopefully, others will learn from it.
Consider the free product given by the following (not necessarily unique) group presentation $$Bbb Z_k+1astBbb Zconglangle x, ymid x^k+1rangle.$$
$endgroup$
This answer was made community wiki due to a technicality. Hopefully, others will learn from it.
Consider the free product given by the following (not necessarily unique) group presentation $$Bbb Z_k+1astBbb Zconglangle x, ymid x^k+1rangle.$$
edited 7 hours ago
community wiki
4 revs
Shaun
$begingroup$
There are infinitely many elements with finite order in this group; there are finitely many up to conjugation.
$endgroup$
– Paul Plummer
7 hours ago
$begingroup$
Ah, yes, of course! I overlooked the possibility of, say, $$underbrace(y^ixy^-1)cdots (y^ixy^-1)_k+1,texttimes$$ for each $i$ and all such products with fewer than $k+1$ terms. Thank you, @PaulPlummer; I should know better by now.
$endgroup$
– Shaun
7 hours ago
add a comment |
$begingroup$
There are infinitely many elements with finite order in this group; there are finitely many up to conjugation.
$endgroup$
– Paul Plummer
7 hours ago
$begingroup$
Ah, yes, of course! I overlooked the possibility of, say, $$underbrace(y^ixy^-1)cdots (y^ixy^-1)_k+1,texttimes$$ for each $i$ and all such products with fewer than $k+1$ terms. Thank you, @PaulPlummer; I should know better by now.
$endgroup$
– Shaun
7 hours ago
$begingroup$
There are infinitely many elements with finite order in this group; there are finitely many up to conjugation.
$endgroup$
– Paul Plummer
7 hours ago
$begingroup$
There are infinitely many elements with finite order in this group; there are finitely many up to conjugation.
$endgroup$
– Paul Plummer
7 hours ago
$begingroup$
Ah, yes, of course! I overlooked the possibility of, say, $$underbrace(y^ixy^-1)cdots (y^ixy^-1)_k+1,texttimes$$ for each $i$ and all such products with fewer than $k+1$ terms. Thank you, @PaulPlummer; I should know better by now.
$endgroup$
– Shaun
7 hours ago
$begingroup$
Ah, yes, of course! I overlooked the possibility of, say, $$underbrace(y^ixy^-1)cdots (y^ixy^-1)_k+1,texttimes$$ for each $i$ and all such products with fewer than $k+1$ terms. Thank you, @PaulPlummer; I should know better by now.
$endgroup$
– Shaun
7 hours ago
add a comment |
$begingroup$
$(mathbbQ,+)$ has finite (none) elements of order $k$ where $k>1$.
answered 5 hours ago
EpsilonDeltaEpsilonDelta
6521 gold badge1 silver badge9 bronze badges
$endgroup$
add a comment |
$begingroup$
$(mathbbQ,+)$ has finite (none) elements of order $k$ where $k>1$.
answered 5 hours ago
EpsilonDeltaEpsilonDelta
6521 gold badge1 silver badge9 bronze badges
$endgroup$
add a comment |
$begingroup$
$(mathbbQ,+)$ has finite (none) elements of order $k$ where $k>1$.
answered 5 hours ago
EpsilonDeltaEpsilonDelta
6521 gold badge1 silver badge9 bronze badges
$endgroup$
$(mathbbQ,+)$ has finite (none) elements of order $k$ where $k>1$.
answered 5 hours ago
EpsilonDeltaEpsilonDelta
6521 gold badge1 silver badge9 bronze badges
answered 5 hours ago
EpsilonDeltaEpsilonDelta
6521 gold badge1 silver badge9 bronze badges
answered 5 hours ago
EpsilonDeltaEpsilonDelta
6521 gold badge1 silver badge9 bronze badges
answered 5 hours ago
EpsilonDeltaEpsilonDelta
6521 gold badge1 silver badge9 bronze badges
6521 gold badge1 silver badge9 bronze badges
add a comment |
add a comment |
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$begingroup$
Got it. Thank you. I wish I could accept multiple answers.
$endgroup$
– Utkarsh Kumar
8 hours ago
$begingroup$
You could at least upvote them, @UtkarshKumar.
$endgroup$
– Shaun
8 hours ago
2
$begingroup$
Yes, I did that. Yours was the latest, and hence I hadn't seen it yet.
$endgroup$
– Utkarsh Kumar
8 hours ago
2
$begingroup$
Wouldn't it be sufficient to consider a group with no elements of finite order except the identity? (E.g. $mathbbZ$) I may have misunderstood the question.
$endgroup$
– Sambo
8 hours ago