The order of the element of a cyclic group divides the order of that cyclic group. [on hold]Proving that in an infinite cyclic group order of every element is infiniteGroup with infinite ordercyclic group and its generatorsDoes every finite group has finite cyclic subgroup?Proving that in an infinite cyclic group order of every element is infiniteIf a cyclic group has an element of infinite order, how many elements of finite order does it have?Abelian group that has power of prime order has an element whose order is power of primeLet $p$ be prime. If an INFINITE group has more than $p-1$ elements of order $p$, why can't the group be cyclic?Order of an element in a cyclic groupInfinite cyclic group isomorphism.Order of element and cyclic group?Proof Help: Order of element divides order of group
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The order of the element of a cyclic group divides the order of that cyclic group. [on hold]
Proving that in an infinite cyclic group order of every element is infiniteGroup with infinite ordercyclic group and its generatorsDoes every finite group has finite cyclic subgroup?Proving that in an infinite cyclic group order of every element is infiniteIf a cyclic group has an element of infinite order, how many elements of finite order does it have?Abelian group that has power of prime order has an element whose order is power of primeLet $p$ be prime. If an INFINITE group has more than $p-1$ elements of order $p$, why can't the group be cyclic?Order of an element in a cyclic groupInfinite cyclic group isomorphism.Order of element and cyclic group?Proof Help: Order of element divides order of group
$begingroup$
If the cyclic group is finite order, by Lagrange's Theorem, it is true. But how about if the cyclic group is of infinite order? Does the statement above still holds true ?
abstract-algebra
edited yesterday
Bernard
122k741116
asked yesterday
Ling Min HaoLing Min Hao
34019
$endgroup$
put on hold as off-topic by Eevee Trainer, Jyrki Lahtonen, Dietrich Burde, Alex Provost, darij grinberg 21 hours ago
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." – Eevee Trainer, Jyrki Lahtonen, Dietrich Burde, Alex Provost
add a comment |
$begingroup$
If the cyclic group is finite order, by Lagrange's Theorem, it is true. But how about if the cyclic group is of infinite order? Does the statement above still holds true ?
abstract-algebra
edited yesterday
Bernard
122k741116
asked yesterday
Ling Min HaoLing Min Hao
34019
$endgroup$
put on hold as off-topic by Eevee Trainer, Jyrki Lahtonen, Dietrich Burde, Alex Provost, darij grinberg 21 hours ago
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." – Eevee Trainer, Jyrki Lahtonen, Dietrich Burde, Alex Provost
$begingroup$
Divisibility is defined in the integers, but not necessarily in the integers with $pm infty$. On the other hand, Lagrange can be formulated also for infinite groups avoiding divisibility, see this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
So you mean by Langrange Theorem , this statement is true (no matter it is finite or infinite?)
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
No, the statement on the order of elements is not true, see this post. But careful, the title claim is wrong. First read the answers to this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
So, for infinite case, Langrange's Theorem do not use the concept of divisibility, instead they use cardinality to link the relation between order of elements with order of the group ?
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
Yes, this is what Dave says in his answer. It is then even equivalent to the axiom of choice.
$endgroup$
– Dietrich Burde
yesterday
add a comment |
$begingroup$
If the cyclic group is finite order, by Lagrange's Theorem, it is true. But how about if the cyclic group is of infinite order? Does the statement above still holds true ?
abstract-algebra
edited yesterday
Bernard
122k741116
asked yesterday
Ling Min HaoLing Min Hao
34019
$endgroup$
If the cyclic group is finite order, by Lagrange's Theorem, it is true. But how about if the cyclic group is of infinite order? Does the statement above still holds true ?
abstract-algebra
abstract-algebra
edited yesterday
Bernard
122k741116
asked yesterday
Ling Min HaoLing Min Hao
34019
edited yesterday
Bernard
122k741116
asked yesterday
Ling Min HaoLing Min Hao
34019
edited yesterday
Bernard
122k741116
edited yesterday
Bernard
122k741116
edited yesterday
Bernard
122k741116
122k741116
asked yesterday
Ling Min HaoLing Min Hao
34019
asked yesterday
Ling Min HaoLing Min Hao
34019
asked yesterday
Ling Min HaoLing Min Hao
34019
34019
put on hold as off-topic by Eevee Trainer, Jyrki Lahtonen, Dietrich Burde, Alex Provost, darij grinberg 21 hours ago
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." – Eevee Trainer, Jyrki Lahtonen, Dietrich Burde, Alex Provost
put on hold as off-topic by Eevee Trainer, Jyrki Lahtonen, Dietrich Burde, Alex Provost, darij grinberg 21 hours ago
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." – Eevee Trainer, Jyrki Lahtonen, Dietrich Burde, Alex Provost
$begingroup$
Divisibility is defined in the integers, but not necessarily in the integers with $pm infty$. On the other hand, Lagrange can be formulated also for infinite groups avoiding divisibility, see this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
So you mean by Langrange Theorem , this statement is true (no matter it is finite or infinite?)
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
No, the statement on the order of elements is not true, see this post. But careful, the title claim is wrong. First read the answers to this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
So, for infinite case, Langrange's Theorem do not use the concept of divisibility, instead they use cardinality to link the relation between order of elements with order of the group ?
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
Yes, this is what Dave says in his answer. It is then even equivalent to the axiom of choice.
$endgroup$
– Dietrich Burde
yesterday
add a comment |
$begingroup$
Divisibility is defined in the integers, but not necessarily in the integers with $pm infty$. On the other hand, Lagrange can be formulated also for infinite groups avoiding divisibility, see this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
So you mean by Langrange Theorem , this statement is true (no matter it is finite or infinite?)
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
No, the statement on the order of elements is not true, see this post. But careful, the title claim is wrong. First read the answers to this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
So, for infinite case, Langrange's Theorem do not use the concept of divisibility, instead they use cardinality to link the relation between order of elements with order of the group ?
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
Yes, this is what Dave says in his answer. It is then even equivalent to the axiom of choice.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
Divisibility is defined in the integers, but not necessarily in the integers with $pm infty$. On the other hand, Lagrange can be formulated also for infinite groups avoiding divisibility, see this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
Divisibility is defined in the integers, but not necessarily in the integers with $pm infty$. On the other hand, Lagrange can be formulated also for infinite groups avoiding divisibility, see this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
So you mean by Langrange Theorem , this statement is true (no matter it is finite or infinite?)
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
So you mean by Langrange Theorem , this statement is true (no matter it is finite or infinite?)
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
No, the statement on the order of elements is not true, see this post. But careful, the title claim is wrong. First read the answers to this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
No, the statement on the order of elements is not true, see this post. But careful, the title claim is wrong. First read the answers to this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
So, for infinite case, Langrange's Theorem do not use the concept of divisibility, instead they use cardinality to link the relation between order of elements with order of the group ?
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
So, for infinite case, Langrange's Theorem do not use the concept of divisibility, instead they use cardinality to link the relation between order of elements with order of the group ?
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
Yes, this is what Dave says in his answer. It is then even equivalent to the axiom of choice.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
Yes, this is what Dave says in his answer. It is then even equivalent to the axiom of choice.
$endgroup$
– Dietrich Burde
yesterday
add a comment |
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$begingroup$
Divisibility is defined in the integers, but not necessarily in the integers with $pm infty$. On the other hand, Lagrange can be formulated also for infinite groups avoiding divisibility, see this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
So you mean by Langrange Theorem , this statement is true (no matter it is finite or infinite?)
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
No, the statement on the order of elements is not true, see this post. But careful, the title claim is wrong. First read the answers to this duplicate.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
So, for infinite case, Langrange's Theorem do not use the concept of divisibility, instead they use cardinality to link the relation between order of elements with order of the group ?
$endgroup$
– Ling Min Hao
yesterday
$begingroup$
Yes, this is what Dave says in his answer. It is then even equivalent to the axiom of choice.
$endgroup$
– Dietrich Burde
yesterday