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Does the order of the subgroup generated by two elements divide the product of the element orders? [duplicate]

Examples and further results about the order of the product of two elements in a groupIn finite groups does counting orders of elements is enough to determine if they are isomorphicListing the orders of the elements and calculating the number of elements in the orderName/notation for the subgroup generated by all stabilizersSubgroup of elements of order at most $2^{m}$Order of product of powers of elements from a groupShow that the set of all elements $a$ of a group $G$ such that $ax=xa$ for every element of $x$ of $G$is a subgroup of $G$.Questions about specific, two-elements generated groupProving order of certain elements in a group is the sameHow to find order of the subgroup?Subgroup generated by all elements with odd order

1

1

$begingroup$

This question already has an answer here:

• 
  Examples and further results about the order of the product of two elements in a group
  
  6 answers
  
  

Let $(G, cdot)$ be a group, $a,bin G$ such that $DeclareMathOperator{ord}{ord}ord(a),ord(b)<infty$.

Do we then have that $left|left<a,bright>right|$ divides $ord(a)ord(b)$? (Where $left< a,b right>$ denotes the subgroup generated by $a$ and $b$)

I managed to show this for the case that $a$ and $b$ commute, however I would be interested if this holds even if they don't.

abstract-algebra group-theory

share|cite|improve this question

asked 16 hours ago

Jakob B.Jakob B.

527111

$endgroup$

marked as duplicate by Asaf Karagila♦ 13 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The order might be infinite.
  $endgroup$
  – Trebor
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Groups have a wonderful amount of flexibility. You might be interested in the free product of groups.
  $endgroup$
  – Santana Afton
  15 hours ago
  

  
  
  
  

add a comment | 

1

1

$begingroup$

This question already has an answer here:

• 
  Examples and further results about the order of the product of two elements in a group
  
  6 answers
  
  

Let $(G, cdot)$ be a group, $a,bin G$ such that $DeclareMathOperator{ord}{ord}ord(a),ord(b)<infty$.

Do we then have that $left|left<a,bright>right|$ divides $ord(a)ord(b)$? (Where $left< a,b right>$ denotes the subgroup generated by $a$ and $b$)

I managed to show this for the case that $a$ and $b$ commute, however I would be interested if this holds even if they don't.

abstract-algebra group-theory

share|cite|improve this question

asked 16 hours ago

Jakob B.Jakob B.

527111

$endgroup$

marked as duplicate by Asaf Karagila♦ 13 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The order might be infinite.
  $endgroup$
  – Trebor
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Groups have a wonderful amount of flexibility. You might be interested in the free product of groups.
  $endgroup$
  – Santana Afton
  15 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Examples and further results about the order of the product of two elements in a group
  
  6 answers
  
  

Let $(G, cdot)$ be a group, $a,bin G$ such that $DeclareMathOperator{ord}{ord}ord(a),ord(b)<infty$.

Do we then have that $left|left<a,bright>right|$ divides $ord(a)ord(b)$? (Where $left< a,b right>$ denotes the subgroup generated by $a$ and $b$)

I managed to show this for the case that $a$ and $b$ commute, however I would be interested if this holds even if they don't.

abstract-algebra group-theory

share|cite|improve this question

asked 16 hours ago

Jakob B.Jakob B.

527111

$endgroup$

share|cite|improve this question

asked 16 hours ago

Jakob B.Jakob B.

527111

share|cite|improve this question

asked 16 hours ago

Jakob B.Jakob B.

527111

asked 16 hours ago

Jakob B.Jakob B.

527111

asked 16 hours ago

Jakob B.Jakob B.

527111

2 Answers
2

active

oldest

votes

3

$begingroup$

$S_3=langle (12), (23)rangle$.

share|cite|improve this answer

answered 16 hours ago

ancientmathematicianancientmathematician

4,6371513

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: $S_n$ is generated by $(12)$ and $(123cdots n)$, yet $n!$ rarely divides $2n$.
  $endgroup$
  – Greg Martin
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And for more fun mathoverflow.net/questions/59213/….
  $endgroup$
  – ancientmathematician
  14 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

The group $$leftlanglebegin{pmatrix}0&-1\1&0end{pmatrix},begin{pmatrix}0&-1\1&1end{pmatrix}rightrangle$$
has infinite order, in spite of the generators having finite order.

share|cite|improve this answer

answered 16 hours ago

Hagen von EitzenHagen von Eitzen

284k23274508

$endgroup$

add a comment | 

3

$begingroup$

$S_3=langle (12), (23)rangle$.

share|cite|improve this answer

answered 16 hours ago

ancientmathematicianancientmathematician

4,6371513

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: $S_n$ is generated by $(12)$ and $(123cdots n)$, yet $n!$ rarely divides $2n$.
  $endgroup$
  – Greg Martin
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And for more fun mathoverflow.net/questions/59213/….
  $endgroup$
  – ancientmathematician
  14 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

$S_3=langle (12), (23)rangle$.

share|cite|improve this answer

answered 16 hours ago

ancientmathematicianancientmathematician

4,6371513

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: $S_n$ is generated by $(12)$ and $(123cdots n)$, yet $n!$ rarely divides $2n$.
  $endgroup$
  – Greg Martin
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And for more fun mathoverflow.net/questions/59213/….
  $endgroup$
  – ancientmathematician
  14 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

$S_3=langle (12), (23)rangle$.

share|cite|improve this answer

answered 16 hours ago

ancientmathematicianancientmathematician

4,6371513

$endgroup$

share|cite|improve this answer

answered 16 hours ago

ancientmathematicianancientmathematician

4,6371513

answered 16 hours ago

ancientmathematicianancientmathematician

4,6371513

answered 16 hours ago

ancientmathematicianancientmathematician

4,6371513

3

$begingroup$

The group $$leftlanglebegin{pmatrix}0&-1\1&0end{pmatrix},begin{pmatrix}0&-1\1&1end{pmatrix}rightrangle$$
has infinite order, in spite of the generators having finite order.

share|cite|improve this answer

answered 16 hours ago

Hagen von EitzenHagen von Eitzen

284k23274508

$endgroup$

add a comment | 

3

$begingroup$

The group $$leftlanglebegin{pmatrix}0&-1\1&0end{pmatrix},begin{pmatrix}0&-1\1&1end{pmatrix}rightrangle$$
has infinite order, in spite of the generators having finite order.

share|cite|improve this answer

answered 16 hours ago

Hagen von EitzenHagen von Eitzen

284k23274508

$endgroup$

add a comment | 

$begingroup$

The group $$leftlanglebegin{pmatrix}0&-1\1&0end{pmatrix},begin{pmatrix}0&-1\1&1end{pmatrix}rightrangle$$
has infinite order, in spite of the generators having finite order.

share|cite|improve this answer

answered 16 hours ago

Hagen von EitzenHagen von Eitzen

284k23274508

$endgroup$

share|cite|improve this answer

answered 16 hours ago

Hagen von EitzenHagen von Eitzen

284k23274508

answered 16 hours ago

Hagen von EitzenHagen von Eitzen

284k23274508

answered 16 hours ago

Hagen von EitzenHagen von Eitzen

284k23274508