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How many permutation groups of order $n$ are in $S_n$, up to isomorphism?

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0

$begingroup$

When $n$ is prime there is only one such group, and it is cyclic group $C_n$.
But when $n$ is not prime there can be several distinct groups. There is an an example of two such groups for $S_4$:

$e$, $(1 0)(3 2)$, $(2 0)(3 1)$, $(2 1)(3 0)$ and $e$, $(1 2 3 0)$, $(2 0)(3 1)$, $(3 2 1 0)$

For larger $n$ more isomorphism classes of groups can be found. How many such groups are in $S_n$ and is there a way to find them all?

UPDATE:

It looks like there is no answer to this question even if take into account only transitive subgroups of $S_n$, took from here:
About transitive subgroups of symmetric group $S_n$

Another similar question:
How many groups exist with order $n$ (two isomorphic groups are treated as the same group)

group-theory permutations

share|cite|improve this question

edited 22 hours ago

dkolmakov

asked yesterday

dkolmakovdkolmakov

1042

New contributor

dkolmakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Every group of order $n$ is in $S_n.$ The general question of how many groups there are, up to isomorphism, of order $n$ is, in general, difficult.
  $endgroup$
  – Thomas Andrews
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thomas Andrews's comment is is the content of Cayley's theorem. en.wikipedia.org/wiki/Cayley%27s_theorem
  $endgroup$
  – Cheerful Parsnip
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's hard to understand why you consider subgroups of $S_n$ since you consider only up to isomorphism. If you consider them as permutation groups there are also non-transitive subgroups.
  $endgroup$
  – YCor
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YCor Am I right that transitive subgroups are those ones that contain transpositions for every pair in {0,1..,n} (if we decompose all permutations to composition of transpositions)? If so then I'm interested only in transitive subgroups which contain exactly n elements.
  $endgroup$
  – dkolmakov
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Simply transitive subgroups on $nge 3$ elements never have transpositions.
  $endgroup$
  – YCor
  yesterday
  

  
  
  
  

 | 
show 3 more comments

0

$begingroup$

When $n$ is prime there is only one such group, and it is cyclic group $C_n$.
But when $n$ is not prime there can be several distinct groups. There is an an example of two such groups for $S_4$:

$e$, $(1 0)(3 2)$, $(2 0)(3 1)$, $(2 1)(3 0)$ and $e$, $(1 2 3 0)$, $(2 0)(3 1)$, $(3 2 1 0)$

For larger $n$ more isomorphism classes of groups can be found. How many such groups are in $S_n$ and is there a way to find them all?

UPDATE:

It looks like there is no answer to this question even if take into account only transitive subgroups of $S_n$, took from here:
About transitive subgroups of symmetric group $S_n$

Another similar question:
How many groups exist with order $n$ (two isomorphic groups are treated as the same group)

group-theory permutations

share|cite|improve this question

edited 22 hours ago

dkolmakov

asked yesterday

dkolmakovdkolmakov

1042

New contributor

dkolmakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Every group of order $n$ is in $S_n.$ The general question of how many groups there are, up to isomorphism, of order $n$ is, in general, difficult.
  $endgroup$
  – Thomas Andrews
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thomas Andrews's comment is is the content of Cayley's theorem. en.wikipedia.org/wiki/Cayley%27s_theorem
  $endgroup$
  – Cheerful Parsnip
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's hard to understand why you consider subgroups of $S_n$ since you consider only up to isomorphism. If you consider them as permutation groups there are also non-transitive subgroups.
  $endgroup$
  – YCor
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YCor Am I right that transitive subgroups are those ones that contain transpositions for every pair in {0,1..,n} (if we decompose all permutations to composition of transpositions)? If so then I'm interested only in transitive subgroups which contain exactly n elements.
  $endgroup$
  – dkolmakov
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Simply transitive subgroups on $nge 3$ elements never have transpositions.
  $endgroup$
  – YCor
  yesterday
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

When $n$ is prime there is only one such group, and it is cyclic group $C_n$.
But when $n$ is not prime there can be several distinct groups. There is an an example of two such groups for $S_4$:

$e$, $(1 0)(3 2)$, $(2 0)(3 1)$, $(2 1)(3 0)$ and $e$, $(1 2 3 0)$, $(2 0)(3 1)$, $(3 2 1 0)$

For larger $n$ more isomorphism classes of groups can be found. How many such groups are in $S_n$ and is there a way to find them all?

UPDATE:

It looks like there is no answer to this question even if take into account only transitive subgroups of $S_n$, took from here:
About transitive subgroups of symmetric group $S_n$

Another similar question:
How many groups exist with order $n$ (two isomorphic groups are treated as the same group)

group-theory permutations

share|cite|improve this question

edited 22 hours ago

dkolmakov

asked yesterday

dkolmakovdkolmakov

1042

New contributor

dkolmakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 22 hours ago

dkolmakov

asked yesterday

dkolmakovdkolmakov

1042

New contributor

dkolmakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 22 hours ago

dkolmakov

asked yesterday

dkolmakovdkolmakov

1042

New contributor

dkolmakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

dkolmakovdkolmakov

1042

New contributor

dkolmakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

dkolmakovdkolmakov

1042