1 | initial version |

The method calls GAP's ConjugacyClassesSubgroups to get the conjugacy classes and then GAP's

`Representative`

to get a representative of each. This does not have the property you want. For example`SymmetricGroup(4).conjugacy_classes_subgroups()`

includes both $K=\langle (1,3)(2,4)\rangle$ and $L=\langle (3,4), (1,2)(3,4) \rangle$ where a conjugate of $K$ is a subgroup of $L$ but $K$ is not a subgroup of $L$.See symmetric group: get back conjugacy class from its generators if you want the actual conjugacy classes as GAP objects; you will be able to call the methods

`Representative()`

and`AsList()`

on them, and on each subgroup you can also call`IsSubgroup(H)`

.I am not a group theorist but your algorithm seems fine to me.

2 | No.2 Revision |

The method calls GAP's ConjugacyClassesSubgroups to get the conjugacy classes and then GAP's

`Representative`

to get a representative of each. This does not have the property you want. For example`SymmetricGroup(4).conjugacy_classes_subgroups()`

includes both $K=\langle (1,3)(2,4)\rangle$ and $L=\langle (3,4), (1,2)(3,4) \rangle$ where a conjugate of $K$ is a subgroup of $L$ but $K$ is not a subgroup of $L$.See symmetric group: get back conjugacy class from its generators if you want the actual conjugacy classes as GAP objects; you will be able to call the methods

`Representative()`

and`AsList()`

on them, and on each subgroup you can also call`IsSubgroup(H)`

.I am not a group theorist but your algorithm seems fine to me.

Edit: Maybe I don't have to tell you this but, caveat: being conjugate in a subgroup is a stronger requirement than being conjugate in the full group (because there are fewer elements to conjugate by).

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