ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sat, 19 Sep 2020 18:56:50 +0200Which subgroups do conjugacy_classes_subgroups() returnhttps://ask.sagemath.org/question/53498/which-subgroups-do-conjugacy_classes_subgroups-return/Starting with a group G and subgroup H (up to conjugacy), my task is to find all subgroups (up to conjugacy) that contains H.
1. Do the subgroups returned by conjugacy_classes_subgroups() satisfy the property that if K and L are two of the returned subgroups such that a conjugate of K is a subgroup of L, then K is a subgroup of L (we can construct a lattice from the subgroups returned)
2. If not, what will be the most optimal way to do this? My current algorithm checks if any conjugate of H is a subgroup of the list returned in conjugacy_classes_subgroups().
Sat, 19 Sep 2020 02:29:46 +0200https://ask.sagemath.org/question/53498/which-subgroups-do-conjugacy_classes_subgroups-return/Answer by rburing for <p>Starting with a group G and subgroup H (up to conjugacy), my task is to find all subgroups (up to conjugacy) that contains H. </p>
<ol>
<li><p>Do the subgroups returned by conjugacy_classes_subgroups() satisfy the property that if K and L are two of the returned subgroups such that a conjugate of K is a subgroup of L, then K is a subgroup of L (we can construct a lattice from the subgroups returned)</p></li>
<li><p>If not, what will be the most optimal way to do this? My current algorithm checks if any conjugate of H is a subgroup of the list returned in conjugacy_classes_subgroups().</p></li>
</ol>
https://ask.sagemath.org/question/53498/which-subgroups-do-conjugacy_classes_subgroups-return/?answer=53506#post-id-535061. The method calls GAP's [ConjugacyClassesSubgroups](https://groupprops.subwiki.org/wiki/GAP: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](https://ask.sagemath.org/question/49641/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)`.
2. 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).Sat, 19 Sep 2020 11:43:07 +0200https://ask.sagemath.org/question/53498/which-subgroups-do-conjugacy_classes_subgroups-return/?answer=53506#post-id-53506Comment by 61plus for <ol>
<li><p>The method calls GAP's <a href="https://groupprops.subwiki.org/wiki/GAP:ConjugacyClassesSubgroups">ConjugacyClassesSubgroups</a> to get the conjugacy classes and then GAP's <code>Representative</code> to get a representative of each. This does not have the property you want. For example <code>SymmetricGroup(4).conjugacy_classes_subgroups()</code> 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 $L$.</p>
<p>See <a href="https://ask.sagemath.org/question/49641/symmetric-group-get-back-conjugacy-class-from-its-generators/">symmetric group: get back conjugacy class from its generators</a> if you want the actual conjugacy classes as GAP objects; you will be able to call the methods <code>Representative()</code> and <code>AsList()</code> on them, and on each subgroup you can also call <code>IsSubgroup(H)</code>.</p></li>
<li><p>I am not a group theorist but your algorithm seems fine to me.</p></li>
</ol>
<p>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).</p>
https://ask.sagemath.org/question/53498/which-subgroups-do-conjugacy_classes_subgroups-return/?comment=53512#post-id-53512Thank you! That is very helpful.Sat, 19 Sep 2020 18:56:50 +0200https://ask.sagemath.org/question/53498/which-subgroups-do-conjugacy_classes_subgroups-return/?comment=53512#post-id-53512