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I am trying to do the following in GAP:

$G$ is a given group of order $2^{19}$ and $H$ is a subgroup (but not normal) of $G$ of order $2^{15}$. My aim is to find a representative, say $A$, for each of the conjugacy classes of subgroups of $G$ such that $A\not \subseteq H$.

I tried the following in GAP:

Filtered(List(ConjugacyClassesSubgroups(G),y->Representative(y)),x-> not IsSubset(H,x));

But since the group $G$ is very large, this is practically consuming all memory and taking infinite time.

Is there any other way I can use in GAP to overcome the situation? Any help will be greatly appreciated.

Pasteable descriptions of the groups are given below:

G:=PcGroupCode(8131228045254952190140857101752853639830107614464014650851209692873866777500298856184821519032105153814704456451033872646102911\ 4739105578388317100628340028962041397792456162362856790287035433614614445724770154742066261363277062070491534257341845886065760\ 9680893404023629145887435702630822420577559446012280802787875613789812926252815799729640075785252102776099420886867267522205122\ 48, 2^19);

H:= PcGroupCode(63259843161787529259663104123028807771949615600420226627717675945661695659171\ 01261938932942099569279276682015405496285786548265341726439956203516708900070\ 21155088492850709644054534706704466392622607233970153096962475626323, 2^15);

Thanks

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  • $\begingroup$ At this size, most likely there will be so many subgroups that you run into storage/bookkeeping problems. $\endgroup$ – ahulpke Mar 16 at 6:44
  • $\begingroup$ Also, the test you sketch depends on the choice of representative (respectively, your condition is not clear): Do you want those representatives which are not contained in $H$, or representatives (if they exist) that are not in $H$ (and might differ from GAP's default choice). $\endgroup$ – ahulpke Mar 16 at 6:46
  • $\begingroup$ Finally, the way you give $H$ does not identify it as subgroup of $G$ but as group of its own. $\endgroup$ – ahulpke Mar 16 at 6:54
  • $\begingroup$ @ahulpke Thanks, yes the size is the problem. So I was thinking if there is any other way to attack the problem. For the representative I should be more clear, you are correct. I actually need those representatives (if exist) which are not in $H$ and surely that might differ from GAP's default choice. $\endgroup$ – usermath Mar 16 at 6:55
  • $\begingroup$ @ahulpke From my info on the question, definitely $H$ is not given as subgroup of $G$,sorry. But $H$ is actually constructed as subgroup of $G$ in my code, but I don't know how to convey that in the question using "PcGroupCode". Is there any way to do that, can you pls help? $\endgroup$ – usermath Mar 16 at 6:58

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