On Tuesday, May 29, 2018 at 9:01:51 PM UTC+10, chris wuthrich wrote: > > > As a simple-minded user I stumbled over exactly this last week. I don't > understand much of what this thread is discussing, but I know what a > simple-minded user wants. > > sage: G = GL(2,13) > sage: G = G.as_matrix_group() > sage: H = G.subgroup([ matrix(GF(13),[[1,0],[1,1]]) ]) > sage: H2 = G.subgroup([ matrix(GF(13),[[1,1],[0,1]]) ]) > sage: H3 = G.subgroup([ matrix(GF(13),[[1,0],[2,1]]) ]) > > I expect H == H3 to say True as they are the same subgroup. > I expect H = H2 to say False, since they are not the same subgroup, even > though they are isomorphic. > Instead > > sage: H == H3 > False > sage: matrix(GF(13),[[1,0],[1,1]]) in H3 > True > sage: matrix(GF(13),[[1,0],[2,1]]) in H > True > sage: H.is_isomorphic(H3) > True > sage: H.is_isomorphic(H2) > True > sage: matrix(GF(13),[[1,0],[1,1]]) in H2 > False > > I though of working around it by checking if they are the same as sets, > but to my surprise: > > sage: Set( h for h in H ) == Set( h for h in H3 ) > False > sage: Set( h.matrix() for h in H ) == Set( h.matrix() for h in H3 ) > True > > In short: I consider this a bug. No matter how this is done at the back, > the user expects == to mean mathematical equality, here equality of > subgroups. Otherwise the function should be called > subgroups_with_given_generators. > > I think it is a bad thing in this case for == to be equality as sets. Imagine if these are two really big, equal, but differently constructed subgroups. This would be a really long and expensive check, whereas in most cases, checking the defining objects are sufficient. I believe we have other places in Sage where == is not strict mathematical equality for similar reasons. -1 on changing the == semantics here.
Now I would say the equality as sets (which really comes down to equality of elements) is a bug. Two matrix group elements should compare by their matrices and not have the coercion system involved otherwise. My guess is the coercion system tries to convert the elements of H into H3 and vice versa, which is cannot do, and hence, results in a False for equality. This probably requires a reimplementation of the comparison method of matrix group elements. Best, Travis -- You received this message because you are subscribed to the Google Groups "sage-devel" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-devel+unsubscr...@googlegroups.com. To post to this group, send email to sage-devel@googlegroups.com. Visit this group at https://groups.google.com/group/sage-devel. For more options, visit https://groups.google.com/d/optout.
