Function: subgrouplist
Section: number_fields
C-Name: subgrouplist0
Prototype: GDGD0,L,
Help: subgrouplist(bnr,{bound},{flag=0}): bnr being as output by bnrinit or
a list of cyclic components of a finite Abelian group G, outputs the list of
subgroups of G (of index bounded by bound, if not omitted), given as HNF
left divisors of the SNF matrix corresponding to G. If flag=0 (default) and
bnr is as output by bnrinit, gives only the subgroups for which the modulus
is the conductor.
Doc: \var{bnr} being as output by \kbd{bnrinit} or a list of cyclic components
of a finite Abelian group $G$, outputs the list of subgroups of $G$. Subgroups
are given as HNF left divisors of the SNF matrix corresponding to $G$.
If $\fl=0$ (default) and \var{bnr} is as output by \kbd{bnrinit}, gives
only the subgroups whose modulus is the conductor. Otherwise, the modulus is
not taken into account.
If \var{bound} is present, and is a positive integer, restrict the output to
subgroups of index less than \var{bound}. If \var{bound} is a vector
containing a single positive integer $B$, then only subgroups of index
exactly equal to $B$ are computed. For instance
\bprog
? subgrouplist([6,2])
%1 = [[6, 0; 0, 2], [2, 0; 0, 2], [6, 3; 0, 1], [2, 1; 0, 1], [3, 0; 0, 2],
[1, 0; 0, 2], [6, 0; 0, 1], [2, 0; 0, 1], [3, 0; 0, 1], [1, 0; 0, 1]]
? subgrouplist([6,2],3) \\@com index less than 3
%2 = [[2, 1; 0, 1], [1, 0; 0, 2], [2, 0; 0, 1], [3, 0; 0, 1], [1, 0; 0, 1]]
? subgrouplist([6,2],[3]) \\@com index 3
%3 = [[3, 0; 0, 1]]
? bnr = bnrinit(bnfinit(x), [120,[1]], 1);
? L = subgrouplist(bnr, [8]);
@eprog\noindent
In the last example, $L$ corresponds to the 24 subfields of
$\Q(\zeta_{120})$, of degree $8$ and conductor $120\infty$ (by setting \fl,
we see there are a total of $43$ subgroups of degree $8$).
\bprog
? vector(#L, i, galoissubcyclo(bnr, L[i]))
@eprog\noindent
will produce their equations. (For a general base field, you would
have to rely on \tet{bnrstark}, or \tet{rnfkummer}.)