Consider the projective special linear group
of order 168. It
has three conjugacy classes of maximal subgroups, two classes of type
and one class of type 7:3 (see e.g. [Conway et al. 1985], p. 3). Assume that we
already know the tables of marks of these groups, that is the corresponding
posets of conjugacy classes of subgroups together with additional incidence
information. The poset structure of
is given in
Figure 1 and the underlying set of elements is:
Figure 1: The poset structures of
and 7:3.
where 2a and
are contained in
. The poset structure of 7:3 is
also given in Figure 1 with underlying set of elements:
In order to construct the poset structure of
we start with the
disjoint union of two copies of the poset structure of
, a yellow (Y)
and a red (R) one, say, and a blue (B) copy of the poset structure of
7:3. They form one colored diagram with vertex set
1(Y), 2a(Y), 2b(Y), 3(Y), 4(Y),
,
,
,
,
,
, 1(R), 2a(R), 2b(R), 3(R), 4(R),
,
,
,
,
,
, 1(B), 3(B), 7(B),
.
Here a symbol like
is just a name for a vertex in the yellow part
of the disjoint union of diagrams. The fact that it denotes an elementary
abelian group of order 4 follows from the information that is stored in the
table of marks, rather than from its name. Of course, the names we are
working with here were carefully chosen as to indicate the type of object
they denote.
According to Lemma 1.3 (i) the table of marks contains information
about the size of each subgroup, and conjugate subgroups of
have the
same size. Therefore, if we split the whole set of vertices into subsets
according to the orders of the corresponding subgroups, then only vertices in
the same subset can be conjugate in
. This yields the following
partition on the set of vertices.
,
,
,
,
,
,
,
,
,
.
Our aim now is to manipulate this colored diagram with the partition of the
vertices step by step until it represents the poset structure of
.
This is achieved by two sorts of manipulations:
The subsets
and
both contain only one element, hence
each of them already determines a unique conjugacy class of subgroups of
.
There is, of course, only one trivial subgroup in
, so we are allowed
to fuse the vertices 1(R), 1(Y) and 1(B) corresponding to the trivial
subgroups of the three maximal subgroups into a single vertex which we simply
denote by 1.
From the table of marks of
we can derive that subgroups named 4
contain one cyclic subgroup of order 2, in contrast to the subgroups named
which contain three cyclic subgroups of order 2 each. Hence these
subgroups can not be conjugate in
and we are allowed to split the
set of groups of order 4 into two subsets accordingly. Now the situation
is as follows.
,
,
,
,
,
,
,
,
,
,
.
By Sylow's theorem, G acts transitively on the set of Sylow p-subgroups
of G for any prime p. Hence we can fuse all the subgroups of type
(for p = 2) and all the subgroups of type 3 (for p = 3).
The normalizer in G of a group of type 3 is a group of type
. (Note
that this can also be decided from the tables of marks, since a cyclic 3
has index two in
and the order of its normalizer in
is 6.)
Together with the subgroups of type 3 their normalizers are conjugate in
G, hence we can fuse the subset corresponding to these groups into a single
vertex. (See Corollary 4.10 for how knowledge about normalizers can
be exploited in the general case.)
Within the Sylow 2-subgroup of type
of
there is only one
-class of subgroups of type 4. Hence, by Sylow's theorem, all
subgroups of this type are conjugate in G and we can fuse the yellow and
the red group named 4. (See Corollary 4.8 for how knowledge
about the Sylow subgroups can be exploited in the general case.)
From the character table of
we can read off that there is only one
class of elements (and hence of subgroups) of order 2, so we can fuse all
vertices named 2a or 2b into a single vertex 2.
We are left with three subsets containing more than one element: those
corresponding to the subgroups of type
,
or
. In order to
show that subgroups of type
which are not conjugate in a maximal
are not conjugate in G either, we examine the permutation character of G
on
. The character table of
admits only one nontrivial
character
of degree 7 that satisfies
for all
, therefore
is the permutation character of G on
. The
restriction of
to
admits two different decompositions into
transitive components, corresponding to the two classes of maximal subgroups
of type
in G. The corresponding sums of rows of the table of marks
of
reveal different values of fixed points for the two different
conjugacy classes of subgroups of type
inside
.
Hence the set containing the groups of this type must split in two subsets.
One of them contains the red
and the yellow
(remember
that the labeling was chosen in such a way that groups of type
are
normal in
). And both subsets now correspond to one conjugacy class of
subgroups of G.
Figure 2: The poset structure of
.
Now the red
and the yellow
lie above different classes of
groups of type
, whence they must correspond to different classes of
subgroups of G. So we split the subset
in two parts.
The same holds for
(well, we knew right from the start that there are
two classes of them in
) and we split the subset
into two parts. (See Corollary 4.5 for how knowledge about
numbers of subgroups can be exploited in the general case.)
Now every subset corresponds to exactly one conjugacy class of subgroups of
G. We finally add the group itself. Thus we have constructed the complete
poset structure of
(see Figure 2), with vertex set
,
,
,
,
,
,
,
,
,
,
,
,
,
,
.
This small example illustrates several aspects of the general procedure. We
have seen essentially two types of steps in the development of the poset of
. Most of the conclusions, like those using Sylow's theorem or the
conjugacy of normalizers, were based on general facts about the structure of
finite groups. Other conclusions, like the existence of exactly one
conjugacy class of subgroups of order 2 or the fusion of the subgroups of
type
, arose from additional knowledge about the particular group
. The next section formalizes the general setting.