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.