lattice of subgroups
For any , define by . Then is a subgroup of and hence an element of . It is not hard to see that is the largest subgroup of both and .
Next, let and define by , the subgroup of generated by . So . Each element in is a finite product of elements from and . Again, it is easy to see that is the smallest subgroup of that has and as its subgroups.
With the two binary operations and , becomes a lattice. It is a bounded lattice, with as the top element and as the bottom element. Furthermore, if is a set of subgroups of indexed by some set , then both
Remark. Finding lattices of subgroups of groups is one way to classify groups. One of the main results in this branch of group theory states that the lattice of subgroups of a group is distributive (http://planetmath.org/DistributiveLattice) iff is locally cyclic.
It is generally not true that the lattice of subgroups of a group determines the group up to isomorphism. Already for groups of order , or , for all primes, there are examples of groups with isomorphic (http://planetmath.org/LatticeIsomorphism) subgroup lattices which are not isomorphic groups.
Example. Note that . Therefore it is possible to from a non-trivial semidirect product . The lattice of subgroups of is the same as the lattice of subgroups of . However, is non-abelian while is abelian so the two groups are not isomorphic.
Similarly, the groups and for any and any primes also have isomorphic subgroup lattices while one is non-abelian and the other abelian. So this is indeed a family of counterexamples.
Upon inspecting these example it becomes clear that the non-abelian groups have a different sublattice of normal subgroups. So the question can be asked whether two groups with isomorphic subgroup lattices including matching up conjugacy classes (so even stronger than matching normal subgroups) can be non-isomorphic groups. Surprisingly the answer is yes and was the dissertation of Ada Rottländer, a student of Schur’s, in 1927. Her example uses groups already discovered by Otto Hölder in his famous classification of the groups of order , , and . With the modern understanding of groups the counterexample is rather simple to describe – though a proof remains a little tedious.
Let where is a prime – that is is the 2-dimensional vector space over the field . Let be another prime. As , so if we write multiplicatively so that we have for every , is an automorphism of . (Note that is often called a primitive -th root of unity in as it spans the subgroup of .) Furthermore, for any we get an automorphism given by
Therefore to every we can define a group , as a subgroup of . That is to say, where the action of on is given by: for every , for set
We are now prepared to give the Rottländer counterexample.
Now let and be integers between and such that is not congruent to modulo . Notice this already forces so our smallest example is and . Then is not isomorphic to (compare the eigenvalues of to – they are not equal so the linear transformations are not conjugate in .) However, and have isomorphic subgroup lattices including matching conjugacy classes.
- 1 Rottländer, Ada, Nachweis der Existenz nicht-isomorpher Gruppen von gleicher Situation der Untergruppen, Math. Z. vol. 28, 1928, 1, pp. 641– 653, ISSN 0025-5874. MR MR1544982,
|Title||lattice of subgroups|
|Date of creation||2013-03-22 15:47:42|
|Last modified on||2013-03-22 15:47:42|
|Last modified by||CWoo (3771)|