lattice of subgroups (original) (raw)

For any H,K∈L⁢(G), define H∧K by H∩K. Then H∧K is a subgroup of G and hence an element of L⁢(G). It is not hard to see that H∧K is the largest subgroup of both H and K.

Next, let X=H∪K and define H∨K by ⟨X⟩, the subgroup of G generated by X. So H∨K∈L⁢(G). Each element in H∨K is a finite productMathworldPlanetmathPlanetmathPlanetmath of elements from H and K. Again, it is easy to see that H∨K is the smallest subgroup of G that has H and K as its subgroups.

Atoms in L⁢(G), if they exist, are finite cyclic groupsMathworldPlanetmath of prime order (or ℤ/p⁢ℤ, where p is a prime), since they have no non-trivial proper subgroupsMathworldPlanetmath.

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 G is distributive (http://planetmath.org/DistributiveLattice) iff G is locally cyclic.

It is generally not true that the lattice of subgroups of a group determines the group up to isomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath. Already for groups of order p3, p>2 or p4, for all primes, there are examples of groups with isomorphic (http://planetmath.org/LatticeIsomorphism) subgroup lattices which are not isomorphic groupsMathworldPlanetmath.

Example. Note that Aut⁡ℤp2≅ℤp-1×ℤp. Therefore it is possible to from a non-trivial semidirect productMathworldPlanetmath ℤp2⋊ℤp. The lattice of subgroups of ℤp2⋊ℤp is the same as the lattice of subgroups of ℤp2×ℤp. However, ℤp2⋊ℤp is non-abelianMathworldPlanetmathPlanetmath while ℤp2×ℤp is abelianMathworldPlanetmath so the two groups are not isomorphic.

Similarly, the groups ℤpi⋊ℤp and ℤpi×ℤp for any i>2 and any primes p 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 groupsMathworldPlanetmath have a different sublattice of normal subgroupsMathworldPlanetmath. So the question can be asked whether two groups with isomorphic subgroup lattices including matching up conjugacy classesMathworldPlanetmathPlanetmath (so even stronger than matching normal subgroups) can be non-isomorphic groups. Surprisingly the answer is yes and was the dissertation of Ada Rottländer[1], 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 p3, p2⁢q, and p4. With the modern understanding of groups the counterexample is rather simple to describe – though a proof remains a little tedious.

We are now prepared to give the Rottländer counterexample.

Now let a and b be integers between 2 and p-1 such that a is not congruent to b modulo p. Notice this already forces p>3 so our smallest example is q=11 and p=5. Then Ga is not isomorphic to Gb (compare the eigenvaluesMathworldPlanetmathPlanetmathPlanetmathPlanetmath of ga to gb – they are not equal so the linear transformations are not conjugate in G⁢L⁢(2,q).) However, Ga and Gb have isomorphic subgroup lattices including matching conjugacy classes.

References