lattice of subgroups

Let G be a group and L(G) be the set of all subgroupsMathworldPlanetmathPlanetmath of G. Elements of L(G) can be ordered by the set inclusion relationMathworldPlanetmathPlanetmathPlanetmath . This way L(G) becomes a partially ordered setMathworldPlanetmath.

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

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

With the two binary operationsMathworldPlanetmath and , L(G) becomes a latticeMathworldPlanetmath. It is a bounded latticeMathworldPlanetmath, with G as the top element and e as the bottom element. Furthermore, if {HiiI} is a set of subgroups of G indexed by some set I, then both

iIHi   and   iIHi

are subgroups of G. So L(G) is a complete latticeMathworldPlanetmath. From this, it is easy to produce a lattice which is not a subgroup lattice of any group.

Atoms in L(G), if they exist, are finite cyclic groupsMathworldPlanetmath of prime order (or /p, 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 ( 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 ( subgroup lattices which are not isomorphic groupsMathworldPlanetmath.

Example. Note that Autp2p-1×p. Therefore it is possible to from a non-trivial semidirect productMathworldPlanetmath p2p. The lattice of subgroups of p2p is the same as the lattice of subgroups of p2×p. However, p2p is non-abelianMathworldPlanetmathPlanetmath while p2×p is abelianMathworldPlanetmath so the two groups are not isomorphic.

Similarly, the groups pip and pi×p 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, p2q, and p4. With the modern understanding of groups the counterexample is rather simple to describe – though a proof remains a little tedious.

Let V=q2 where q is a prime – that is V is the 2-dimensional vector spaceMathworldPlanetmath over the field q. Let p|q-1 be another prime. As p|q-1, pq× so if we write p=ω multiplicatively so that we have for every nq, nωn is an automorphismPlanetmathPlanetmathPlanetmath of q. (Note that ω is often called a primitive p-th root of unity in q as it spans the p subgroup of q×.) Furthermore, for any 0ip-1 we get an automorphism fi:qq given by


Therefore to every 0ip-1 we can define a group Gi=V,gi, gi=f1fi as a subgroup of AGL(V). That is to say, Gi=Vgi where the action of gi on V is given by: for every vV, v=[nm] for n,mq set


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 GL(2,q).) However, Ga and Gb 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
Canonical name LatticeOfSubgroups
Date of creation 2013-03-22 15:47:42
Last modified on 2013-03-22 15:47:42
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 14
Author CWoo (3771)
Entry type Definition
Classification msc 20E15
Synonym subgroup lattice