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If we have a two lattices (partially ordered) - one for subgroups, one for factor groups, and we know order of the group we want to have these subgroup and factor group lattices, is such a group unique up to isomorphism (if exists)? Or is there a counterexample?

If that's true, are sufficient conditions on the order and subgroup lattices to guarantee uniqueness? Another way - what if we now lattice for subgroup and group of automorphism of group; is that group uniquely determined by that information?

Thanks for help. (sorry for English)

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@tomas.lang: Are you really asking the subgroup and factor group lattices to be totally ordered? Or just partially ordered? Because there are very few groups with totally ordered subgroup lattices, and in the finite case, they must be cyclic. – Arturo Magidin Dec 16 '10 at 21:39
@tomas.lang: this might be relevant: – Arturo Magidin Dec 16 '10 at 21:44
@Arturo Migdin: Oh - partially ordered, mistake :-) Thanks for link... – tomas.lang Dec 16 '10 at 21:49
You really are just looking for groups that have isomorphic subgroup lattice, isomorphic normal subgroup lattice, and same order. If I had to guess, I would guess that you will find examples of nonisomorphic groups with the same order and isomorphic lattices among the $p$-groups, just because these kinds of invariants almost always seem to not suffice to distinguish $p$-groups; same for replacing the lattice of normal subgroups with the automorphism group. Perhaps someone can check with GAP for some small exponents. – Arturo Magidin Dec 17 '10 at 2:57
I am curious if SmallGroup(243,8) and SmallGroup(243,9) work. There is a block of 36 hard to distinguish elements in each subgroup lattice, and I cannot tell if they can be mapped to each other. – Jack Schmidt Dec 17 '10 at 15:55

1 Answer 1

up vote 16 down vote accepted

No, the lattice of subgroups, the lattice of normal subgroups, the order of the group, and the automorphism group do not (even taken together) determine the isomorphism type of a finite group.

Take G = SmallGroup(243,19) and H = SmallGroup(243,20). There is a bijection f:L(G)→L(H) between their lattices of subgroups such that:

  • |X| = |f(X)|
  • X ≅ f(X) unless X = G
  • X ≤ Y iff f(X) ≤ f(Y)
  • X ⊴ G iff f(X) ⊴ f(G) = H
  • G/X ≅ H/f(X) whenever X≠1 is normal

Additionally Aut(G) ≅ Aut(H). The fourth bullet shows in particular, that f induces an isomorphism between the lattice of quotient groups of G and the lattice of quotient groups of H. The second and fifth bullets show the isomorphism respects everything about the subgroups' properties as abstract groups.

The groups have presentations:

\begin{align*} G &= \bigl\langle a,b,c \mid a^{27} = b^{3} = c^{3} = 1,\ ba = abc,\ ca = acz,\ cb = bcz \bigr\rangle\text{ where }z = a^9\\ H &= \bigl\langle a,b,c \mid a^{27} = b^3 = c^{3} = 1,\ ba = abc,\ ca = acz,\ cb = bcz \bigr\rangle\text{ where }z = a^{-9} \end{align*}

The function is induced by a bijection of the underlying sets:

  • f(ai bj ck) = ai bj ck

There are no such groups of order dividing 64 (even just having an isomorphism of subgroup lattices respecting normal subgroups).

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Thanks for doing the legwork! I'm glad to know my intuition was right and that counterexamples would be found among $p$-groups. – Arturo Magidin Dec 18 '10 at 0:29
Sorry, I am not familiar with the SmallGroup terminology. Can you point me to a source where it is introduced? – Andrés Caicedo Dec 18 '10 at 2:34
@Andres Caicedo: They are the nomenclature of GAP's "SmallGroup" library.; – Arturo Magidin Dec 18 '10 at 2:42
@Arturo : Many thanks! – Andrés Caicedo Dec 18 '10 at 2:44
I added some less computer dependent details. You can see how G and H are basically identical, and since the lattice isomorphism preserves group orders, it is induced by a bijection of the underlying sets of group elements. I gave presentations where this bijection is the "obvious" one. – Jack Schmidt Dec 18 '10 at 4:41

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