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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.2010-12-16
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    @tomas.lang: this might be relevant: http://www.jstor.org/stable/19903752010-12-16
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    @Arturo Migdin: Oh - partially ordered, mistake :-) Thanks for link...2010-12-16
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    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.2010-12-17
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    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.2010-12-17
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    #8 and #9 definitely don't work. There are still few left of order 243 to check.2010-12-18

1 Answers 1

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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.2010-12-18
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    Sorry, I am not familiar with the SmallGroup terminology. Can you point me to a source where it is introduced?2010-12-18
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    @Andres Caicedo: They are the nomenclature of GAP's "SmallGroup" library. http://www.gap-system.org/Gap3/Datalib3/small.html; http://www-public.tu-bs.de:8080/~hubesche/small.html2010-12-18
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    @Arturo : Many thanks!2010-12-18
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    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.2010-12-18