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)

  • $\begingroup$ @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. $\endgroup$ Dec 16, 2010 at 21:39
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    $\begingroup$ @tomas.lang: this might be relevant: jstor.org/stable/1990375 $\endgroup$ Dec 16, 2010 at 21:44
  • $\begingroup$ @Arturo Migdin: Oh - partially ordered, mistake :-) Thanks for link... $\endgroup$
    – tomas.lang
    Dec 16, 2010 at 21:49
  • $\begingroup$ 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. $\endgroup$ Dec 17, 2010 at 2:57
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    $\begingroup$ 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. $\endgroup$ Dec 17, 2010 at 15:55

1 Answer 1


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 $G = \mathrm{SmallGroup}(243, 19)$ and $H = \mathrm{SmallGroup}(243, 20)$. There is a bijection $f \colon L(G) \to L(H)$ 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, $\operatorname{Aut}(G) ≅ \operatorname{Aut}(H)$. 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 $G$ and $H$ 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 $f$ is induced by a bijection of the underlying sets:

  • $f(a^i b^j c^k) = a^i b^j c^k$.

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

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    $\begingroup$ Thanks for doing the legwork! I'm glad to know my intuition was right and that counterexamples would be found among $p$-groups. $\endgroup$ Dec 18, 2010 at 0:29
  • $\begingroup$ Sorry, I am not familiar with the SmallGroup terminology. Can you point me to a source where it is introduced? $\endgroup$ Dec 18, 2010 at 2:34
  • $\begingroup$ @Andres Caicedo: They are the nomenclature of GAP's "SmallGroup" library. gap-system.org/Gap3/Datalib3/small.html; www-public.tu-bs.de:8080/~hubesche/small.html $\endgroup$ Dec 18, 2010 at 2:42
  • $\begingroup$ 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. $\endgroup$ Dec 18, 2010 at 4:41

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