TITLE: Does the order, lattice of subgroups, and lattice of factor groups, uniquely determine a group up to isomorphism?
QUESTION [16 upvotes]: 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)

REPLY [25 votes]: 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).