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