Subgroup Lattices and Dimension

230 Views Asked by At

I apologize in advance in the case that this question is nonsensical. If the idea isn't clear, I can perhaps explain more below.

In the fall I am taking an undergraduate abstract algebra course, and wanted to get a head start on the material. He recommended me Dummit and Foote's Abstract Algebra, and I've been working through the problems and recently came to the section on lattices of subgroups and noticed that some of them are quite simple, others quite complex. In particular, I noted that the lattice of subgroups for something like $\Bbb Z/2^n \Bbb Z$ for any integer $n$ can be written down as a straight line, while the lattice of the Klein Group or $\Bbb Z/6 \Bbb Z$ are drawn in a plane. Then there were particularly complicated lattices like those of $D_{16}$ that couldn't even be drawn in a plane but required more "dimensions" so that the lines didn't overlap or intersect anywhere. Is it possible to engineer a group that is complicated enough to require more dimensions yet?

So I'm wondering if it makes sense to refer to the dimension of a lattice when talking about the subgroups. By dimension I mean the minimal number of vectors it would take if we could impose some kind of coordinates on the lattice. I know this number would not be ambiguous, since isomorphic groups have isomorphic subgroup lattices, and furthermore that different representations of the same group could not have different lattice dimension.

Such a number might give insight into the relationship between subgroups simply, in terms of the relationships between the generators. Is this an idea that is fleshed out in any more detail? Is it actually useful, and if so where can I find any additional reading?

2

There are 2 best solutions below

7
On BEST ANSWER

The lattice of subgroups of a finite group form a finite graph, where subgroups are vertices and edges are formed between inclusions $H\subset K$ with no intermediate subgroups.

Every finite graph can be embedded into $\Bbb R^3$. Googling around: in particular, if we map the vertex set into some subset of $\Bbb R^3$ in which any four points are not coplanar, then we can simply draw straight line segments between vertices for the edges. This can be achived by putting the vertices anywhere on the helix $\{(\sin t,\cos t,t):t\in\Bbb R\}$ for example. So this notion of "dimension" is not an interesting or useful concept for lattices in general, let alone lattices of subgroups.

I suppose it would make more sense to talk about order-embeddings. This can also be achived inductively. First, enumerate the subgroups so that for any inclusion $H\subset K$, the group $K$ comes after $H$. This can be achived by writing the trivial group, then the minimal subgroups, then the subgroups that contain the minimal subgroups with no intermediates, and so on. Then go through this list $H_1,H_2,H_3,\cdots$ and put a point in $\Bbb R^3$ that is higher than all of the other points but is not contained in any plane containing any three previous points. This is possible because the set of previous points is finite so the set of planes to avoid is finite and of measure zero while the half-space lying above all points has infinite measure.

There is some literature on the order-theoretic properties of subgroup lattices. From Wikipedia:

Lattice theoretic information about the lattice of subgroups can sometimes be used to infer information about the original group, an idea that goes back to the work of Øystein Ore (1937, 1938). For instance, as Ore proved, a group is locally cyclic if and only if its lattice of subgroups is distributive. Lattice-theoretic characterizations of this type also exist for solvable groups and perfect groups (Suzuki 1951).

0
On

One way to formalize what you're looking for is order dimension. It should be possible to write down finite groups whose subgroup lattices have arbitrarily high order dimension; I would look at $(\mathbb{Z}/2\mathbb{Z})^n$ for starters.