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Think of a simple cubic planar graph $G$ with no triangles (let's call the set of such graphs plus the empty graph, $[G]$). Now pick a vertex and remove it and all its edges. You're left with a graph having 3 vertices of degree 2. Replace the edges with these left-over vertices by plain edges and you get back a cubic planar graph. It doesn't matter in which order you remove vertices in that way, so the operation is commutative.

The situation changes when you like to add vertices: Imagine you'd like to add two vertices inside the same face of a graph. It's easy to see that the order how you add vertices, in the opposite way as described above, matters, when the face, where you put the new vertex has only 4 vertices: To add a vertex you need to choose 3 edges that will be connected to the new vertex and by that it might not be possible to add the second vertex without violating planarity.

For larger face degrees there might be situations, where the addition of two vertices still commutes.

This brings us to an interesting situation:

  • All resulting elements after the operation are cubic planar graphs
  • Every addition of a vertex has an inverse element: its removal
  • Removal is commutative
  • Addition is not (in general)

What kind of algebraic structure is that? Is it a (free) group?

EDIT

You might also put another cubic planar graph $G'$ inside a given face $f$ of $G$ , choose three outer edge $(a,b,c)$ of $G'$ and now join the middle of these outer edges and connect them to the middle of three edges $(A,B,C)$ of the face where you put $G'$. This shows the closure of our operation. I'll denote this sum of graphs as $G'(a,b,c) \oplus_f G(A,B,C)$.

Associativity is shown when you realize that $$ G''(x,y,z)\oplus_{f'}\biggr(G'(a,b,c)\oplus_f G(A,B,C)\biggr)=\biggr(G''(x,y,z) \oplus_{f'} G'(a,b,c)\biggr)\oplus_fG(A,B,C), $$ where $f'$ is a face inside $G'$.

We already had some examples of inverse elements at the beginning: We inserted points and subtracted them. If we revert the operation $G'(a,b,c)\oplus_fG(A,B,C)$, we get a free-floating component $G'$ inside $G$, that can subsequently be reduce to either a tetrahedron or triangular prism graph and finally to single point, that we just drop out.

The identity element is the empty graph.

-> So all properties of a group are, more than less, fulfilled.

draks ...
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    I guess you could define the partially ordered set of simple cubic planar graph with no triangles, with relation $G \leq H$ if $G$ can be obtained from $H$ by removing a non-negative amount of vertices the way you described. An algebraic structure looks more complicated to define since the operation you are describing depends on the graph. What would be the identity, etc? – MBW Aug 17 '15 at 22:52

1 Answers1

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It's not closed, because you are working on the set of cubic planar graphs with no triangles, but once you remove a vertex you end up with a triangle, if I understand your operation correctly. Also, your "addition of vertices" operation does not seem well defined.

I don't see any way to interpret it as a group, because neither of the operations you have defined can be seen as binary operations on any set (for removal, your operation takes a graph and a vertex of the graph and gives a new graph; but I cannot compose that vertex with some other graph in any well-defined way. For addition of vertices, you are composing a graph with... some choice of three vertices that are part of a common face. And you get a new graph).

To have a group, you would need to have a set along with a binary operation on the set (as a start, of course).

xxxxxxxxx
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  • "No Triangles" was not meant that strict. What I meant was that triangles can, most of the time, easily be removed, once the appear. Also a 2-regular vertices, that might appear during the operation can be ignored. I'll update the part "addition of vertices": Think of a tetrahedron (yes I know, it has triangles) that is put inside a face $f$. Connect its outer faces to the three edges of $f$ and remove one vertices of the tetrahedron. Some 2-regular vertices drop out and your left with a single added... – draks ... Nov 16 '15 at 11:15
  • Okay, but it still is not a binary operation on planar graphs, even with these modifications. Because to combine graphs in this way, say we are putting $B$ into a face of $A$, this isn't just an operation involving the two graphs but a face needs to be specified as well. You have the same problem removing a graph $B$ from $A$: we can only remove it if $A$ has a subgraph isomorphic to $B$. Even then, $A$ may have several such subgraphs, leading to more confusion. – xxxxxxxxx Nov 16 '15 at 12:06
  • So let's break up the addition into several parts. Put B inside A and add only one new edge, connecting b of B and a of A. Now only two edges are involved. Is this binary? – draks ... Nov 16 '15 at 12:28
  • It is not, for the same reason (you still need to specify a face, and which vertices you are joining). https://en.wikipedia.org/wiki/Binary_operation#Binary_operations_as_ternary_relations – xxxxxxxxx Nov 16 '15 at 13:01
  • I only need two edges (a&b), the ones I'd like to connect (maybe I should draw it). In the first place the result graph is 1-connected so the added graph (A) can lie in one of two faces (adjacent to b). Let's call the two drawing equivalent. – draks ... Nov 16 '15 at 13:07
  • A binary operation on a set $S$ is a function $f: S \times S \to S$ (to be a function requires the map to be well-defined, among other things). So to be a binary operation on the set of planar graphs, it needs to give a new graph from the pair of graphs $(A,B)$ that does not depend on any further decisions (such as choosing two vertices to join, choosing a face to put $B$ inside, etc.), only on the graphs $A$ and $B$$ and the order they show up in the pair. – xxxxxxxxx Nov 16 '15 at 13:12
  • The set has changed to the set of edges. $f$ connects two edges $(a,b)$ Why is this not binary? Planarity is a by product. – draks ... Nov 16 '15 at 13:58
  • I don't entirely understand your new description, but it does not sound like a binary operation. You are claiming it is, on what set? The set of edges of a graph? But that is not a well-defined set if your operations are changing the set of edges. Anyhow, this discussion is getting very distant from your original question. – xxxxxxxxx Nov 19 '15 at 09:30
  • yeah true, sorry, I think I'll think again and setup a new one once I clarified some things on my own. Thanks anyway... – draks ... Nov 19 '15 at 10:10
  • My idea was to create something what I cite here called the "Generation of Barnette Graphs"... – draks ... Nov 26 '15 at 21:47