The poset ( \mathcal{P}_n ) of isomorphism classes of graphs with ( n ) vertices (for ( n \geq 5 )) ordered by subgraph inclusion is analyzed to determine if it has nontrivial automorphisms. An automorphism must preserve the partial order, meaning it maps each graph class to another class with identical inclusion relations.

Key observations:

1. Grading by Edge Count: The poset is grade...

Short answer: This is open in general. It would indeed follow from the (strictly stronger) statement that a graph on n vertices is determined by the set (not multiset) of isomorphism types of its edge-deleted subgraphs, but that stronger statement is false, so a different approach would be needed.

A few relevant points.

• Automorphisms preserve rank. The poset is graded by the number of edges, w...

This is a very interesting question about the rigidity of the poset structure on graphs. Let me work through this problem.

First, let me clarify what we're asking: we want to know if every automorphism of (i.e., every order-preserving bijection ) is trivial (the identity).

Key Observations

1. What automorphisms preserve: Any automorp...

Is the Poset of Graphs Automorphism-free?

This is an interesting question about the structure of the poset of isomorphism classes of graphs with vertices ordered by the subgraph relation.

Why Edge Reconstruction Would Imply No Automorphisms

First, let me clarify why being able to recognize a graph from its edge-deleted subgraphs would imply the poset has no non-trivial ...

Nobody knows the answer at the moment.

1. What would be needed.
   An order–automorphism of the poset
   Pn = {isomorphism classes of graphs on n vertices},
   G ≤ H ⇔ G is a (not-necessarily-induced) subgraph of H
   is a bijection φ : Pn → Pn such that
   G ≤ H ⇔ φ(G) ≤ φ(H).
   Let Ĝ be the set of isomorphism classes of graphs obtained from G by deleting a s...

<think> I need to determine if the poset of graphs for n vertices, with n at least 5, has no nontrivial automorphisms when ordered by the subgraph relation. The poset is made up of isomorphism classes of graphs, so we're dealing with unlabeled graphs.

First, recall that an automorphism of a poset is a bijection that preserves the order. So, a nontrivial automorphism would be one that isn't the id...

No complete answer is known – at the present time nobody knows whether the poset

        Pn  =  {isomorphism classes of n–vertex graphs}  
               ordered by the relation “is an (un-induced) subgraph of”

is rigid for every n ≥ 5. In other words, it is not known whether every order–automorphism of Pn must be the identity.

What is known

1. The problem is at least as hard as th...

This is an excellent question that connects poset theory with a deep problem in graph theory, the reconstruction conjecture. The answer to your question is yes.

Summary of the Answer

Yes, for , the poset has no nontrivial automorphisms. Your proposed line of reasoning is correct and is indeed the standard way to prove this. It relies on a strong version of the edge r...

Here is what is known (and not known) about automorphisms of the subgraph‐poset

Pₙ := {isomorphism–classes of graphs on n vertices},
ordered by G ≤ H ⇔ G is (spanning) subgraph of H,

when n ≥ 5.

1. Any poset–automorphism φ of Pₙ must preserve the unique minimum ∅ and maximum Kₙ, and must carry a graph with e edges to another graph with e edges. Hence φ preserves the “rank’’ e = |E(...