Show that no asymmetric graph $G$ exists with $1 < big|V(G)big| leq 5.$
$begingroup$
Show that no asymmetric graph $G$ exists with $$1 < big|V(G)big| leq 5,.$$
I tried listing all the possibilities for $big|V(G)big| leq 5$ to prove this statement. I did all for $2$ and $3$, and then it’s getting complicated. Is there any elegant, simple solution to show all $G$ of five vertices or less must have some automorphism other than identity mapping?
combinatorics discrete-mathematics graph-theory algebraic-graph-theory automorphism-group
edited Dec 13 '18 at 1:10
Batominovski
1
asked Dec 12 '18 at 22:35
The Driven manThe Driven man
205
$endgroup$
add a comment |
$begingroup$
Show that no asymmetric graph $G$ exists with $$1 < big|V(G)big| leq 5,.$$
I tried listing all the possibilities for $big|V(G)big| leq 5$ to prove this statement. I did all for $2$ and $3$, and then it’s getting complicated. Is there any elegant, simple solution to show all $G$ of five vertices or less must have some automorphism other than identity mapping?
combinatorics discrete-mathematics graph-theory algebraic-graph-theory automorphism-group
edited Dec 13 '18 at 1:10
Batominovski
1
asked Dec 12 '18 at 22:35
The Driven manThe Driven man
205
$endgroup$
$begingroup$
You could try breaking into cases: What's the longest path in the graph? Then how do the other vertices interact with this path? For at most $5$ vertices, this is at least not too many cases.
$endgroup$
– platty
Dec 12 '18 at 22:45
add a comment |
$begingroup$
Show that no asymmetric graph $G$ exists with $$1 < big|V(G)big| leq 5,.$$
I tried listing all the possibilities for $big|V(G)big| leq 5$ to prove this statement. I did all for $2$ and $3$, and then it’s getting complicated. Is there any elegant, simple solution to show all $G$ of five vertices or less must have some automorphism other than identity mapping?
combinatorics discrete-mathematics graph-theory algebraic-graph-theory automorphism-group
edited Dec 13 '18 at 1:10
Batominovski
1
asked Dec 12 '18 at 22:35
The Driven manThe Driven man
205
$endgroup$
Show that no asymmetric graph $G$ exists with $$1 < big|V(G)big| leq 5,.$$
I tried listing all the possibilities for $big|V(G)big| leq 5$ to prove this statement. I did all for $2$ and $3$, and then it’s getting complicated. Is there any elegant, simple solution to show all $G$ of five vertices or less must have some automorphism other than identity mapping?
combinatorics discrete-mathematics graph-theory algebraic-graph-theory automorphism-group
combinatorics discrete-mathematics graph-theory algebraic-graph-theory automorphism-group
edited Dec 13 '18 at 1:10
Batominovski
1
asked Dec 12 '18 at 22:35
The Driven manThe Driven man
205
edited Dec 13 '18 at 1:10
Batominovski
1
asked Dec 12 '18 at 22:35
The Driven manThe Driven man
205
edited Dec 13 '18 at 1:10
Batominovski
1
edited Dec 13 '18 at 1:10
Batominovski
1
edited Dec 13 '18 at 1:10
Batominovski
1
1
asked Dec 12 '18 at 22:35
The Driven manThe Driven man
205
asked Dec 12 '18 at 22:35
The Driven manThe Driven man
205
asked Dec 12 '18 at 22:35
The Driven manThe Driven man
205
205
$begingroup$
You could try breaking into cases: What's the longest path in the graph? Then how do the other vertices interact with this path? For at most $5$ vertices, this is at least not too many cases.
$endgroup$
– platty
Dec 12 '18 at 22:45
add a comment |
$begingroup$
You could try breaking into cases: What's the longest path in the graph? Then how do the other vertices interact with this path? For at most $5$ vertices, this is at least not too many cases.
$endgroup$
– platty
Dec 12 '18 at 22:45
$begingroup$
You could try breaking into cases: What's the longest path in the graph? Then how do the other vertices interact with this path? For at most $5$ vertices, this is at least not too many cases.
$endgroup$
– platty
Dec 12 '18 at 22:45
$begingroup$
You could try breaking into cases: What's the longest path in the graph? Then how do the other vertices interact with this path? For at most $5$ vertices, this is at least not too many cases.
$endgroup$
– platty
Dec 12 '18 at 22:45
add a comment |
1 Answer
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oldest
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$begingroup$
It is not hard to prove that a simple graph (on more than one vertex) in which all vertices have degree at most $2$ has a nontrivial automorphism. So any counterexample must contain the following subgraph:
This immediately deals with all simple graphs $G$ with $|V(G)|leq3$.
Every disconnected graph has automorphisms from its connected components. So it suffices to prove that every connected simple graph with $4leq|V(G)|leq5$ has a nontrivial automorphism.
A graph has a nontrivial automorphism if and only if its complement does. So you can restrict to graphs with connected complements. In particular, to graphs $G$ that have no vertex of degree $|V(G)|-1$, so any counterexample must contain the following subgraph:
where $A$ and $B$ cannot share an edge. This leaves only a few graphs $G$ with $|V(G)|=5$. More specifically, those with at least one vertex of degree $3$ and no vertices of degree $4$. There are not many such graphs, you can check this yourself. In fact there are only $5$ such graphs.
answered Dec 12 '18 at 22:48
ServaesServaes
23.6k33893
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1 Answer
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$begingroup$
It is not hard to prove that a simple graph (on more than one vertex) in which all vertices have degree at most $2$ has a nontrivial automorphism. So any counterexample must contain the following subgraph:
This immediately deals with all simple graphs $G$ with $|V(G)|leq3$.
Every disconnected graph has automorphisms from its connected components. So it suffices to prove that every connected simple graph with $4leq|V(G)|leq5$ has a nontrivial automorphism.
A graph has a nontrivial automorphism if and only if its complement does. So you can restrict to graphs with connected complements. In particular, to graphs $G$ that have no vertex of degree $|V(G)|-1$, so any counterexample must contain the following subgraph:
where $A$ and $B$ cannot share an edge. This leaves only a few graphs $G$ with $|V(G)|=5$. More specifically, those with at least one vertex of degree $3$ and no vertices of degree $4$. There are not many such graphs, you can check this yourself. In fact there are only $5$ such graphs.
answered Dec 12 '18 at 22:48
ServaesServaes
23.6k33893
$endgroup$
add a comment |
$begingroup$
It is not hard to prove that a simple graph (on more than one vertex) in which all vertices have degree at most $2$ has a nontrivial automorphism. So any counterexample must contain the following subgraph:
This immediately deals with all simple graphs $G$ with $|V(G)|leq3$.
Every disconnected graph has automorphisms from its connected components. So it suffices to prove that every connected simple graph with $4leq|V(G)|leq5$ has a nontrivial automorphism.
A graph has a nontrivial automorphism if and only if its complement does. So you can restrict to graphs with connected complements. In particular, to graphs $G$ that have no vertex of degree $|V(G)|-1$, so any counterexample must contain the following subgraph:
where $A$ and $B$ cannot share an edge. This leaves only a few graphs $G$ with $|V(G)|=5$. More specifically, those with at least one vertex of degree $3$ and no vertices of degree $4$. There are not many such graphs, you can check this yourself. In fact there are only $5$ such graphs.
answered Dec 12 '18 at 22:48
ServaesServaes
23.6k33893
$endgroup$
add a comment |
$begingroup$
It is not hard to prove that a simple graph (on more than one vertex) in which all vertices have degree at most $2$ has a nontrivial automorphism. So any counterexample must contain the following subgraph:
This immediately deals with all simple graphs $G$ with $|V(G)|leq3$.
Every disconnected graph has automorphisms from its connected components. So it suffices to prove that every connected simple graph with $4leq|V(G)|leq5$ has a nontrivial automorphism.
A graph has a nontrivial automorphism if and only if its complement does. So you can restrict to graphs with connected complements. In particular, to graphs $G$ that have no vertex of degree $|V(G)|-1$, so any counterexample must contain the following subgraph:
where $A$ and $B$ cannot share an edge. This leaves only a few graphs $G$ with $|V(G)|=5$. More specifically, those with at least one vertex of degree $3$ and no vertices of degree $4$. There are not many such graphs, you can check this yourself. In fact there are only $5$ such graphs.
answered Dec 12 '18 at 22:48
ServaesServaes
23.6k33893
$endgroup$
It is not hard to prove that a simple graph (on more than one vertex) in which all vertices have degree at most $2$ has a nontrivial automorphism. So any counterexample must contain the following subgraph:
This immediately deals with all simple graphs $G$ with $|V(G)|leq3$.
Every disconnected graph has automorphisms from its connected components. So it suffices to prove that every connected simple graph with $4leq|V(G)|leq5$ has a nontrivial automorphism.
A graph has a nontrivial automorphism if and only if its complement does. So you can restrict to graphs with connected complements. In particular, to graphs $G$ that have no vertex of degree $|V(G)|-1$, so any counterexample must contain the following subgraph:
where $A$ and $B$ cannot share an edge. This leaves only a few graphs $G$ with $|V(G)|=5$. More specifically, those with at least one vertex of degree $3$ and no vertices of degree $4$. There are not many such graphs, you can check this yourself. In fact there are only $5$ such graphs.
answered Dec 12 '18 at 22:48
ServaesServaes
23.6k33893
edited Dec 12 '18 at 23:32
answered Dec 12 '18 at 22:48
ServaesServaes
23.6k33893
answered Dec 12 '18 at 22:48
ServaesServaes
23.6k33893
answered Dec 12 '18 at 22:48
ServaesServaes
23.6k33893
23.6k33893
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$begingroup$
You could try breaking into cases: What's the longest path in the graph? Then how do the other vertices interact with this path? For at most $5$ vertices, this is at least not too many cases.
$endgroup$
– platty
Dec 12 '18 at 22:45