New answers tagged automorphism-group
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Smallest graph with automorphism group $\mathbb{Z}_3 \times \mathbb{Z}_3$ ($\mathbb{Z}_3^k$)
Actually, (*) includes a construction which is based on a $3$ 'double hats' per hat and an additional 'tangling' of the respective 'tips of the hats'
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Next to minimal graph with automorphism group $\mathbb{Z}_3$
Indeed, by replacing the central triangle with a star $S_4$, a graph with $10$ vertices, $15$ edges and $\mathbb{Z}_3$ automorphism group can be constructed
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Smallest graph with automorphism group $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 = \mathbb{E}_8$ (elementary abelian group of order $8$)
Many thanks!!
Wrote a small Mathematica code to depict the case of the graph with $2k$ vertices and $4k-3$ edges and $\mathbb{Z}_2^k$ automorphism group
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Smallest graph with automorphism group $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 = \mathbb{E}_8$ (elementary abelian group of order $8$)
Generalizing Jyrki Lahtonen's example, for each positive integer $k$ there is a connected graph $G_k$ with $2k$ vertices and $4k-3$ edges whose automorphism group is $\mathbb Z_2^k$, the elementary ...
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Smallest graph with automorphism group $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 = \mathbb{E}_8$ (elementary abelian group of order $8$)
Sorry I don't know how to draw pictures but here are two examples with $7$ vertices and $9$ edges. Start with vertices $A,B,C,D,E,F,G$ and edges $AB,AC,AD, BC,CD,EF,EG$.Make it connected by adding two ...
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Smallest graph with automorphism group $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 = \mathbb{E}_8$ (elementary abelian group of order $8$)
My first attempt was irrepairable. This second attempt was inspired by user14111's example. Please upvote that.
In this graph there are two vertices of valence $4$ (blue), two of valence $3$ (red) and ...
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Accepted
Are there infinitely many groups whose smallest graph with that automorphism group has more vertices than the group has elements?
The answer to the first question (also the one in the title) is yes. If $p$ is an odd prime, then $\text{Agr}(C_p)<1$. This is not hard to show.
(Sketch: Suppose otherwise, the automorphism group ...
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Graph with many automorphisms
I could not find any example of such a graph.
If there is one, I believe this requires an approach leveraging the classification of graphs or of groups or both.
Having said that, let's play a bit with ...
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On proving that $\operatorname{Aut} A_n \cong \operatorname{Aut} S_n$?
This answer starts in the spirit of this proof that $S_n$ has only inner automorphisms for $n\ne6$ (that is, any automorphism is conjugation by some element of $S_n$). That proof involves two ideas:
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On proving that $\operatorname{Aut} A_n \cong \operatorname{Aut} S_n$?
Consider the set $T$ of all ordered triples of distinct numbers from ${1, 2, \dots, n}$. Define a function $F: A_n \to \text{Perm}(T)$ where each $\alpha \in A_n$ acts on a triple $(x, y, z) \in T$ by
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Accepted
In what sense is $\operatorname{Aut}(\mathbb P^1)$ a scheme?
Here's a rough sketch, I'm afraid I don't know where to direct you for more details. The sense in which $\text{Aut}(\mathbb{P}^1)$ is a scheme is that it has a functor of points; it's not important ...
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Accepted
Does $G$ centerless and $G\cong \operatorname{Aut}(G)$ imply $G$ complete
Let $D=\langle s,t|s^2=t^2=1\rangle=C_2*C_2$ be the infinite dihedral group. Then this answer shows that
$$\mathrm{Aut}(D)=D\rtimes C_2$$
where $C_2$ exchanges the two factors of $C_2$. Explicitly,
$$...
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Accepted
If $f \circ f$ is the identity, then $f(x) = x^{-1}$ for every $x \in G$
If $f\circ f = \operatorname{id}_G$ then you have
$$f(h(x)) = f(x^{-1}f(x)) = f(x)^{-1} x = (x^{-1}f(x))^{-1} = h(x)^{-1}$$
But since $h$ is bijective, you have
$$f(x) = f(h(h^{-1}(x))) = h(h^{-1}(x))^...
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