I have a combinatorial graph with vertex set [itex]\mathbb{Z}[/itex] such that the set [itex]\{\{i,i+1\}\,:\,i\in\mathbb{Z}\}[/itex], which are the geometric edges, is its geometric realization. The graph is basically the real line with vertices on the integers, and where every edge connects an integer i with i-1 and i+1.(adsbygoogle = window.adsbygoogle || []).push({});

As well as having this graph, I have been given two automorphisms

[tex]s(i):=-i[/tex]

[tex]t(i):=-i+1[/tex]

defined on the vertices of the graph. My task is to show that these two automotphisms generate the automorphism group of the graph.

Questions:

[1] It is my understanding that an automorphism group contains automorphisms as "elements" (I'm used to thinking of groups as collections of objects that satisfy the four group axioms). Is this right?

[2] To show that my two automorphisms generate an automorphism group of this graph must I show the following four things:

i) Closure: The composition of the given 2 automorphisms in any way returns a composition of automorphisms.

ii) Associativity: [tex](s\circ t) \circ s = s\circ (t\circ s)[/tex] and [tex](s\circ t)\circ t = s\circ (t \circ t)[/tex].

iii) Identity: This is easy, right? Is the identity automorphism simply [itex]s\circ s[/itex]? Or [itex]t\circ t[/itex]? Both composition automorphisms map i to i. Perhaps Im getting this confused with s(0) = 0? I dont know!

iv) Inverse: Surely s(i) is the inverse automorphism. Since s(i) maps i to -i.

If I show all four of these then have I shown that s and t generate the automorphism group for the graph?

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# Automorphism Group

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