Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26
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Cayley-graph neighbourhoods are equipotent, and local finiteness is equivalent to finiteness of the symmetrised subset

Statement

Let G be a group, S⊆G, and S±:=(S∪S−1)∖{e}. Left translation gives a bijection between the neighbourhoods of any two vertices of Cay⁡(G,S). The graph is locally finite exactly when S± is finite; in that case it is regular of finite degree ∣S±∣.

Facts & Assumptions

Given: A group G, a subset S⊆G, and S±=(S∪S−1)∖{e}.

[F1]

The Cayley graph of a group G with respect to a subset S has vertex set G and edge set {{g,gs}:g∈G, s∈(S∪S−1)∖{e}} (The Cayley graph of a group with respect to a subset).

[L1]

A graph is locally finite when every vertex has finitely many neighbours (Locally finite graphs and vertex degree without a finiteness hypothesis).

[L2]

The degree of v is deg⁡G(v):=∣NG(v)∣, equivalently the number of edges incident with v. A graph is r-regular when every vertex has degree r; it is cubic when it is 3-regular. (Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

[L3]

A set A is finite when A≈n for some n∈N. (The cardinality ∣A∣ of a finite set).

Proof

technique · direct
1.1F1L2

The neighbours of g are the elements gs with s in the symmetrised set minus the identity, and left multiplication by hg−1 is a bijection from the neighbours of g to those of h.

2.1F1L1L2L3step 1.1∎

Thus one neighbourhood is finite exactly when all are, which occurs exactly when S± is finite. In that case the degree is defined at every vertex and equals ∣S±∣, so the graph is regular of that finite degree.

Depends on

Used by

Dependency tree · two levels

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Sources