I've managed to answer a few of the latter questions. <b>Theorem.</b> There is an uncountable graph Γ that is not a Cayley graph in the set-theoretic universe V, but which is a Cayley graph in a larger set-theoretic universe, obtained by forcing. Proof. Let Γ be a directed graph that is a tree, such that all vertices have infinite in-out degree, but such that some of these degrees are countable and some uncountable. For example, perhaps all the in-degrees are countable and all out-degrees are uncountable. It is not difficult to construct such a graph. Clearly, Γ cannot be a Cayley graph, since the degrees don't match. Denote the (current) set-theoretic universe by V, and perform forcing to a forcing extension V[G] in which Γ is countable. (It is a remarkable fact about forcing that any set at all can become countable in a forcing extension.) In the extension V[G], the graph Γ is the countable tree in which every vertex has countably infinite in-out degree. Thus, in the forcing extension, Γ is the Cayley graph of the free group on countably many generators.QED <b>Theorem.</b> There is a graph Γ, which is not a Cayley graph, but every finite subgraph of Γ is part of a Cayley graph. Indeed, every countable subgraph of Γ is part of a Cayley graph. Every countable subgraph of Γ can be extended to a larger countable subgraph of Γ that is a Cayley graph. Proof. The same graph Γ as above works. Every countable subgraph of Γ involves only countably many edges, and can be placed into a countable subgraph of Γ that is the Cayley graph of the free group on countably many generators. QED The conclusion is that you cannot tell if an uncountable graph is a Cayley graph by looking only at the finite subgraphs, or indeed, by looking only at its countable subgraphs. These observations show that in the uncountable case, one should just restrict the question at the outset to graphs satisfying the degree-matching condition. -------- Let me also give the fuller details for the finite-degree case, following the suggestion of François Dorais (and my tree-of-attempts argument). <b>Theorem.</b> There is a finitistic condition on countable graphs Γ that holds exactly of the finite-degree Cayley graphs. Specifically, the set of finite-degree countable Cayley graphs has complexity at most Σ<sup>0</sup><sub>3</sub> in the arithmetic hierarchy. Proof. Let us suppose that the graph Γ is given to us simply as a binary relation on ω, that is, an element of 2<sup>ωxω</sup>. The assertion that the graph is connected has complexity Π<sup>0</sup><sub>2</sub>, since you must only say that every two vertices are connected by a finite path. The assertion that the graph has finite degree and all in-out degrees are the same has complexity Σ<sup>0</sup><sub>3</sub>, since you can say "there is a natural number k such that for all vertices v there are k vertices w<sub>1</sub>,...,w<sub>k</sub> pointing at v, such that no other vertex points at v (and similarly for pointing out). Now, suppose that Γ is a connected directed graph and all in-out degrees are k. Fix a node e that will represent the group identity (if Γ is a Cayley graph, any node will do since they all look the same, so take e=0). Let us refer to the k nodes pointed at from e as the set of generators (since if Γ really is a Cayley graph, these will be the generators). Define that p is a *partial labeling* of the edges of Γ with generators, if p is a function defined in a finite distance ball of e in Γ, where p labels each arrow in that ball with a generator. Such a labeling is *coherent* if first, for every node every generator is used once going out from that node and once going into that node, and second, if for every node v, if there is a loop starting and ending at v, then the word w obtained from that loop works as a loop from every node to itself. We only enforce the requirements on coherence of p as far as p is defined (since it is merely a partial function). Thus, a coherent labeling is an attempt to make a labeling of the the graph into a Cayley graph, that has not run into trouble yet. Let T be the collection of finite coherent labelings, labeling all edges on the first n nodes for some n. This is a finitely branching tree under inclusion. I claim that Γ is a Cayley graph if and only if there are arbitrarily such large coherent labelings. The forward direction is clear, since we may restrict a full coherent labeling to any finite subgraph. Conversely, if we can find a coherent labeling for any finite subgraph of Γ, then the tree T is infinite and finitely branching. Thus, by Konig's lemma there is an infinite branch. Such a branch labels all the edges in Γ and satisfies the coherence condition. Such a labeling exactly provides a group presentation with Cayley graph Γ. The complexity of saying that every initial segment of the nodes admits a coherent labeling is Π<sup>0</sup><sub>2</sub>, since you must say for every n, there is a labeling p of the first n nodes such that p is coherent. Being coherent is a Δ<sup>0</sup><sub>0</sub> requirement on p. QED In particular, to recognize when a graph is a finitely generated Cayley graph does not require one to quantify over infinite objects, and so for finite degree graphs, this is an answer to the main question. I was surprised that the part of the description driving the complexity is the assertion that the degrees must all match, rather than the assertion that there are coherent labelings on the finite subgraphs. I suspect that my complexity calculation is optimal, since it seems difficult to simplify the description that the degrees match. This still leaves the case of countably many generators wide open. Also, this still doesn't answer the question about having a computable graph Γ, which is a Cayley graph, but which has no computable presentation. Such an example would be very interesting. Even in the finite degree case, as I mentioned in the question, the best the argument above produces is a low branch.