Cycles in Cayley graphs and group actions I have recently begun studying geometric group theory and have a couple of basic questions about Cayley graphs.
Given a (finitely generated) group $G$ with presentation $\langle x_{1},\ldots,x_{n}\mid R_{1},\ldots,R_{m}\rangle$ I know that the relations $R_{i}$ correspond to loops at the identity (and every other element). 
Lets say that a loop is a finite sequence of vertices $g_{1},\ldots,g_{k}$ where there is an edge between $g_{k}$ and $g_{1}$ and no $g_{i}$ is repeated. Does every loop in the Cayley graph corresponding to the presentation above arise as one of the relations in the presentation? 
Somewhat related is my second question, is left translation by a group element $g$ necessarily an isometry? 
 A: There are in general many loops that do not arise as one of the relations in the presentation. You can see this best by an example, take the group $\mathbb Z \oplus \mathbb Z$ with presentation $\langle a,b \mid aba^{-1}b^{-1}\rangle$. The Cayley graph can be visualized as the union of the horizontal and vertical axes in $\mathbb R^2$ with integer coordinates, that is, the union of all the lines $x=m$ and $y=n$ for $m,n \in \mathbb Z$. The only loops that correspond to the relator $aba^{-1}b^{-1}$ are ones that go around a unit square of the Cayley graph, but there are as you can surely see many, many, many other loops in the Cayley graph.
You didn't ask, but I'll add that what one can say is that any loop in the Cayley graph can be "decomposed" in some appropriate sense as a product of loops corresponding to the relators.
For your second question, left translation by $g$ is actually a graph isomorphism of the Cayley graph, and therefore an isometry. This is consequence of the definition of the Cayley graph, which has a vertex $v_h$ for each $h \in G$, and an edge $v_h$ --- $v_{hx_i}$ for each $h \in G$ and each generator $x_i$. The group $G$ acts by bijections on the vertices: for each $g,h \in G$ the left action by $g$ takes the vertex $v_h$ to $v_{gh}$. Also $G$ acts by bijections on edges: for each $g,h \in G$ and for each generator $x_i$ the left action by $g$ takes the edge $v_h$ --- $v_{hx_i}$ to the edge 
$$\text{$v_{gh}$ --- $v_{(gh)x_i}=v_{g(hx_i)}$}
$$
