Why is this Cayley diagram valid? I'm reading A Book of Abstract Algebra by Charles C. Pinter to give me an introduction to the topic, but one of the exercises has me confused. The following Cayley diagram is shown: 
But I can't understand why this isn't invalid - isn't each vertex connected to two vertices which represent the same element of of the group since the lines are all of the same type and undirected?
I'm sure the diagram is valid since it features on the front cover of the book!
 A: This depends a lot on the book's definition of "Cayley diagram". 
If the pair $(G,S)$ is such that $S=S^{-1}$ (meaning if $s\in S$ then $s^{-1}\in S$), then the information contained in the direction of the edges is redundant (since for every arrow $a\to b$ there is an arrow $b\to a$). Also although coloring each edge obtained by $s\in S$ makes it apparent which generator is responsible for the edge, a lot of the times that information is not very useful.
For those reasons in case $S=S^{-1}$ sometimes (actually almost always) people call the undirected, uncolored version, the "Cayley diagram".
Now if you take $G=\mathbb{Z}_6$ and $S=\{1,-1\}$, then your diagram is the Cayley diagram of $(G,S)$ in the undirected uncolored sense.
However, if you insist on either the colored or directions then that diagram, since it has neither direction nor color cannot be a valid Cayley diagram. 
A: Yes -- you're partially correct! Each edge represents multiplying by the same generator (element) but each vertex represents a different element of the group. So this particular diagram represents the group $C_6$ (some people write it as $(\mathbb{Z}/6\mathbb{Z})$ or $\mathbb{Z}_6$). This group consists of six elements:
$$C_6=\{ [0], [1], [2], [3], [4], [5] \}$$
and has the binary operation of addition modulus 6, that is, $[1]+[3]=[4]$ but $[2]+[4]=[0]$, since $1+3 \equiv 4 \ (\text{mod} \ 6)$ but $2+4 \equiv 0 \ (\text{mod} \ 6)$.
Anyways, this group has a special property concerning the element $[1]$. Particularly, we can represent every element in the group with $[1]$! Let's let $[1]$ be $x$. Then:
$x^0 = [1]^0 = [0] = e$
$x^1 = [1]$
$x^2 = [1] + [1] = [2]$
$x^3 = [1] + [1] + [1] = [3]$
...
$x^5 = [1] + [1] + [1] + [1] + [1] = [5]$
$x^6 = [1] + [1] + [1] + [1] + [1] + [1] = [6] = [0] = e = x^0$.
Now let's get back to the diagram. In the diagram, let one vertex be $e$. Then, traveling (lets say "counterclockwise" here) along a path is the same as multiplying by $x$ (that is, by $[1]$). As a result, here is what our diagram looks like:

So going picking any point in this diagram and going counterclockwise represents multiplying by $x$ whenever you go over an edge, that is, starting at $x^3$ for example and moving counterclockwise over two edges represents the element $x^3 \cdot x \cdot x = x^5 = [5] \in C_6$. Similarly, moving clockwise is the same as multiplying by $x^{-1}$. 
A: The diagram you showed has bidirectional edges.  That implies that the generator they represent is its own inverse. 
But if the generator were its own inverse it couldn’t generate the 6 different states depicted on its own.
Therefore the 6-cycle you’ve shown needs unidirectional edges.

The key here is that edges in Cayley diagrams only represent generators and any generator of a cycle bigger than two must be unidirectional even though the group actions are all invertible.
(If we showed all the edges that connect nodes we’d have to connect each node [a “complete graph”, of your familiar with that vocabulary], which would be difficult to parse.)
