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!

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 $x^{-1}$. 
