I came across this exercise while learning about group theory and I find it hard to understand it.
Definitions: A coloured oriented graph $\mathfrak{S}$ is a 6-tuple $(V,A,C,h,t,m)$ where $V$ is the set of vertices, $A$ is the set of arrows (oriented edges), $C$ is the set of colours, $h,t:A \rightarrow V$ are head and tail functions, $m:A \rightarrow C$ is the mark function. One draws an arrow $\alpha \in A$ like
$t(\alpha) \bullet \rightarrow \bullet h(\alpha)$
Now a path in $\mathfrak{S}$ is a sequence of vertices such that any two consecutive vertices are connected by an arrow without any regard for colour and orientation. For instance, $\bullet \rightarrow \bullet \leftarrow \bullet$ is a path. Finally, for any $v \in V$, let $U(v)$ be the set of all the vertices connected to $v$ by a path. In other words, $U(v)$ are the vertices of the connected component of $\mathfrak{S}$ containing $v$.
Problem: Given a group $G$ and a subset $X \subseteq G$, define the Cayley graph $\mathfrak{S}(G,X)$ as a coloured oriented graph with
$V=G,\ A=G \times X, \ C=X, \ h(g,x)=gx, \ t(g,x)=g, \ m(g,x)=x$
How do I draw the following three Cayley graphs: $\mathfrak{S}(C_n,(g))$ , $(g$ is a cyclic generator of $C_n$), $\mathfrak{S}(D_{10}, (ab,b))$, $\mathfrak{S}(D_{10}, (a,b))$ ($a$ is a rotation by $2\pi/5$, $b$ is a reflection).
Also prove that $U(e_G)=<X>$ is in the Cayley graph $\mathfrak{S}(G,X)$. Use this fact to conclude that $X$ generates $G$ if and only if $\mathfrak{S}(G,X)$ is connected.
So, the vertices of your graph are the elements of the group, with edges for each of your chosen generators, say $x$, pointing from each element $g$ to the product $gx$ (and the edge is labeled $x$).
So, for the cyclic graph $C_n$ and a single generator $g$, you would have a cycle of $n$ vertices $(e, g, g^2, \dots, g^{n-1})$ with edges labeled $g$ pointing from one to the next.
For $D_{10}$, lay out all the elements of the group, and put edges labeled $b$ or $ab$ from each element to the result of multiplying by that generator; or edges for $a$ and $b$ if that's your chosen set of generators.
Since every element $w$ of $\langle X \rangle$ can be written as a product of elements of $X$, starting from $e_G$ and following the edges labeled by those elements gets you to the vertex for $w$. (And vice versa.) So the connected component containing $e_G$ also contains all the elements of $\langle X \rangle$.