Difference between the Cayley Graph and the Cayley Sum Graph. Could someone help me visualize the difference between the following graphs?
Take $G$ to be a group generated by the symmetric generating set $S$. Take $g, h$ to be elements of $G$. We define the following graphs:
Cayley Graph: The vertices $g, h$ in $G$ are adjacent iff $g = hs$ for some $s\in S$.
Cayley Sum Graph: The vertices $g, h$ in $G$ are adjacent iff $g = h^{-1} s$ for some $s\in S$.
 A: Cayley Graph
Since you have chosen $S$ to be symmetric, the Cayley Graph is naturally an undirected graph since the relation $h^{-1}g\in S$ is actually symmetric. Also there are no issues with self-loops on vertices since either $e\in S$ so all loops are there, or $e\not\in S$ so no loops are there. I guess one usually doesn't put $e$ in their generating set, for instance if $S$ is minimal. Another property of the Cayley Graph is connectivity since if $x,y\in G$ then we can write
$$
y^{-1} x = s_{1} s_{2} \ldots s_{2}
$$
for some elements of $S$. Now there is a path from $y$ to $x$ along the vertices $y, ys_1, ys_1s_2, \ldots,ys_1s_2\ldots s_n=x$.
Cayley Sum Graph
Your Cayley Sum Graph is based on the relation $hg\in S$. This might not be a symmetric relation even with the assumption that $S$ is symmetric. Here is an example:
Example 1. In the group $S_3$ we can take the generating set of adjacent transpositions $\{(1 \space 2),(2\space  3)\}$. Then there is an edge from $(1\space 2 \space 3)$ to $(2\space 3)$ since the product $(1 \space 2)$ is in the generating set. But there is no edge from $(2\space 3)$ to $(1\space 2\space 3)$ because the product $(1\space 3)$ is not in the generating set.
But since you're calling it a SUM graph then maybe you mostly care about these in the abelian case. Of course for abelian groups the edge relation in the Cayley Sum Graph is symmetric. So let's just assume we're working with abelian groups. (Now we don't even need symmetry of $S$ to get a symmetric graph relation. But symmetry of $S$ is important for connectivity as we shall see soon.)
But then you still have the issue of self loops since for any $x\in G$ if $2x\in S$ then there is an edge from $x$ to itself. So things could get weird. Here is an example.
Example 2. Consider $\mathbb{Z}_5$ with symmetric generating set $\{1,4\}$. Then the Cayley Graph is just a $5$-cycle. But the Cayley Sum Graph looks like the path $(2,4,0,1,3)$ with self loops at the end on $2$ and $3$. A similar situation happens with $\mathbb{Z}_n$ for any odd $n$. For instance if $G$ has odd size then every element $s\in S$ has a "square root" $t$ for which $s=t+t$ and so there will be a self loop on $t$ in the Cayley Sum Graph.
However the Cayley Sum Graph is still connected.  To prove this we just need to demonstrate a path from any element of $G$ to $0$. So let $x$ be in $G$. We can write
$$
x = s_{1}+ s_{2} + \ldots + s_{n}
$$
for some elements of $S$. If $n$ is even then there is a path
$$
0,\space -s_{1}, \space s_{1}+s_{2},\space -s_{1}-s_{2}-s_{3},\space s_{1}+s_{2}+s_{3}+s_{4},\space \ldots, \space s_{1}+s_{2}+\ldots +s_{n} = x
$$
If $n$ is odd then there is a path
$$
0,\space s_{1}, \space -s_{1}-s_{2},\space s_{1}+s_{2}+s_{3},\space \ldots, \space  s_{1}+s_{2}+\ldots +s_{n} = x
$$
The proof is finished.
In the last proof the fact that $S$ is symmetric was important. For example the Cayley Sum Graph of $\mathbb{Z}_n$ with respect to the generating set $S = \{1\}$ is disconnected as long as $n>1$.
