Let 
be a Cayley graph of group $G$ ( $e$ is the neutral element ).
Is the group commutative? How do I see that from the graph ?
I am not sure if my thinking is correct, the red arrows mean multiplying from the right by $a$, and the blue lines mean multiplying on both sides by $x$? Now I took for example the element $c^2$ and wrote it as $ c^2 = x*a*a$ then when multiplied on the right by $a$ by reading the graph it is $ {c}^{2} * a = x$ , but when multiplying from the left by $a$ I get $ a* {c}^{2} = a*x*a*a $, which is $z$ by looking at the graph.That would mean it is not commutative. Is this thinking correct?
How do I find for example ${c}^{-1}$ ? And in general cosets from the graph? For example $ b * < ax> $, where $< ax> $ is the orbit of $ax$?
From the graph, $x$ is an involution and $a$ is of order $3$. We see that $ax = c$ but $xa = b\ne c$. Hence $G$ is not commutative. For $c^{-1}$, we have $c^{-1} = (ax)^{-1} = x^{-1}a^{-1} = xa^2 = c^2$ from the graph.
Thus, there are at least two subgroups of order $3$ in $G$, namely $\langle a\rangle$ and $\langle c\rangle$. Therefore, we have 4 Sylow $3$-subgroups. $G$ is then isomorphic to $(\mathbb{Z}_2\times\mathbb{Z}_2):\mathbb{Z}_3$, the semi-direct product, which is actually $A_4$.
Indeed, use the standard notation in $A_4$ we have \begin{equation*} \begin{cases} a = (132)\\ b = (143)\\ c = (234)\\ x = (12)(34)\\ y = (14)(23)\\ z = (13)(24) \end{cases} \end{equation*} and the set $S$ of the Cayley graph is $S = \{x,a\} = \{(12)(34),(132)\}$.