A group with six elements which are given partially by relations. In a textbook I saw the following Group 
$$
  G = \{ 1, x, x^2, y, xy, x^2y \}
$$
and it was said that it is the $S_3$, surely the $S_3$ is a model of this group, but when I set $x^3 = y$ (in $S_3$ the relation would be $x^3 = 1$) I get that
$$
  G = \{ 1, x, x,^2, x^3, x^4, x^5 \}
$$
and this group is cyclic, hence commutative, so could not be the $S_3$. I cannot see were this would be a contradiction to $ G = \{ 1, x, x^2, y, xy, x^2y \}$ so I think this group has at least two models, is this true? So why did my textbook said the group $S_3$ is $G = \{ 1, x, x^2, y, xy, x^2y \}$?
 A: Your list of expressions for group elements has the nice property that if the expression $u \cdot v$ is in the list, then so are $u$ and $v$. Such a list turns the Cayley graph of the group into a directed acyclic graph with a single source (the identity), or perhaps better a spanning tree with direction. The multiplication in the group is given by the (labeled) edges omitted from the spanning tree.
A wonderful way to create such a directed graph is to decide on a “reduction ordering” on expressions. Many expressions will yield the same group element, so we desire a way to choose the “simplest” expression. If $u\cdot v$ is simplest, then both $u$ and $v$ should be simplest (otherwise why not simplify them).
Such a reduction ordering determines a special set of relations that suffice to actually simplify expressions. These relations are called a “rewriting system” and they are the natural thing: if $u$ is not a simplest form (it equals $v$ where $v$ is simplest), but every proper sub-expression of $u$ is a simplest, then we need the rule $u \mapsto v$.
Doing this for your list of expressions finds the rules $x^3 \mapsto ?$, $y^2 \mapsto ?$ and $yx \mapsto ?$. All other missing edges of the Cayley graph can be found from these by a sort of “translation” or “substitution”.
I believe these are the only possibilities:
$\begin{array}{ccc|c}
x^3 & y^2 & yx & G \\ \hline
1 & 1 & xy & C_6 \\
1 & 1 & x^2y & S_3 \\
y & 1 & xy & C_6 \\
1 & x & xy & C_6 \\
1 & x^2 & xy & C_6 
\end{array}$
As you can see the choices are limited, and not independent. Additionally $C_6$ appears in several disguises.
