A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $ I know that it is duplicated. But I'm confusing some step of this proof. Please help me.
pf) Let $ G $ be a nontrivial group of order $ 6 $.
Since $ G $ is non-abelian, no elements in $ G $ have the order $ 6 $.
Assume that every element except $ e $ is of order 2.
If $ x $ and $ y $ are of order $ 2 $ and not equal. Then $ \langle x, y \rangle $ has the order $ 4 $. It is contradiction, since $ 4 $ does not divide the order of $ G $.
So, $ G $ must contain an element of order $ 3 $, say $ y $. Let $ \langle y \rangle$ and $x \langle y \rangle$ be two cosets.
Consider $ yx $.
Since $ x\notin\langle y \rangle$ and $ y\neq x$, $yx = xy$ or $yx=xy^2$ .
In the case of $yx = xy$, consider the order of $ xy$.
If the order of $ xy$ is $2$, then $y=x^2$. And so the order of $ x $ is $ 6 $ . Then it tis contradiction.
(I don't understand why it consider only when the order of $ xy $ is $ 2$ .)
Hence $yx=xy^2$. Moreover, since $ x^2 \in x \langle y \rangle$ , $ x^2 \in \langle y \rangle$ . 
(This is another confusing part. Why $ x^2$ need to be in $  x \langle y \rangle$? And even if it is true, why it means $ x^2 \in \langle y \rangle$? )
Since $ x \neq y, y^2 $, $ x = e $. Hence $ G $ is isomorphic to $ S_3 $.
 A: Here is an alternative answer:
By Cauchy's theorem, a group of  order 6 has an element $x$ of order $2$ and an element $y$ of order $3$. These two elements generate the group. The 6 elements $e$, $y$, $y^2$, $x$,$xy$, $xy^2$ must all be different from each other, hence this is the list of all elements of the group. Therefore, $yx$ must be somewhere on this list. 
Checking each element: we know that $yx\neq e$ because $x\neq y^{-1}$, $yx\neq y$ because $x\neq e$, $yx\neq y^2$ because $x\neq y$, $yx\neq x$ because $y\neq e$, and $yx\neq xy$ because by assumption the group is not abelian. Thus $yx=xy^2$, hence our group is the symmetric group $S_3$.
A: Assume (for contradiction) that $xy = yx$.
When you consider the order of $xy$, it can only be equal to $2$ or $3$, because you've already argued that the group has no elements of order $6$ and $xy = e \Rightarrow x \in \langle y \rangle$, a contradiction.
If $xy$ had order $2$, then $e = (xy)^2 = x^2 y^2$ implies that $y = y^{-2} = x^2$ (using that $y$ has order $3$), forcing $x$ to have order $6$, a contradiction.  To spell this out, clearly $x^6 = y^3 = e$, while if $x^k = e$, then $k \neq 2, 4$ because $y, y^2 \neq e$ while $k$ cannot be odd because in that case $x^k \in x \langle y \rangle$ which does not contain $e$.
Now suppose that the order of $xy$ is $3$.  Then $e = (xy)^3 = x^3 y^3 = x^3$, so $x$ has order $3$.  But $x^3 \in x \langle y \rangle$, a contradiction by the same argument as above.
This proves that $xy = y^2x$.  If $x^2 \in x \langle y \rangle$, then $x \in \langle y \rangle$, which is false by hypothesis.  So $x^2 \in \langle y \rangle$.  We wish to show that $x^2 = e$, so we must rule out $x^2 = y$ and $x^2 = y^2$.  If $y = x^2$, then as above, we argue that $x$ has order $6$, a contradiction.  If $x^2 = y^2$, then $x^2$ has order $3$, so $x = x^4 = y^4 = y$, also a contradiction.
Once you know that $x^2 = y^3 = e$ and $xy = y^2x$, it is possible to construct an explicit isomorphism from $G$ to $S_3$.
