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 $.
Best Answer
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 $S_3$.