A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $

Question

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 $.

Answer 1

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 $S_3$.

Answer 2

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 $S_3$.

Answer 3

Below I give a different proof, which uses Cauchy's Theorem and the Cayley homomorphism:

By Cauchy's Theorem, there exists $a,b\in G$ of order $2$ and $3$, respectively. Then, let $H=\{1,a\}$ to be the subgroup of $G$ generated by $a$ and consider the Cayley homomorphism $$ \phi:G\to S(G/H); \ g\mapsto \lambda_{g},$$ where $\lambda_{g}(xH)=gxH$ and $S(G/H)$ is the permutation group of the quotient set $G/H$. It can be easily checked that $\ker{\phi}\subseteq H$. Therefore, either $\ker\phi=\{1\}$ or $\ker\phi=H$. If the latter holds then $H$ is a normal subgroup of $G$ and so, since $a\in H$ we have that $bab^{-1}\in H$. But if $bab^{-1}=1$ then $a=1$, which cannot hold and if $bab^{-1}=a$ then $ba=ab$, which implies that the order of $ab$ equals $2\cdot 3=6$, and so $G$ is a cyclic group with generator $6$ and thus abelian, a contradiction. Therefore, $\ker\phi=\{1\}$ and so $\phi$ is an isomorphism $G\cong S(G/H)\cong S_{3}$, since $|G|=|S_{3}|=6$.