Non-abelian Order of $6$ is isomorphic to $S_3$

Question

I know that it's duplicate, but , How can I prove it?

I know that must be element "$a$" of order 2, and element "$b$" of order 3. What is the next step? In fact, from what I search it is claimed that if $ba=ab^2$, then the group is $S_3$. Can someone explain to me why?

Answer 1

You have six elements: $e, a, b, b^2, ab, ab^2$. Prove that they are all different.

Then consider what $ba$ is.

$ba$ can only be $ab$ or $ab^2$ because it cannot be any of the others. (Prove it!)

If $ba=ab$, then the group is abelian.

If $ba=ab^2$, then $a \mapsto (12), b \mapsto (123)$ induces an isomorphism with $S_3$.

Answer 2

As $G$ is not Abelian $Z(G)=\{e\}$ and hence $C_G(x) = \langle x\rangle$ for all $e\ne x\in G$. So the class equation reads $6=|G|=1+2n+3m$ where $n$ and $m$ are the numbers of order $3$ and $2$ elements respectively. It follows that $n=m=1$. Pick $x,y\in G$ with $|x|=3$ and $|y|=2$. Assuming that $H=\langle y\rangle$ be a normal subgroup of $ G$. we have $y^x=y$ for all $x\in G$ ,i.e $G$ is Abelian in contradiction to the assumption. So $H$ is not normal and hence $H\cap H^x=\{e\}$ for some $x \in G$ . Then $G\to S_{G/H}\simeq S_3$