Every non abelian group of order $6$ is isomorphic to $\displaystyle S_{3}$ (the symmetry group of order $6$).
I tried to prove it but got stuck mid-way.
Let $\displaystyle G$ be a non-abelian group of order $6$. In $\displaystyle G,$ not all elements (non-identity) have order 2 because then the group would be abelian ($\displaystyle \forall \ a,b\in G\Longrightarrow ( ab) =( ab)^{-1} \Longrightarrow ab=ba)$.
Similarly, not all non identity elements can have order $3$ (order $3$ groups come in multiples of $2$).
We conclude there is at least one element $\displaystyle a_{1} \in G$ of order $3$ and one element $\displaystyle b_{1} \in G$ of order $2$. We have now two cases:
Case 1: $\displaystyle G$ has two order $3$ elements and three order $2$ elements.
$\displaystyle S_{3}$ also has three order $2$ elements (viz. the cycles $\displaystyle ( 12) ,( 23) ,( 31)$) and two order 3 elements (viz. the cycles $\displaystyle \left(( 123) ,( 123)^{-1}\right)$.
$\displaystyle G=\{e,a_{1} ,a_{1}^{-1} ,b_{1} ,b_{2} ,b_{3} \}$. $\displaystyle \phi :\ G\rightarrow S_{3}$ defined by $\displaystyle \phi ( e) =( 1) ,\ \phi ( a_{1}) =( 123) ,\phi \left( a_{1}^{-1}\right) =( 132) ,\ \phi ( b_{1}) =( 12) ,\phi ( b_{2}) =( 23)$ and $\displaystyle \phi ( b_{3}) =( 31)$ is an isomorphism. So $\displaystyle G\cong \ S_{3}$.
Case 2: $\displaystyle G$ has four order 3 elements and one order 2 element.
This is where I am not getting any contradiction (unless I'm trying to prove a wrong statement).
$\displaystyle G=\{e,a_{1} ,a_{1}^{-1} ,a_{2} ,a_{2}^{-1} ,b\}$. Now $\displaystyle ba_{1} \notin \{e,a_{1} ,b,a_{1}^{-1} \}$.
So we have two cases:Either $\displaystyle ba_{1} =a_{2}$ or $\displaystyle ba_{1} =a_{2}^{-1}$
If $\displaystyle ba_{1} =a_{2}$ then $\displaystyle b=a_{2} a_{1}^{2} =a_{1}^{-2} a_{2}^{-1} =a_{1} a_{2}^{-1} \Longrightarrow a_{2} a_{1}^{-1} =a_{1} a_{2}^{-1}$
How does this give any contradiction? Thanks.
PS: I am not allowed to use group actions or Sylow's theorems.
Best Answer
Let us get a contradiction in Case 2:
Suppose there are four distinct elements of order $3$. Then, they are of the form $a, a^2, b, b^2$ with $a^2 = a^{-1}$ and $b^2 = b^{-1}$. Let $c$ be the unique element of order $2$.
Consider the element $ab$. What are its possible orders? Clearly, $1$ and $6$ are not possible. What about $2$?
Well, if $ab$ has order $2$, then $ab = c$. Note that the coset $a \cdot \{1, b, b^2\}$ is equal to $G \setminus \{1, b, b^2\} = \{c, a, a^2\}$. Thus, we see that $ab^2 = a^2$.
Now, cancelling $a$ gives $b^2 = a$, which is a contradiction since $a$ and $b^2$ were distinct.
Thus, $ab$ has order $3$. But the only elements of order $3$ are $a, a^2, b, b^2$. Equating $ab$ to each of them and cancelling appropriately gives a contradiction.
As pointed out in the comments, your solution of Case 1 is not complete.