Existence of an element of order $2$ and an element of order $6$ in a non-Abelian group of order $6$ Let $G$ be a non-Abelian group of order $6$. I want to prove that $G$ isomorphic to $S_ 3$.
in order to do that I first need to prove that there exist an element $a$ of order $2$ in $G$ and an element $b$ of order $3$ in $G$ such that $\langle b\rangle \vartriangleleft G$. I don't really know where to start.
P.S: I can't use sylow theorems.
 A: Every group $G$ of even order has an element of order $2$. The trick is to pair each element of $G$ with its inverse; the elements which are not of order $\leq 2$ come in pairs, so the number elements which are of order $\leq 2$ must be even. Since the identity has order $1$, the number of elements which are of order $2$ must be $\geq 1$. (I can explain this in more detail if necessary). 
Let $G$ be your group of order $6$. Note that by Lagrange, the elements of $G$ must have order $1, 2, 3$ or $6$. Only the identity of $G$ has order $1$. As noted above, $G$ must have an element of order $2$. $G$ can't have an element of order $6$, otherwise $G$ would be cyclic, hence abelian. Suppose then that every nonidentity element of $G$ has order $2$, and no element of $G$ has order $3$. Complete the proof by showing that if every nonidentity element of $G$ has an order $2$, then $G$ is abelian.
A: Any non-identity element in  a non-abelian group $G$ can have order either $2,3$.Note that all non-identity elements can't have order $2$ since if so then $a^2=e\implies a=a^{-1}$.Then $(ab)^{-1}=b^{-1}a^{-1}\implies ab=ba$ .Hence   $G$ becomes abelian.
Again not all can have order $3$ otherwise if so then if $H=\langle a\rangle $ and $K=\langle b\rangle $ where $o(a)=o(b)=3$ then $H$ is normal and $o(HK)=\dfrac{o(H)o(K)}{o(H\cap K}=9$ which can't be.
So it has one element of order $2$ say $b$ and another of order $3$ say $a$. Let $H=\langle a\rangle $ where $o(a)=3 $ then $H$ is normal.Thus $bab^{-1}\in H$ ;then $bab^{-1}=a$ or $bab^{-1}=a^2$.If $bab^{-1}=a\implies ba=ab$.Thus $o(ab)=6$ and $G$ is abelian.
Thus $bab^{-1}=a^2\implies ba=a^2b$ Thus $G=\{<a,b>:a^3=e,b^2=e\}$
A: Let $G$ be a group of order 6. Then there exists an element $x$ of order $3$. Since $\langle x\rangle$ is subgroup of order $3$, i.e. of index $2$ in $G$, it is normal. There also exists element $y$ of order $2$. Since $\langle x\rangle$ is normal, $yxy^{-1}\in \langle x\rangle=\{1,x,x^2\}$. Since $x$ and $yxy^{-1}$ have same order, so $yxy^{-1}\neq 1$. 


*

*If $yxy^{-1}=x$, i.e. $y,x$ commute, then $G$ is abelian.

*If $yxy^{-1}=x^2$ then $G=\langle x,y\colon x^3=y^2=1, yxy^{-1}=x^2\rangle$, which is isomorphic to $S_3$ 
