Non-abelian group of order $6$ If $G$ is a non-abelian group of order $6$, prove that $G\cong S_3$.
I have met this problem in forum but it's solution is somewhat brief and not detailed and I cannot understand some its moments.
My efforts: Looking at the structure of $S_3$ I tried to draw up similarities between non-ableian group $G$ and symmetric group $S_3$.
Since $G$ has order $6$ then none of the elements have order $6$, otherwise it would be cyclic then abelian. Hence, all elements of $G$ except $e$ have order $2$ or $3$.
The case when all elements have order $2$ is not possible. I do not know why it is true. Please explain that.
Then one of the element has order $3$. And what to do after that I do not know.
It would be great and very useful if somebody will demonstrate the whole and detailed proof.
Remark: This problem from Herstein's book and please do not use Sylow and Cauchy's theorem  and group actions.
 A: This can be dealt with by completely elementary means. Assume that $G$ is non-abelian and has order $6$. Hence we can find two elements $a,b \in G$ with $ab \neq ba$ and $1 \notin \{a,b\}$ (here $1$ denotes the identity element of $G$ and note that $a$ is not a power of $b$ and $b$ is not a power of $a$). So the subset $\{1,a,b,ab,ba\}$ of $G$ consists of exactly five different elements. Let us have a look at $a^2$. Then an easy check gives $a^2 \notin \{a,b,ab,ba\}$. Hence $a^2=1$ or $a^2$ is a "new" element $\neq 1$.
Assume that $a^2 \neq 1$, so $G=\{1,a,b,ab,ba,a^2\}$. Then $a^3 \notin \{a,b,ab,ba,a^2\}$. So in this case we must have $a^3=1$. And also $b^2 \notin \{a,b,ab,ba,a^2\}$ (for the last element in this set: if $b^2=a^2$ then $ab^2=a^3=1$, hence $a^{-1}$ and thus $a$ is a power of $b$, contradiction), so $b^2=1$. Now the map $a \mapsto (1 2 3)$ and $b \mapsto (1 2)$ yields an isomorphism with $S_3$.
Because of symmetry, we can finally assume that $a^2=1=b^2$, that is $a=a^{-1}$ and $b=b^{-1}$. It is again easily checked that $aba \notin \{1,a,b,ab,ba\}$. Then $aba$ is the sixth element and by symmetry, $aba=bab$, so $ba=(ab)^2$ and $(ab)^3=1$. Now the map $ab \mapsto (1 2 3)$ $b \mapsto (12)$ gives the desired isomorphism with $S_3$.
A: If you do not allow the use of Sylow theorem, Cauchy's theorem or group actions, then you must construct by hand the multipilcation table of a group of order $6$, assuming it is not abelian (which rules out the cyclic case). Then, you must compare your multiplication table to that of $S_3$ and see that they are the same.
The question is: do you really want to do that?
