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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.