This is what I have done so far (the prev question).
@BalarkaSen, although I have missed a short part of the proof, here is what I have done far. I am trying to fully complete it (the missed part).
Let $a \in G$, with $o(G)=6$. Then either $o(a)=3$, or $o(a)=2$. The number of elements of order $2$ in $G$ must be odd. Therefore, the number of elements of order $2$ is either $3$ or $1$.
It cannot be the case where $|\{a \in G: o(a)=2\}|=1$ (I missed the proof here, will come back to it). So, let the elements be $ p, q, r , x, y, e$, with $ o(p)=o(q)=o(r)=2$ and $o(x)=o(y)=3$. Now, we ass…