I've seen some proofs of this fact but they seemed more complicated than necessary, so I've tried to come up with my own, which is hopefully simpler. Can you kindly tell me if it's correct?
Proof. $G$ cannot have an element of order 6, else it would be cyclic and thus abelian. So the elements of $G$ different from $e$ can only have order 3 or 2. If all $g \in G$ have order 2 then $G$ is abelian, so there must be at least one element $a \in G$ of order 3. The subgroup $H= \langle a \rangle$ is normal in $G$ because its index is 2. $G/H$ is a cyclic group of order 2. Let $b \not \in H$. We have $(bH)^2=H \implies b^2 \in H$. If $o(b)=3$, then $ b=b^4=b^2b^2 \in H$, a contradiction. So we must conclude that every $g \in bH$ has order 2.
We have $G= H \cup bH = \{e, a, a^2, b, ba, ba^2 \}$. Moreover, $ab \in Hb = bH$ so $(ab)^2=e$. Then $ab=b^{-1}a^{-1}=ba^2$, which is exactly the relation in $S_3$. $\square$
from Hot Weekly Questions - Mathematics Stack Exchange
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