Why choose $ab$ and $ab^2$ for group with $6$ elements?

I'm referring to this proof (section (c)) which shows that a group with $6$ elements is isomorphic to either $\mathcal{C}_6$ or $\mathcal{S}_3$.

I don't quite get the part why we can pick $ab$ and $ab^2$. I guess we can choose any $c,d$ as elements that have order $2$ (meaning $c^2=e=d^2$), right? So I think the author chose $ab$ and $ba = ab^2$?

Can anyone confirm/explain?

Solution 1:

Yes, this is correct.

To see this, if $xy=yx^2$, it suffices to understand that $S_3\cong D_3$ is the semidirect product $\Bbb Z_3\rtimes \Bbb Z_2$, which, when viewed as a presentation, is

$$\langle x,y\mid x^3, y^2, xy=yx^2\rangle.$$

If $xy=yx$, the semidirect product becomes the direct product $\Bbb Z_3\times\Bbb Z_2$; and then just apply the Chinese remainder theorem.