Basic Group Theory Question

Question

Consider the dihedral group $D_3$ of order $|D_3|=6$ comprised of the $6$ symmetries $D_3=\{e,r,r^2,rs,r^2s\}$ of an equilateral triangle ($e=r^0s^0$ is the do-nothing symmetry, $r$ is a rotation by $2\pi/3$, and $s$ is a reflection about an axis through the center of the triangle and one of its $3$ vertices). It's clear that $r^3=s^2=e$. However, it's not as immediately apparent that $D_3$ is a non-abelian group with $sr=r^{-1}s\neq rs$. Right now though, the only way I'm able to "derive" the identity $sr=r^{-1}s$ is to think geometrically about how $D_3$ acts on the vertices of the equilateral triangle. Instead, I was wondering if there's a purely algebraic way to derive $sr=r^{-1}s$ using only the identity $r^3=s^2=e$ together with the axioms for a group.

Answer 1

Using only the identities you've listed ($r^3=s^2=e$), it's still possible that your group is $\mathbb Z_2 \times \mathbb Z_3$.

And in that group, $sr = rs$, not $r^{-1} s$ (and $r^{-1} \ne r$), so we have in fact proved that we cannot derive $rs = r^{-1}s$ using purely algebraic methods, for the same proof would work in $\mathbb Z_2 \times \mathbb Z_3$ too. But it can't, because it's not true there.

Answer 2

The group that is generated by $s,r$ and satisfies only $r^3=s^2=e$ is the free product $G=\mathbb{Z}_2\ast \mathbb{Z}_3$. If you wish to have more relations, like $rs=sr$ or $rs=sr^{-1}$, then you have to take a quotient of $G$ with the normal subroup generated by $rsr^{-1}s^{-1}$ or $rsrs^{-1}$, respectively.