# Do the elements $rs$ and $r^2s$ generate the dihedral group $D_n$?

The dihedral group $D_n$ is generated by the elements $r$ and $s$. Is it possible for the elements $rs$ and $r^2s$ generate the group as well?

In the free group with basis $\{r,s\}$ we have $r = (r^2s)(rs)^{-1}$ and $s = (rs)(r^2s)^{-1}(rs)$.

Hence every group generated by two elements $r$ and $s$ is also generated by $r^2s$ and $rs$.

• Much more elegant than my blabbering. Thank you Zoe H. – Jacopo Stifani Jan 20 '16 at 11:20

Yes, they do indeed generate $D_n$. We already know that $\langle r,s\rangle =D_n$, so if we can show that the elements $r$ and $s$ can be written as words in $rs$ and $r^{2}s$, then $\langle rs,r^{2}s\rangle =D_n$.

First note that $sr=r^{-1}s$ implies $rs=sr^{-1}$. Therefore, $$(rs)(r^{2}s)=(rs)(sr^{-2})=rs^{2}r^{-2}=r^{-1}$$ Also, recall $(r^{k}s)^{-1}=r^{k}s$. We use these facts to express $r$ and $s$ in terms of $rs$ and $r^{2}s$: $$r=[(rs)(r^{2}s)]^{-1}=(r^{2}s)^{-1}(rs)^{-1}=(r^{2}s)(rs)$$ $$s=r^{-1}(rs)=(rs)(r^{2}s)(rs)$$

• Any feedback on my attempt is appreciated. – Jacopo Stifani Jan 20 '16 at 10:15
• Your proof is right! – Josh Chen Jan 20 '16 at 10:23
• Seems good. I corrected a small typesetting detail (it looks bad when you use <> as delimiters), but otherwise it is correct. – Arthur Jan 20 '16 at 10:25
• Thanks Arthur. Where can I find a list of all symbols? – Jacopo Stifani Jan 20 '16 at 10:39