Is $G=\langle r, f\,:\, r^3=e , f^3=e, fr=r^2f\rangle$ well known? In trying to construct a non-commutative group with exactly $1$ nilpotent of degree $2$, I ended up constructing one with the following presentation: $$G=\langle r, f\,:\, r^3=e ,  f^3=e, fr=r^2f\rangle$$ (which is noncommutative and has no nilpotents of degree $2$ so $\mathbb{Z}_2 \times G$ satisfies what I wanted).
This group $G$ reminds me a lot of dihedral groups. Is it a well known group?
 A: If $fr= r^2f$ and $r^3=e$, then you have $frf^{-1}=r^{-1}$. But then $f^2rf^{-2}=r$ and $r = f^3rf^{-3}=r^{-1}$. So $r^2=e$; since $r^3=e$, then $r=e$, and so your group is just the cyclic group of order $3$ generated by $f$.
A: Looking at the presentation, one might expect that the group you've constructed is order $9$ since every element can be written as $r^af^b$ where $0\leq a,b\leq 2$.
There are only two groups of order $9$, but they are both abelian (this follows since $9=3^2$ and all groups of order $p^2$ are abelian when $p$ is a prime).  In an abelian group, $fr=rf$, but this also equals $r^2f$, so $r=r^2$ and $r$ is the identity, which is not possible.
Therefore, this group is of order less than $9$.  Unless it is the trivial group, one of $r$ and $f$ must be nonidentity, so the group must be be of order a multiple of $3$, i.e., either order $3$ or $6$.
This leaves three groups: $\mathbb{Z}/3$, $\mathbb{Z}/6$, and $S_3$.  $S_3$ is not generated by elements of order $3$ (they are all even), so this group is abelian.  Now, in abelian groups, $fr=rf$, so $r=r^2$ and so $r$ is the identity.
Therefore, this group is the cyclic group of order $3$.
