Can anyone give any insight on this group given these generators and relations? $G = \langle x,y | x^3 = 1, y^3 = 1, (xy)^3 = 1, (xy^2)^n = 1 \rangle$
I am studying this group and I can't seem to get anywhere with it.  I've tried making a Cayley Table but it's getting pretty big.  This makes me think I'm doing something wrong.  Which doesn't have to be the case.  Maybe the group is larger than I expected.
I am assuming it's nonabelian.  My specific questions are:
Is this a relatively common group?  Does it have a name?
What is the order of G?
(Note: an earlier version of this question accidentally left out the $n$ in $(xy^2)^n$.)
 A: In $G$, we have $xy^2=1$, hence $x=y^{-2}$. Since $y^3=1$, we also have $y=y^{-2}$, so $x=y$. Thus $G$ is generated by a single element of order at most $3$. It's easy to see that the cyclic group of order $3$ satisfies all the given relations, so $G$ is that group.
A: Set $a=xy^2$ and $b=a^x = y^2x$, so that $ab= xyx$ and $ba =y^2 x^2 y^2$.  But $xyxyxy = 1$ so $xyx = y^{-1} x^{-1} y^{-1} = y^2 x^2 y^2$, so $a$ and $b$ commute, and so form a normal abelian subgroup $A$ that is a quotient of $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$, and which together with either $x$ or $y$ generates $G$.
Check that the semi direct product of $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ with $x(a) = b$, $x(b)= (ab)^{-1}$ satisfies the relations so that $G$ is a generalization of the alternating group of order 12, having in general order $3n^2$ instead of $3\cdot 2^2$.
If you omit the last relation (set $n=0$), then you get the following faithful integral matrix representation of the group.  To include the last relation, just interpret the matrices mod $n$ to get a faithful matrix rep over $\mathbb{Z}/n\mathbb{Z}$.
$$
x = \left[\begin{smallmatrix} 0 & 1 & 0 \\ -1 & -1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right]
\quad
y = \left[\begin{smallmatrix} 0 & 1 & -1 \\-1 & -1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right]
\quad
a = \left[\begin{smallmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right]
\quad
b = \left[\begin{smallmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{smallmatrix}\right]
$$
