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$G = \langle x,y | x^3 = 1, y^3 = 1, (xy)^3 = 1, (xy^2)^n = 1 \rangle$

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 $(xy^2)^n$.)

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    Ok. I surely will. Sorry for the inconvenient mistake.2012-06-20

2 Answers 2

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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] $

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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.

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    I see. Thanks @Jack2012-06-20