# Show group is isomorphic to finite Heisenberg group

Show that the group $\langle x,y,z$ $|$ $z = xyx^{-1}y^{-1}$, $zx=xz$, $zy = yz$, $x^n = \mathbb{I},$ $y^n = \mathbb{I}$, $z^n = \mathbb{I} \rangle$, $(n \in \mathbb{Z_{>0}})$ is isomorphic to $H_n = \left\{ [x,y,z]:= \left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array} \right) | x,y,z \in \mathbb{Z_n} \right\}$ i.e. the finite Heisenberg group.

Now my first thought was to simply define the following maps $\phi: x \rightarrow X, y \rightarrow Y, z \rightarrow Z$ where $X = \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right), Y = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right), Z = \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$ for which I know that the above relations hold. Is the idea here simply that I have to show that these relations really hold for the given matrices (i.e. computing some matrix multiplications) or is there more to this exercise? I'm a bit confused by how easy it seems to be and think I'm missing something.

Cheers!

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You have to show, that these matrices don't have more relations, too. Let's call your group $G$. With your observation, you get a group homomorphism $\phi\colon G\rightarrow H_n$. Is it injective? Is it surjective? – archipelago Mar 24 '13 at 16:12
Any hints how to see whether or not these are all relations there are? – Howdy Ho Mar 24 '13 at 16:30
Well, looking at things a slightly different way, the group $H_{n}$ clearly has order $n^{3}.$ The abstract group $G$ as defined has $H_{n}$ as a homomorphic image. If you could show that $|G| \leq n^{3},$ you would be home. – Geoff Robinson Mar 24 '13 at 16:35
Tbh I don't think that's what they want from us. We didn't even define what a homomorphic image is. Any ideas how to do it with more straight forward calculations? – Howdy Ho Mar 24 '13 at 16:40
I don't believe that there is any other way of doing it. To show that two groups are isomorphic, you have to find a bijective homomorphism from one group to the other! Showing that $|G| \le n^3$ is not particularly hard. Just use the group relations to show that every group element can be written as $x^ij^jz^k$ with $0 \le i,j,k, < n$. – Derek Holt Mar 24 '13 at 16:42