Proving the Heisenberg Group of modulo $p$ is isomorphic to $D_8$.

Question

Prove the Heisenberg Group of modulo $p$ is isomorphic to $D_8$.

I'm having trouble specifically figuring out the way $D_8$ can be related to the unipotent upper triangle matrices with entries in $\mathbb{F}_p$ to prove isomorphism.

Answer 1

Well, presumably you need to prove that the Heisenberg group modulo two is isomorphic to $D_8$, because otherwise you have a group of order $p^3\neq 8$, and $D_8$ of course has order $8$. So let's assume you mean $p=2$.

Next, don't think about what the elements look like, but about how they behave. In $D_8$, you have an element of order $4$, $r$, and an element of order $2$, $s$, and $sr=r^3s$ ($r$ is the rotation and $s$ is the reflection, if you think of $D_8$ as the rigid motions of a square).

What about the group of matrices of the form $$\left(\begin{array}{ccc} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1 \end{array}\right),\qquad a,b,c\in\mathbb{F}_2\text{?}$$ Presumably, you've played with the Heisenberg group, so you know that $$\left(\begin{array}{ccc} 1 & a & c\\ 0 & 1 & b 0 & 0 & 1 \end{array}\right) \left(\begin{array}{ccc} 1 & x & z\\ 0 & 1 & y\\ 0 & 0 & 1 \end{array}\right) = \left(\begin{array}{ccc} 1 & a+x & c+z+ay\\ 0 & 1 & b+y\\ 0 & 0 & 1 \end{array}\right).$$ Is there an element $\mathbf{x}$ of order $4$ that can be a candidate of corresponding to $r$? Is there an element $\mathbf{y}$ of order $2$ that satisfies $\mathbf{yx} = \mathbf{x}^3\mathbf{y}$? Can you make that into an isomorphism?

(Alternatively, if you've classified the groups of order $8$, you know there are three abelian groups, namely $C_8$, $C_4\times C_2$, and $C_2\times C_2\times C_2$; and two nonabelian groups, namely $D_8$ and $Q_8$, so you would just need to check to see whether the Heisenberg group is abelian or not, and whether it can be $Q_8$ or not.)