Exercise 2.2.14 (Center of Heisenberg group.)

Proof. Let $X=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill c\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$ and $Y=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill d\hfill & \hfill e\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill f\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$ be elements of $H(F)$. Then

Therefore

Hence

Suppose that $Y\in Z$, then $\mathit{af}=\mathit{dc}$ for all $a\in F$ and $c\in F$. In particular, for $a=1$ and $c=0$, we obtain $f=0$, and for $a=0$ and $c=1$, we obtain $d=0$.

Conversely, if $d=f=0$, then $\mathit{af}=\mathit{dc}(=0)$, therefore $Y\in Z$. Hence

Consider the map

Then $\phi$ is surjective by (1), and $\phi (e)=\phi (e^{\prime })\Rightarrow e=e^{\prime }$, so $\phi$ is injective.

Moreover, for all $e,e^{\prime }\in F$,

therefore $\phi$ is an isomorphism, and

In conclusion, $Z(H(F))$ is isomorphic to the additive group $F$. □