**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.2 Exercise 2.2.14**

Let $F$ be a field and let $H(F)$ denote the Heisenberg group over $F$ as defined in Exercise 1.4.11. Compute the center of $H(F)$ and show that $Z(H(F))$ is isomorphic to the additive group $F$.

Solution: We claim that $$Z(H(F)) = \left\{ \begin{bmatrix} 1 & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \ |\ b \in F \right\}.$$ ($\supseteq$) Let $$\begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{bmatrix}$$ be an arbitrary element of $H(F)$, and let $b \in F$. Note that $$\begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & x & y+b \\ 0 & 1 & z \\ 0 & 0 & 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & x & y+b \\ 0 & 1 & z \\ 0 & 0 & 1 \end{bmatrix},$$ so that $$\begin{bmatrix} 1 & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ is indeed in the center of $H(F)$.

($\subseteq$) Let $$\begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{bmatrix}$$ be an arbitrary element of $H(F)$, and fix $$\begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}$$ in $Z(H(F))$. Now $$\begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & x+a & b+y+cx \\ 0 & 1 & z+c \\ 0 & 0 & 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & x+a & b+y+az \\ 0 & 1 & z+c \\ 0 & 0 & 1 \end{bmatrix}.$$ We demand that these results be equal. Since two matrices are equal precisely when their dimensions and corresponding entries are equal, looking at the (1,3) entries gives us that $cx = az$. With $a$ and $c$ fixed and $x$ and $y$ arbitrary in $F$, we see that $a = c = 0$.

For the second part, define a mapping $\varphi : F \rightarrow Z(H(F))$ by $$b \mapsto \begin{bmatrix} 1 & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$ It is easy to see that this mapping is an additive group homomorphism. Moreover, it is bijective. Thus $\varphi$ is an isomorphism.