Center of Heisenberg group- Dummit and Foote, pg 54, 2.2

Let $H(F)$ be the Heisenberg group over the field $F$ introduced in Exercise 11 of Section 1.4. Determine which matrices lie in the center of $H(F)$ and prove that $Z(H(F))$ is isomorphic to the additive group $F$.

Section 1.4 defines the Heisenberg group $H(F)$ over the field $F$ to be the group of all upper-unitriangular $3\times 3$-matrices $\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$ with entries in $F$.

When I worked out the elements in the center, I got its elements such that the diagonal was multiplied by a scalar and the element in the first row,third column possibly non-zero. So, you see that my problem was compounded as I just did not see how the center was isomorphic to only F.

Thanks for any help that I might receive.

Let $$ Y = \begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix} \in Z \big( H(F) \big) $$ and $$ A = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} \in H(F). $$ We need to determine conditions on $Y$ so that the commutator vanishes for arbitrary $A$. $$ 0 = AY - YA = \begin{pmatrix} 0 & 0 & az - cx \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ CLAIM: $x = z = 0$.

Why? If $x \ne 0$, then with $a = 0$ and $c \ne 0$, $AY - YA \ne 0$. Similarly for $z$.

Notice that there are no restrictions on $y$ since it doesn't appear in the commutator. Thus, the center consists of matrices of the form $$ Y = \begin{pmatrix} 1 & 0 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$

You can check that they form a subgroup isomorphic to the additive group underlying $F$: $$ \begin{pmatrix} 1 & 0 & y_1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & y_2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & y_1 + y_2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$