# The Center of the Heisenberg Group Over a Field $F$ is Isomorphic to the Additive Group $F$

## Problem 283

Let $F$ be a field and let
$H(F)=\left\{\, \begin{bmatrix} 1 & a & b \\ 0 &1 &c \\ 0 & 0 & 1 \end{bmatrix} \quad \middle| \quad \text{ for any} a,b,c\in F\, \right\}$ be the Heisenberg group over $F$.
(The group operation of the Heisenberg group is matrix multiplication.)

Determine which matrices lie in the center of $H(F)$ and prove that the center $Z\big(H(F)\big)$ is isomorphic to the additive group $F$.

## Proof.

Suppose that the matrix
$M=\begin{bmatrix} 1 & x & y \\ 0 &1 &z \\ 0 & 0 & 1 \end{bmatrix}$ is in the center of the Heisenberg group $H(F)$.
Let
$A=\begin{bmatrix} 1 & a & b \\ 0 &1 &c \\ 0 & 0 & 1 \end{bmatrix}$ be an arbitrary element in $H(F)$.

Since $M$ is in the center, we have $AM=MA$, that is,
$\begin{bmatrix} 1 & a & b \\ 0 &1 &c \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & x & y \\ 0 &1 &z \\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & x & y \\ 0 &1 &z \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & a & b \\ 0 &1 &c \\ 0 & 0 & 1 \end{bmatrix}.$

Computing the products, we obtain
$\begin{bmatrix} 1 & x+a & y+az+b \\ 0 &1 &z+c \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & a+x & b+cx+y \\ 0 &1 &c+z \\ 0 & 0 & 1 \end{bmatrix}.$ Comparing $(1,3)$-entries, we have
$az=cx.$ This equality must be true for any $a, c \in F$.

We claim that $x=z=0$.
Taking $a=0, c=1$ (Note that since $F$ is a field, $0, 1 \in F$),
we have $x=0$. Also if $a=1, c=0$, then we have $z=0$.
Thus $x=z=0$ and the matrix $M$ becomes
$M=\begin{bmatrix} 1 & 0 & y \\ 0 &1 &0 \\ 0 & 0 & 1 \end{bmatrix}.$

It is clear from the computation of $AM=MA$ that this matrix is in the center for any $y$.
Therefore we have determined the center of the Heisenberg group:
$Z\big(H(F)\big)=\left\{\, \begin{bmatrix} 1 & 0 & y \\ 0 &1 &0 \\ 0 & 0 & 1 \end{bmatrix} \quad \middle| \quad \text{ for any } y\in F \, \right \}.$

To prove that the center $Z\big(H(F)\big)$ is isomorphic to the additive group $F$, consider the map
$\phi: Z\big(H(F)\big) \to F$ which sends
$M=\begin{bmatrix} 1 & 0 & y \\ 0 &1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \in Z\big(H(F)\big)$ to $y\in F$.
We prove that the map $\phi$ is a group isomorphism.

Let
$M=\begin{bmatrix} 1 & 0 & y \\ 0 &1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, M’=\begin{bmatrix} 1 & 0 & y’ \\ 0 &1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ be any two elements in the center $Z\big(H(F)\big)$.
Then we have
$MM’=\begin{bmatrix} 1 & 0 & y+y’ \\ 0 &1 & 0 \\ 0 & 0 & 0 \end{bmatrix}.$ Therefore we have
$\phi(MM’)=y+y’=\phi(M)+\phi(M’).$ Thus, $\phi$ is a group homomorphism.

From the definition of $\phi$, it is easy to see that the homomorphism $\phi$ is injective and surjective, and hence $\phi$ is a group isomorphism.
Therefore, the center $Z\big(H(F)\big)$ of the Heisenberg group is isomorphic to the additive group of $F$.

## Related Question.

The inverse element of the matrix
$\begin{bmatrix} 1 & x & y \\ 0 &1 &z \\ 0 & 0 & 1 \end{bmatrix}$ is given by
$\begin{bmatrix} 1 & -x & xz-y \\ 0 & 1 & -z \\ 0 & 0 & 1 \end{bmatrix}.$ For a proof, see the post The inverse matrix of an upper triangular matrix with variables.

Prove that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.