# 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.

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