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 $F$.

 
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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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