10. Prove that the set of matrices of the form $$ \left(\begin{array}{lll}1 & x & y \ 0 & 1 & z \ 0 & 0 & 1\end{array}\right) $$ is a group under matrix multiplication. This group, known as the Heisenberg group, is important in quantum physics. Matrix multiplication in the Heisenberg group is defined by $$ \left(\begin{array}{lll}1 & x & y \ 0 & 1 & z \ 0 & 0 & 1\end{array}\right)\left(\begin{array}{ccc}1 & x^{\prime} & y \ 0 & 1 & z^{\prime} \ 0 & 0 & 1\end{array}\right)=\left(\begin{array}{cc}1 & x+x^{\prime} \ 0 & 1 \ 0 & 0\end{array}\right) $$

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We wish to show that the set

$$
H = \left\{ \begin{pmatrix} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{pmatrix} : x,y,z\in \mathbb{R} \right\}
$$

forms a group under the usual matrix multiplication. We prove this by verifying the group axioms.

---

1. **Closure**

Let

$$
A = \begin{pmatrix} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{pmatrix} \quad\text{and}\quad B = \begin{pmatrix} 1 & x' & y' \ 0 & 1 & z' \ 0 & 0 & 1 \end{pmatrix}
$$

be two arbitrary elements of $$ H $$. Their product is computed by standard matrix multiplication. We have

$$
A B = \begin{pmatrix} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{pmatrix}
\begin{pmatrix} 1 & x' & y' \ 0 & 1 & z' \ 0 & 0 & 1 \end{pmatrix}.
$$

Multiply the matrices entry by entry:

- The $$(1,1)$$ entry:
  $$
  1\cdot 1 + x\cdot 0 + y\cdot 0 = 1.
  $$
- The $$(1,2)$$ entry:
  $$
  1\cdot x' + x\cdot 1 + y\cdot 0 = x + x'.
  $$
- The $$(1,3)$$ entry:
  $$
  1\cdot y' + x\cdot z' + y\cdot 1 = y' + xz' + y.
  $$
- The $$(2,1)$$ entry:
  $$
  0\cdot 1 + 1\cdot 0 + z\cdot 0 = 0.
  $$
- The $$(2,2)$$ entry:
  $$
  0\cdot x' + 1\cdot 1 + z\cdot 0 = 1.
  $$
- The $$(2,3)$$ entry:
  $$
  0\cdot y' + 1\cdot z' + z\cdot 1 = z+ z'.
  $$
- The $$(3,1)$$ entry:
  $$
  0\cdot 1 + 0\cdot 0 + 1\cdot 0 = 0.
  $$
- The $$(3,2)$$ entry:
  $$
  0\cdot x' + 0\cdot 1 + 1\cdot 0 = 0.
  $$
- The $$(3,3)$$ entry:
  $$
  0\cdot y' + 0\cdot z' + 1\cdot 1 = 1.
  $$

Thus, we obtain

$$
A B = \begin{pmatrix}
1 & x+x' & y + xz' + y' \[1mm]
0 & 1 & z+z' \
0 & 0 & 1 
\end{pmatrix}.
$$

Since $$x+x'$$, $$y + xz' + y'$$, and $$z+z'$$ are real numbers, the product matrix is again in $$H$$. This establishes closure.

---

2. **Associativity**

Matrix multiplication is associative. That is, for any matrices $$A$$, $$B$$, and $$C$$ (of appropriate sizes), we have

$$
(A B) C = A (B C).
$$

Since the elements of $$H$$ are $$3\times 3$$ matrices and our multiplication is the standard one, associativity holds automatically.

---

3. **Identity Element**

The identity element in the group of $$3 \times 3$$ matrices is the identity matrix

$$
I = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}.
$$

Notice that $$I$$ is of the form

$$
\begin{pmatrix} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{pmatrix}
$$

with $$x = 0$$, $$y = 0$$, and $$z = 0$$. Thus, the identity matrix belongs to $$H$$ and serves as the identity element because for any

$$
A = \begin{pmatrix} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{pmatrix} \in H,
$$

we have

$$
I A = A I = A.
$$

---

4. **Inverses**

Let

$$
A = \begin{pmatrix} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{pmatrix} \in H.
$$

We wish to find a matrix

$$
A^{-1} = \begin{pmatrix} 1 & a & b \ 0 & 1 & c \ 0 & 0 & 1 \end{pmatrix} \in H
$$

such that

$$
A A^{-1} = I.
$$

Multiplying $$A$$ and $$A^{-1}$$, we have

$$
A \, A^{-1} =
\begin{pmatrix}
1 & x & y \
0 & 1 & z \
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & a & b \
0 & 1 & c \
0 & 0 & 1
\end{pmatrix}
=
\begin{pmatrix}
1 & x+a & y + xc + b \[1mm]
0 & 1 & z+c \[1mm]
0 & 0 & 1
\end{pmatrix}.
$$

For this product to equal $$I$$, we require:

- $$x+a = 0$$, so $$a = -x$$.
- $$z+c = 0$$, so $$c = -z$$.
- $$y + xc + b = 0$$. Since $$c=-z$$, we have
  $$
  y + x(-z) + b = y - xz + b = 0,
  $$
  so $$b = xz - y$$.

Thus, the inverse of $$A$$ is

$$
A^{-1} = \begin{pmatrix}
1 & -x & xz - y \
0 & 1 & -z \
0 & 0 & 1
\end{pmatrix}.
$$

Since $$-x$$, $$xz - y$$, and $$-z$$ are real numbers, $$A^{-1}$$ is in $$H$$. One can verify that

$$
A A^{-1} = A^{-1} A = I.
$$

---

Since all four axioms (closure, associativity, identity, and inverses) are satisfied, we conclude that $$H$$ is a group under matrix multiplication.
The set of matrices of the form $$ \begin{pmatrix} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{pmatrix} $$ forms a group under matrix multiplication. This group, known as the Heisenberg group, satisfies the group axioms: closure, associativity, has an identity element, and every element has an inverse within the group.

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