Exercise 1.4.11 (Heisenberg group over $F$)

Question

Let $H(F) = \left\{ \begin{pmatrix}1 & a & b\0 & 1 & c\ 0 & 0 & 1 \end{pmatrix} \mid a,b,c \in F ight\}$ --- called the {\bf Heisenberg group} over $F$. Let $X = \begin{pmatrix}1 & a & b\0 & 1 & c\ 0 & 0 & 1 \end{pmatrix}$ and $Y = \begin{pmatrix}1 & d & e\0 & 1 & f\ 0 & 0 & 1 \end{pmatrix}$ be elements of $H(F)$.
\begin{enumerate}
\item[{\bf (a)}] Compute the matrix product $XY$ and deduce that $H(F)$ is closed under matrix multiplication. Exhibit matrices such that $XY 
e YX$ (so that $H(F)$ is always non-abelian).
\item[{\bf (b)}] Find an explicit formula for the matrix inverse $X^{-1}$ and deduce that $H(F)$ is closed under inverses.
\item[{\bf (c)}] Prove the associative law for $H(F)$ and deduce that $H(F)$ is a group of order $|F|^3$.

(Do not assume that matrix multiplication is associative.)
\item[{\bf (d)}] Find the order of each element of the finite group $H(\mathbb{Z}/2\mathbb{Z})$.
\item[{\bf (e)}] Prove that every nonidentity element of the group $H(\mathbb{R})$ has infinite order.
\end{enumerate}

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

\begin{proof} Let
$$H(F) = \left\{ \begin{pmatrix}1 & a & b\0 & 1 & c\ 0 & 0 & 1 \end{pmatrix} \mid a,b,c \in F ight\}.$$
\begin{enumerate}
\item[{\bf (a)}] Let 
$X = \begin{pmatrix}1 & a & b\0 & 1 & c\ 0 & 0 & 1 \end{pmatrix}$ and $Y = \begin{pmatrix}1 & d & e\0 & 1 & f\ 0 & 0 & 1 \end{pmatrix}$ be elements of $H(F)$. Then
$$XY =   \begin{pmatrix}1 & d+a & e + af + b\0 & 1 & f+c\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix}1 & A & B\0 & 1 & C\ 0 & 0 & 1 \end{pmatrix},$$
where $(A,B,C) = (d+a, e+ af + b, f+c) \in F^3$.
Thus $XY \in H(F)$, so $H(F)$ is closed under matrix multiplication.

Put $X = \begin{pmatrix}1 & 1 & 0\0 & 1 & 1\ 0 & 0 & 1 \end{pmatrix}, Y = \begin{pmatrix}1 & 0 & 0\0 & 1 & 1\ 0 & 0 & 1 \end{pmatrix}$. Then
$$XY = \begin{pmatrix}1 & 1 & 1\0 & 1 & 1+1\ 0 & 0 & 1 \end{pmatrix} 
e  \begin{pmatrix}1 & 1 & 0\0 & 1 & 1+1\ 0 & 0 & 1 \end{pmatrix} = YX,$$
since $1 
e 0$ in any field. So $H(F)$ is always non-abelian.

\item[{\bf (b)}] If $C$ is the matrix of cofactors of $X$, then $X^{-1} = \frac{1}{\det(X)}\,  C^{T}$, where $C$ is the cofactor matrix. This gives
$$X^{-1} = \begin{pmatrix}1 & -a & ac - b\0 & 1 & -c\ 0 & 0 & 1 \end{pmatrix}.$$
Therefore $X^{-1} \in H$, so $H(F)$ is closed under inverses.

\item[{\bf (c)}] $H (F) \subset \mathrm{GL}_n(F), H(F) 
e \varnothing$, and $H(F)$ is closed under matrix multiplication and under inverses, thus $H(F)$ is a subgroup of $\mathrm{GL}_n(F)$. So $H(F)$ is a group.

Write $M(a,b,c) = \begin{pmatrix}1 & a & b\0 & 1 & c\ 0 & 0 & 1 \end{pmatrix}$. Then
$$M
\left\{
\begin{array}{ccl}
F^3 & 	o & H(F)\
(a,b,c) &\mapsto &M(a,b,c) 
\end{array}
ight.
$$
is bijective (surjective by definition of $H(F)$, and injective because $M(a,b,c) = M(c,d,e) \Rightarrow (a,b,c) = (d,e,f)$). Therefore $|H(F)|= |F^3|$.

By part (a) and (b),
$$M(a,b,c) M (a',b',c') = M(a+ a', b+b'+ac', c + c'), \qquad M(a,b,c)^{-1} = M(-a, ac -b, -c).$$

The authors insist that we check associativity directly:
\begin{align*}
[M(a,b,c)M(a',b',c')] M (a'', b'',c'') &= M(a+a', b + b' + ac', c+c') M (a'', b'',c'')\
&=M(a+a'+a'', b+b'+ b'' +  ac' + a c'' +a'c'', c + c' + c'')\
M(a,b,c)[M(a',b',c') M (a'', b'',c'') ] &= M(a,b,c)M(a'+a'', b' + b''+ a'c'', c'+c'')\
&= M(a + a' + a'', b + b' + b'' +a'c''+ ac'+ac'', c+c'+c'').
\end{align*}
This shows that $[M(a,b,c)M(a',b',c')] M (a'', b'',c'') = M(a,b,c)[M(a',b',c') M (a'', b'',c'')]$, so the associativity of multiplication is checked.

\item[{\bf (d)}]The elements of $H(\F_2) $ are
\begin{align*}
 \begin{pmatrix}1 & 0& 0\0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix},
 \begin{pmatrix}1 & 0 & 0\0 & 1 & 1\ 0 & 0 & 1 \end{pmatrix},
 \begin{pmatrix}1 & 0 & 1\0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix},
 \begin{pmatrix}1 & 0 & 1\0 & 1 & 1\ 0 & 0 & 1 \end{pmatrix},\
 \begin{pmatrix}1 & 1 & 0\0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix},
 \begin{pmatrix}1 & 1 & 0\0 & 1 & 1\ 0 & 0 & 1 \end{pmatrix},
 \begin{pmatrix}1 & 1 & 1\0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix},
 \begin{pmatrix}1 & 1 & 1\0 & 1 & 1\ 0 & 0 & 1 \end{pmatrix}.
\end{align*}
More shortly, $H(\F_2) =$
$$ \{ M(0,0,0), M (0,0,1), M (0,1,0), M(0,1,1),M(1,0,0), M (1,0,1), M (1,1,0), M(1,1,1)\}$$
The orders of these elements are
\begin{center}
\begin{tabular}{|c|c|}
\hline
$M$ & $|M|$     \
\hline
M(0, 0, 0) & 1\
M(0, 0, 1) & 2\
M(0, 1, 0) & 2\
M(0, 1, 1) & 2\
M(1, 0, 0) & 2\
M(1, 0, 1) & 4\
M(1, 1, 0) & 2\
M(1, 1, 1) & 4\
\hline
\end{tabular}
\end{center}
For instance, 
$$\begin{pmatrix}1 & 1 & 0\0 & 1 & 1\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}1 & 1 & 0\0 & 1 & 1\ 0 & 0 & 1 \end{pmatrix} =  \begin{pmatrix}1 & 0 & 1\0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix},$$
and
$$\begin{pmatrix}1 & 0 & 1\0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}1 & 0 & 1\0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix} =  \begin{pmatrix}1 & 0& 0\0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix},$$
so $M(1,0,1)^2 = M(0,1,0)$, and $M(0,1,0)^4 =I_3$, so $|M(1,0,1)| = 4$.

\bigskip
Note: If we write $r= M(1,1,1)$ and $s = M(0,0,1)$, then $r^4 = s^2 = e$, and $rs = sr^3$. Moreover $H(\F_2)$ has $8$ elements. Therefore $H(\F_2) \simeq D_8$.

\item[{\bf (e)}] Let $M = M(a,b,c)$ be any nonidentity element of $H(R)$, so that $(a,b,c) 
e (0,0,0)$.

Assume that $\delta = |M(a,b,c)|$ is finite.
By induction, if $(a,b,c) \in \R^3$, for all $n$, there is some real $b_n$ such that
$$M(a,b,c)^n = M(na, b_n ,nc),$$
therefore $M(0,0,0) = M(a,b,c)^\delta = M(\delta a, b_\delta ,\delta c)$. Thus $\delta a = \delta c = 0$. Since $\delta$ is a positive integer, $a = c =0$, so $M = M(0, b, 0)$.

A new induction shows that for all $n \in \N$,
$$M(0,b,0)^n = M(0,nb,0).$$
Therefore $M(0,0,0) = M(0,b,0)^\delta = M(0, \delta b , 0)$, thus $\delta b = 0$ and $b = 0$. Therefore $M = M(0,0,0) = I_3$.

So every nonidentity element of the group $H(\R)$ has infinite order.
\end{enumerate}
\end{proof} 
Verification of (d) with Sagemath:
\begin{verbatim}
sage: def M(a,b,c):
....:     return matrix([[1, a, b],[0, 1 , c],[0, 0, 1]])
....: 
sage: F2 = GF(2); F2
Finite Field of size 2
sage: G = [ M(a,b,c) for a in F2 for b in F2 for c in F2]; G

[
[1 0 0]  [1 0 0]  [1 0 1]  [1 0 1]  [1 1 0]  [1 1 0]  [1 1 1]  [1 1 1]
[0 1 0]  [0 1 1]  [0 1 0]  [0 1 1]  [0 1 0]  [0 1 1]  [0 1 0]  [0 1 1]
[0 0 1], [0 0 1], [0 0 1], [0 0 1], [0 0 1], [0 0 1], [0 0 1], [0 0 1]
]
sage: def order(g):
....:     m = g; 
....:     counter = 1
....:     while m != G[0]:
....:         counter += 1
....:         m = g * m
....:     return counter
....: 
sage: l = [((g[0,1],g[0,2],g[1,2]), order(g)) for g in G]; l

[((0, 0, 0), 1),
 ((0, 0, 1), 2),
 ((0, 1, 0), 2),
 ((0, 1, 1), 2),
 ((1, 0, 0), 2),
 ((1, 0, 1), 4),
 ((1, 1, 0), 2),
 ((1, 1, 1), 4)]
\end{verbatim}

Exercise 1.4.11 (Heisenberg group over $F$)

Answers

Comments


$M$	$\| M \|$

M(0, 0, 0)	1
M(0, 0, 1)	2
M(0, 1, 0)	2
M(0, 1, 1)	2
M(1, 0, 0)	2
M(1, 0, 1)	4
M(1, 1, 0)	2
M(1, 1, 1)	4

Exercise 1.4.11 (Heisenberg group over $F$)

Answers

Comments

Add answer