Exercise 2.4.12 (The Heisenberg group over $\mathbb{F}_2$ is isomorphic to $D_8$)

Question

Prove that the subgroup of upper triangular matrices in $\mathrm{GL}_3(\mathbb{F}_2)$ is isomorphic to the dihedral group of order $8$ (cf. Exercise 16, Section 1). [First find the order of this subgroup.]

Richard Ganaye · Accepted Answer

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

\begin{proof} Let $A $ be an upper triangular matrix in  $\mathrm{GL}_3(\mathbb{F}_2)$. Since $\det(A) 
e 0$ is the product of the diagonal elements, all these elements are equal to $1$, so $A$ is of the form
$$A = \begin{pmatrix} 1 & a & b\0 & 1 & c\0 & 0 & 1 \end{pmatrix},\qquad (a,b,c) \in \F_2^3,$$
thus the subgroup of upper triangular matrices in $\mathrm{GL}_3(\mathbb{F}_2)$ is
$$  G = H(\F_2) \subset  \mathrm{GL}_3(\mathbb{F}_2) )$$
that is $G$ is the Heisenberg group over $\F_2$ (cf. Exercise 2.2.14).

Therefore the map 
$$\psi\ 
\left\{
\begin{array}{ccc}
 \F_2^3 &	o& G\ 
 (a,b,c) &\mapsto &  \begin{pmatrix} 1 & a & b\0 & 1 & c\0 & 0 & 1 \end{pmatrix}
 \end{array}
ight.
$$
is bijective, so 
$$|G| = |H(\F_2)| = |\F_2^3| = 8.$$

The dihedral group $D_8$ is given by the presentation
$$D_8 = \langle r , s \mid r^4 = s^2 = e,\ sr = r^3 s angle$$
(Let us denote $r, s$ the generators of $D_8$, with some abuse of language: the generators are $\overline{r} = rN, \overline{s} = sN \in F(t,s)/N$, where $N$ is the normal subgroup of the free group $F(r,s)$ generated by  $r^4, s^2, s^{-1} r^{-3}sr$.)

Put
$$ S =  \begin{pmatrix} 1 & 1 & 0\0 & 1 & 0\0 & 0 & 1 \end{pmatrix}, \qquad R =  \begin{pmatrix} 1 & 1 & 1\0 & 1 & 1\0 & 0 & 1 \end{pmatrix}.$$
Then $R, S \in  G$ and 
\begin{align*}
\langle R, S angle &\supseteq \{I , R , R^2, R^3, S, RS, RS^2,RS^3 \}\
=\{&
\left(\begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 1
\end{array}ight), \left(\begin{array}{rrr}
1 & 1 & 1 \
0 & 1 & 1 \
0 & 0 & 1
\end{array}ight), \left(\begin{array}{rrr}
1 & 0 & 1 \
0 & 1 & 0 \
0 & 0 & 1
\end{array}ight), \left(\begin{array}{rrr}
1 & 1 & 0 \
0 & 1 & 1 \
0 & 0 & 1
\end{array}ight),\
& \left(\begin{array}{rrr}
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 1
\end{array}ight), \left(\begin{array}{rrr}
1 & 0 & 1 \
0 & 1 & 1 \
0 & 0 & 1
\end{array}ight), \left(\begin{array}{rrr}
1 & 1 & 1 \
0 & 1 & 0 \
0 & 0 & 1
\end{array}ight), \left(\begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & 1 \
0 & 0 & 1
\end{array}ight)\},
\end{align*}
where these $8$ matrices are distinct.

Therefore $\langle R, S  angle \supseteq G$ (and $\langle R, S  angle \subseteq G$), so 
$$G = \langle R, S angle.$$

Moreover  $R^4= S^2 = I_2$ and $S  R = R^3  S$.

By van Dyck's Theorem (see Note in Exercise 2.4.7), there is a surjective homomorphism $\varphi : D_8 	o  G$ such that $\varphi(r) = R,\ \varphi(s) = S$. Since $|D_8| = |G| = 8$, $\varphi$ is an isomorphism, so
$$H(\F_2) \simeq D_8.$$
 
\end{proof}
With Sagemath:
\begin{verbatim}
sage: F = GF(2)
sage: S =  matrix(F, 3, [1, 1, 0, 0, 1,0, 0,0,1 ]); 
sage: R =  matrix(F, 3, [1, 1, 1, 0, 1,1, 0,0,1 ]); 
sage: gens = [R,S]
sage: G = MatrixGroup(gens)
sage: G.order()
8
sage: R^4 , S^2

(
[1 0 0]  [1 0 0]
[0 1 0]  [0 1 0]
[0 0 1], [0 0 1]
)
sage: S * R, R^3 * S

(
[1 0 0]  [1 0 0]
[0 1 1]  [0 1 1]
[0 0 1], [0 0 1]
)
sage: l = [R^k for k in range(4)] + [R^k * S for k in range(4)]; l

[
[1 0 0]  [1 1 1]  [1 0 1]  [1 1 0]  [1 1 0]  [1 0 1]  [1 1 1]  [1 0 0]
[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: latex(l)
(...)
\end{verbatim}

Exercise 2.4.12 (The Heisenberg group over $\mathbb{F}_2$ is isomorphic to $D_8$)

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Exercise 2.4.12 (The Heisenberg group over $\mathbb{F}_2$ is isomorphic to $D_8$)

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