Problem 5.5.25 (The Heisenberg group as semidirect product)

Question

Let $H(\mathbb{F}_p)$ be the Heisenberg group over the finite field $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ (cf. Exercise 20 in Section 4). Prove that $H(\mathbb{F}_2) \simeq D_8$, and that $H(\mathbb{F}_p)$ has exponent $p$ and is isomorphic to the first non-abelian group in Example 7.

Richard Ganaye · Accepted Answer

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\begin{proof} 
By Exercise 2.4.12,
$$H(\mathbb{F}_2) \simeq D_8.$$

Let $X = \begin{pmatrix}1 & a & b\0 & 1 & c\ 0 & 0 & 1 \end{pmatrix} \in H(\mathbb{F}_p)$. By induction, for all integers $n\geq 0$,
$$ X^n =  \begin{pmatrix}1 & na & nb + \frac{n(n-1)}{2} ac\0 & 1 & nc\ 0 & 0 & 1 \end{pmatrix}$$
The characteristic of $\F_p$ is $p$, so
$$ X^p =  \begin{pmatrix}1 & 0 & \frac{p(p-1)}{2} ac\0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix}.$$
If $p$ is odd, $\frac{p(p-1)}{2}  = p \, \frac{p-1}{2} \equiv 0 \pmod p$, thus
$$X^p = 1.$$
If $X 
e I$, the order of $X$ is $p$, therefore the exponent of $H(\mathbb{F}_p)$ is $p$ if $p
e 2$.

\bigskip
Since $H(\mathbb{F}_p)$ is non-abelian, it is isomorphic to one of the two groups in Example 7: $(Z_p 	imes Z_p) times Z_p$ or $Z_{p^2} times Z_p$. In the second group, there are elements of order $p^2$, such as $(1,0)$, therefore $H(\mathbb{F}_p) 
ot \simeq  Z_{p^2} times Z_p$. This shows
$$H(\mathbb{F}_p) \simeq (Z_p 	imes Z_p) times Z_p.$$
\end{proof}

Problem 5.5.25 (The Heisenberg group as semidirect product)

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Problem 5.5.25 (The Heisenberg group as semidirect product)

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