Heisenberg Group Over Prime Fields (5.5.25)

Dummit and Foote Abstract Algebra, section 5.5, exercise 25:

Prove that $H(\mathbb{F}_2)≅D_8$ and for $p$ an odd prime prove that $H(\mathbb{F}_p)$ is of exponent $p$ and isomorphic to the first nonabelian group presented in example 7: $$\langle x,a,b~|~x^p=a^p=b^p=1,~ab=ba,~xa=abx,~xb=bx \rangle$$

Proof: One can easily check that the groups generated by $\left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)$ and $\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)$ are of orders 4 and 2 and have trivial intersection and thus generate $H(\mathbb{F}_2)$ of order 8, and indeed the generators fulfill the qualities of $r$ and $s$ in the presentation of $H(\mathbb{F}_2)$.

We can see $\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$ and $\left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)$ don't commute in any Heisenberg group over a prime field, so that by previous investigations of groups of order $p^3$ the center of $H(\mathbb{F}_p)$ is of order $p$. Now, $Z=\langle \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \rangle$ is of order $p$ and is clearly within (and thus completely comprises) the center. Prove using induction that $\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)^n=\left( \begin{array}{ccc} 1 & n & n \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$ and that $\left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)^n=\left( \begin{array}{ccc} 1 & 0 & n \\ 0 & 1 & n \\ 0 & 0 & 1 \end{array} \right)$ for all $n∈\mathbb{Z}^+$, and let $A=\langle \left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \rangle$ and $B=\langle \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right) \rangle$ be subgroups of order $p$. Due to the normality of $Z$ we can tell $AZ$ is a group, and due to its index it is normal in $H(\mathbb{F}_p)$. Now, $B∩AZ = 1$ and $BAZ=H(\mathbb{F}_p)$ so that $H(\mathbb{F}_p)=AZ \rtimes_φ B$ for some homomorphism $φ : B → Aut(AZ)$. Since $Z=Z(H(\mathbb{F}_p))$, we know that the exponent of $AZ$ is $p$, so that $AZ≅Z_p \times Z_p$ and thus $|Aut(AZ)|=p(p-1)^2(p+1)$, and in particular the Sylow p-subgroups of $Aut(AZ)$ are of order $p$ and already established to be conjugate, so that by exercise 6 there is only one unique nonabelian construction of $AZ \rtimes_φ B$, which must be $H(\mathbb{F}_p)$. Mapping the generator of $B$ to the element of $GL_2(\mathbb{F}_p)≅Aut(AZ)$ described in the example, we obtain the presentation mentioned (with the generator of $B$ in place of $x$, and the generator of $Z$ in place of $b$, and the generator of $A$ in place of $a$).

As for the exponent, under their notation we can see $b∈Z(G)$, leading to $(x^{n_1}a^{n_2}b^{n_3})^p=(x^{n_1}a^{n_2})^p$. Migrating the $a$ terms to the left of the block using the quality $xa=abx$ and then migrating the $b$ terms similarly since $b$ is in the center, we obtain $(x^{n_1}a^{n_2})^p=a^{pn_2}b^{m}x^{pn_1}$, where $m$ is determined by the process. Note that we migrate the first $a^{n_2}$ term through 1 block of $x^{n_2}$'s to obtain $n_1n_2$ resultant $b$'s, and we migrate the second $a^{n_2}$ term through 2 blocks of $x^{n_2}$'s to obtain $2n_1n_2$ resultant $b$'s, and so forth. Counting up all of these, we obtain $n_1n_2+2n_1n_2+...+pn_1n_2=n_1n_2(1+2+...+p)=n_1n_2 \dfrac{p(p+1)}{2}$ resultant $b$'s after the whole process, which is divisible by $p$ - the order of $b$ - when $p$ is an odd prime. $\square$