At $p=2$ the Heisenberg construction produces $\mathrm{Dih}(C_4)$, not a group of exponent $2$

Given: The objects and hypotheses in the Example.

The Heisenberg group of order $p^3$ is the set $(\mathbb{Z}\mathrm{/}p{)}^3$ with $(a,b,c)(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})=(a+a^{\mathrm{\prime }},b+b^{\mathrm{\prime }},c+c^{\mathrm{\prime }}+a{b}^{\mathrm{\prime }})$ (The Heisenberg group of order $p^3$ over $\mathbb{Z}\mathrm{/}p$).

The Heisenberg multiplication makes $(\mathbb{Z}\mathrm{/}p{)}^3$ a nonabelian group of order $p^3$ (The Heisenberg multiplication is a group law, nonabelian, on a set of $p^3$ elements).

The Heisenberg group of order $p^3$ is extraspecial, and for odd $p$ it has exponent $p$ (The Heisenberg group of order $p^3$ is extraspecial, and for odd $p$ it has exponent $p$).

Every nonabelian group of order $p^3$ is extraspecial (A nonabelian group of order $p^3$ is extraspecial).

The generalized dihedral group $\mathrm{Dih}(C_4)$ and the quaternion group $Q_8$ are extraspecial of order $8$, with six and two solutions of $x^2=1$ respectively ($\mathrm{Dih}(C_4)$ and $Q_8$ are extraspecial of order $8$, with six and two solutions of $x^2=1$ respectively).

The two nonabelian groups of order eight are $\mathrm{Dih}(C_4)$ and $Q_8$ (For each prime there are exactly two nonabelian groups of order $p^3$ up to isomorphism).

The order of a finite group. Let $G$ be a group whose underlying set is finite, so that $G\approx n$ for some $n\in \mathbb{N}$. (The order $\mathrm{\mid }G\mathrm{\mid }$ of a finite group and the order $\mathrm{ord}(g)$ of an element, with $\mathrm{ord}(g)=\mathrm{\infty }$ when no positive power of $g$ is the identity).