Corollary 4.14.
There exists a projective plane of order $p^n$.
Proof.
(b) Let $\mathbb{F}_{p^n}$ be a finite field of order $p^n$ and $V$ be a three-dimensional vector space over $\mathbb{F}_{p^n}$. Consider the projective plane $\mathbb{P}\left(\mathbb{F}_{p^n}\right)$. Fix a two dimensional subspace $U$ in $V$. Then $U=$ $\left\{a u+b v: a, b \in \mathbb{F}_{p^n}\right\}$ for some linearly independent vectors $u, v \in U$. Notice that when $a \neq 0$, $a u+b v$ and $u+\frac{b}{a} v$ define the same point in the projective plane, hence it suffices to count the number of choices of $u+c v$ for $c \in \mathbb{F}_{p^n}$, which is $p^n$. Adding the case when $a=0$, corresponding to the point $\langle v\rangle$, gives $p^n+1$ points passing through the line $U$. Hence $\mathbb{P}\left(\mathbb{F}_{p^n}\right)$ has order $p^n$.