= Solution
The <group order> $|G|$ is the number of elements of the finite group. A <normal subgroup> $H\leq G$ satisfies $gHg^{-1}=H$ for every $g\in G$, equivalently its left and right cosets agree.
For the <square>, let $R$ be counterclockwise rotation by $\pi/2$, and $s$ reflection in the horizontal axis. The <dihedral group> is
$$
D_4=\{I,R,R^2,R^3,s,Rs,R^2s,R^3s\},\qquad R^4=s^2=I,\quad sRs=R^{-1}.
$$
The first four elements rotate by $0,90,180,270$ degrees. The other four are <Euclidean reflections> in the horizontal axis, the line $y=x$, the vertical axis and the line $y=-x$, respectively, for a square centred at the origin with sides parallel to the axes.
The eight \b[nontrivial proper subgroups] are
$$
\boxed{\begin{gathered}
\{I,R^2\},\qquad\{I,R^js\}\quad(j=0,1,2,3),\\
\langle R\rangle=\{I,R,R^2,R^3\},\\
\{I,R^2,s,R^2s\},\qquad\{I,R^2,Rs,R^3s\}.
\end{gathered}}
$$
To check completeness, <Lagrange's theorem> allows proper orders one, two and four. The order-two subgroups are generated by the five involutions displayed. An order-four subgroup containing $R$ or $R^3$ is the rotation subgroup. Otherwise it must contain reflections; their product is a rotation, which must be $R^2$, so its reflections differ by that half-turn, giving the two listed Klein subgroups. The trivial subgroup $\{I\}$ is an additional proper subgroup. Thus the PDF's count of eight uses “proper” to exclude the trivial subgroup; with the usual inclusive definition there are nine proper subgroups.
A <normal subgroup of order two is central>: conjugation fixes its identity and must also fix its only other member. The half-turn $R^2$ commutes with every square symmetry, whereas the reflection subgroups are exchanged by conjugation. Hence the unique order-two normal subgroup is
$$
\boxed{\{I,R^2\}.}
$$
For the general polygon, write $D_n=\{R^j,R^js:0\leq j<n\}$ with $R^n=s^2=I$ and $sRs=R^{-1}$. A rotation $R^j$ commutes with $s$ exactly when $R^j=R^{-j}$, or $2j\equiv0\pmod n$. A reflection cannot commute with $R$ when $n\geq3$, because
$$
R(R^js)R^{-1}=R^{j+2}s\ne R^js.
$$
Therefore the <center of a group> here is $\{I\}$ for odd $n$, and $\{I,R^{n/2}\}$ for even $n$. The order-two normal subgroups correspond exactly to nonidentity central involutions, proving
$$
\boxed{\begin{cases}\text{no order-two normal subgroup},&n\text{ odd},\\\{I,R^{n/2}\}\text{ is the unique one},&n\text{ even}.
\end{cases}}
$$
This agrees with the <involutions in an even dihedral group> and uses the convention $|D_n|=2n$ throughout.
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