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The involutions in an even dihedral group are the half-turn $g^3$ and the six reflections $g^{j}h$. Thus the seven order-two subgroups areThe half-turn is central, so only $⟨g^3⟩$ is normal. Indeed $g(g^{j}h)g^{-1}=g^{j+2}h$, a different reflection; an order-two subgroup has only one nonidentity element, so each reflection subgroup fails the normal subgroup condition.

Use the dihedral group presentation $G=⟨r,s:r^{2n}=s^2=1,srs=r^{-1}⟩$. Every element is $r^j$ or $s{r}^j$. A nonidentity rotation has order two only when its exponent is $n$, while $(s{r}^j{)}^2=1$ for every $j$. Thus the complete list isThe half-turn $r^n$ is central, so its conjugacy class is the singleton $\{r^n\}$ and its normal closure is $\{1,r^n\}$.

For reflections, $r^k(s{r}^j)r^{-k}=s{r}^{j-2k}$ and $s(s{r}^j)s^{-1}=s{r}^{-j}$. Hence the conjugacy class is exactly the $n$ reflections whose exponents have the same parity as $j$:The subgroup generated by this class contains $r^2$, as the product of two consecutive listed reflections, and contains $s{r}^j$. Conversely those two generators contain every listed reflection. Their subgroup is invariant under conjugation by both $r$ and $s$, so the smallest normal subgroup containing $s{r}^j$ isof order $2n$ and index two. This involutions in an even dihedral group classification distinguishes the two reflection classes from the central half-turn.

The group order $|G|$ is the number of elements of the finite group. A normal subgroup $H\le G$ satisfies $gH{g}^{-1}=H$ 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 isThe 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 nontrivial proper subgroups areTo 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 isFor the general polygon, write $D_n=\{R^j,R^{j}s:0\le j<n\}$ with $R^n=s^2=I$ and $sRs=R^{-1}$. A rotation $R^j$ commutes with $s$ exactly when $R^j=R^{-j}$, or $2j\equiv 0(\mathrm{mod}n)$. A reflection cannot commute with $R$ when $n\ge 3$, becauseTherefore 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, provingThis agrees with the involutions in an even dihedral group and uses the convention $|D_n|=2n$ throughout.

All six reflections square to the identity, and the rotation $R^3$ is the half-turn. These exhaust the nonidentity involutions: $R^k$ has square one only when $2k\equiv 0(\mathrm{mod}6)$, giving $k=3$ apart from the identity. Thus the order-two subgroups areA reflection cannot have order three. The rotations of order three are $R^2$ and $R^4$, and they generate the same subgroup:All these subgroups are cyclic, because a group of prime order is cyclic; the displayed nonidentity elements supply their generators. This gives seven subgroups of order two and one of order three, consistently with involutions in an even dihedral group and the classification of subgroups of a dihedral group.

Write $r=g_1$. The matrix $D(s)$ is a reflection by part (a). Reflection reverses a rotation angle, giving $D(s)D(r)D(s)=D(r{)}^{-1}$. Faithfulness lifts this matrix identity back to the group:Moving each $s$ past a rotation reduces every word to $r^i$ or $r^{i}s$, $0\le i<p$. These $2p$ elements are distinct: the rotations are distinct by the order of $r$, and reflections have determinant $-1$, so cannot equal a rotation. ThereforeHere the subscript denotes group order. This is the reflection case of finite subgroups of the plane orthogonal group.

Every $r^{i}s$ has square $r^{i}s{r}^{i}s=r^i{r}^{-i}=I$, so there are $p$ reflection subgroups of order two. Within the rotations, the even-$p$ assumption from part (b) supplies the additional half-turn $r^{p/2}$. Distinct involutions generate distinct two-element subgroups. ThusFor comparison, without the evenness requirement the corresponding odd-$p$ count would be $p$. The even case agrees with involutions in an even dihedral group.

For $p=4$, takeDirect multiplication gives $SRS=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)=R^3$, so $srs=r^3=g_3$. The product rules arewith exponents modulo four. They yield the complete multiplication table, where the row element multiplies the column element on the left:This is the symmetry group of a square, the dihedral group of order eight, often written $D_4$ when its subscript counts vertices instead. Its five order-two subgroups are generated by $r^2,s,rs,r^{2}s,r^{3}s$.