= Solution
The <two-involution characterization of a dihedral group> uses the extended convention that includes $C_2$ and the <infinite dihedral group>; the two generators need not be distinct. Put $a=xy$ and $b=y$. The inversion relation gives
$$
a^2=xyxy=x(yxy^{-1})y^2=xx^{-1}=e,\qquad b^2=e,\qquad x=ab.
$$
If $a\ne e$, both $a,b$ are <involutions> and generate the original <group>. If $a=e$, then $x=y$ and the <group> is $C_2$; choose the same nonidentity element twice. Thus \b[two elements of order two generate the group], including the degenerate case allowed here. The distinction between square equal to the identity and <order of a group element> exactly two is essential.
Back to article page
