The dihedral group $D_{2n}$ is generated by $x$ and $y$ such that $x^n = y^2 = xyxy = e$. Show (algebraically) that elements not in the subgroup $<x>$ is a reflection and find the line (geometrically) through which it is a reflection.
Since $<x>$ is $\{x, x^2, ... x^{n-1}\}$, $y$ is not in this set, and so I thought $y$ represents a reflection. I don't understand what it means to show that an element not in $<x>$ is a reflection, besides that elements not in $<x>$ have order $2$.
On
The dihedral group $D_{2n}$ can be realised as the group of symmetries of the regular $n$-gon. Note that every symmetry is either a rotation or a reflection. As $x$ represents a rotation (with maximal order), the subgroup $\langle x \rangle$ is just the set of all rotations and as every element is either a rotation or a reflection then every element outside of $\langle x \rangle$ must be a reflection.
Considering orders, the reflections all have order $2$ since when we reflect back we get the identity and a rotation of order $2$ exists if and only if $n$ is even (and is then $x^{n/2}$. Since we have $n=5$ in your case, no such rotation exists so the reflections are precisely the elements of order $2$ in your group.
Here they define reflection means an element $y_1 \in D_2n$ s.t $y_1^2=1$ Now for any $s\notin <x>$ imply $s=x^ry$ now see $s^2= x^ry x^ry=yx^{-r}x^ry=1$. So all the elements not is $<x>$ are reflections.