Let $l_1$ and $l_2$ be the lines through the origin in $R^2$ that intersect in an angle π/n and let $r_i$ be the reflection about $l_i$. Prove the $r_1$ and $r_2$ generate a dihedral group $D_n$.
Attempt:
So $D_n$ is the dihedral group of order 2n generated by $ρ_θ$, where θ = 2π/n, and a reflection r' about a line l through the origin.
The product $r_1r_2$ of these reflections preserves orientation and is a rotation about the origin. Its angle of rotation is ±2θ.
Could I just say that by the definition of the dihedral group, we need to find a reflection and a rotation which fit the descriptions?
Best Answer
The dihedral group is generated by $a$ and $b$ where $b^n=1$ and $a^2=1$ and $(ab)^2=1$.
In your case, let $a=r_1$ and $b=r_2r_1$. You have of course $a^2=(r_1)^2=1$, and $(ab)^2 = (r_2)^2 = 1$ (compose from the right: $(ab)(x) = b(a(x)) = r_2r_1r_1(x)$).
Now, $b^n = (r_2r_1)^n$ as you correctly observe is a complete rotation, so it's 1 too. You can prove that using for instance basic arithmetics in polar coordinates.