Suppose $F_1$ and $F_2$ are distinct reflections in $D_n$ such that $F_1$$F_2=F_2F_1$, prove that $F_1F_2=R_{180}$.
I'm stumped on where to even start. Up to the point where I've gotten the book I am using has barely focused on Dihedral groups, yet this is in the chapter questions for basic group properties…
Best Answer
An important relation to keep in mind for dihedral groups is $$ sr^n=r^{-n}s $$ where $r$ and $s$ are elementary rotation and reflection, respectively, and $D_n$ is generated by $r$ and $s$. (You can prove this from the $n=1$ case, $sr=r^{-1}s$.) This can be used to show that the product of reflections is a rotation, since we can write any two reflections as $sr^a$ and $sr^b$ (where $0\leq a,b<n$) and compute: $$ F_1F_2=(sr^a)(sr^b)=(sr^a)(r^{-b}s)=s(r^{a-b})s=ss(r^{b-a})=r^{b-a} $$ using the fact that $s$ has order $2$.
Finally note that $$ (F_1F_2)^2=(F_1F_2)(F_2F_1)=F_1F_2^2F_1=F_1^2=1 $$ since $F_1$ and $F_2$ are reflections and thus have order $2$. We know that $ F_1F_2=F_1F_2^{-1}\neq 1 $ using the fact that each reflection is its own inverse, since $F_1$ and $F_2$ are distinct. This means that $F_1F_2$ is a non-identity rotation whose square is $1$, so $F_1F_2=R_{180}$.