[Math] In dihedral group $FR=R^{-1}F$

Question

Let F be any reflection (flip ) about axis of symmetry .
And R be rotation by $\frac{2\pi}{n} $ radian counterclockwise .(n is the number of vertex).

Then $FR=R^{-1}F$

I looked at some example and it works .But how can we prove this thing rigorously ( I need this in proving subgroup -set of all rotation is normal )

I have no idea on proving this rigorously. Please help. Thanks

Answer 1

Note that $F$ is a reflection and $R$ is a rotation, so $FR$ is a reflection and note that any reflection has order $2$.

Now

$(FR)^2=e$ implies $FRFR=e=R^{-1}R$ , cancellation gives $$FRF=R^{-1}=R^{-1}.e=R^{-1}(F^{-1}F)$$ and again by cancellation $$FR=R^{-1}F^{-1}=R^{-1}F$$