Question

consider a regular \(2n-gon\)
show that a horizontal flip followed by a vertical flip is equivalent to 180 degree rotation

Answer 1

what are you thinking of, interrupting it with geometrical coordinates or just prove symmetric on 2n-gon in group way ?

Answer 2

I think I have it using coordinate geometry(transformation rules) :

Horizontal flip (x coordinate remains same, y coordinate changes sign) :
\((x,y)\longrightarrow (x,-y)\)

after that, we do Vertical flip(x coordinate changes sign, y coordinate remains same) :
\((x,-y)\longrightarrow (-x,-y)\)

Answer 3

next recalling that the rule for rotating 180 degrees is \((x,y)\longrightarrow (-x,-y)\) ends the proof

Answer 4

Also consider the fact that the angle between the lines of your two reflections determines the angle of rotation (if you wanted to go that route)

Answer 5

i.e. Two reflections are equivalent to a specific angle rotation. And the angle between the lines determines the rotation about their intersection point.

Answer 6

Oh, and I forgot to mention that it's twice the angle between them. This means that since there is a 90 degree angle between the x- and y-axis, there will be a 180 degree rotation.

Answer 7

what im trying to understand is why u have chosen a regular \(2n-gon\) xD i believe it applies on each n-gon, this is confusing me since im trying to know what is special about it but in general all rotation and fillips are alike for Dihedral groups xD
or maybe i'm missing something out.

Answer 8

@jabberwock thanks! that works perfectly!

Answer 9

@ikram002p
when the number of sides is not even, we will not be haveing the element \(R_{180}\) in the dihedral group right ?

Answer 10

well that part is confusing me, since basically rotation in abstract algebra (R_something) mean identity and u need it to take origin form, how ever in geometry u can rotate anything u want as much as u want without such restrict xD if you know what i mean.