Why is a reflection followed by another reflection is a rotation?

Question

I just started abstract algebra and we are working with dihedral groups. I've made Cayley tables for D3 and D4 but I can't explain why two reflections are the same as a rotation

Answer 1

Consider the dihedral group $D_5$, and consider its action on the pentagon. In particular, every element of the group can be thought of as some combination of rotations and reflections of a pentagon whose corners are labeled $1,2,3,4,5$ going clockwise.

First, notice that no matter what we do, the numbers will be in the order $1,2,3,4,5$ in either the clockwise (cw) or counterclockwise (ccw) direction.

If our change switches the order from ccw to cw (or vice versa), then we must have reflected the image. On the other hand, if no such change occurs, then we must have rotated the image.

Note that reflecting twice results in switching from ccw to cw, then to ccw. So, the numbers still go $1,2,3,4,5$ in the ccw direction. So, we must have rotated the image.

Answer 2

It all depends on what you mean by "reflection/rotation."

Quite often you say that a rotation is an orthogonal transformation with determinant $1$, and a reflection is an orthogonal transformation with determinant $-1$. By multiplicatively of determinant, this explains why the product of two reflections is a rotation. This works if you consider your dihedral group as a subgroup of linear transformations on $\mathbb R^2$.

Just thinking in terms of the structure of the dihedral group, the fact that the subgroup of rotations has index $2$ explains why the product of any two reflections (in the sense of a dihedral group) is a rotation. If $R$ is the rotation subgroup and $x,y$ are reflections, then $xR=yR$ and $xR•xR=R$ imply $xR•yR=xyR=R$, that is, $xy\in R$.

Answer 3

Note that the mirror axis for both reflections passes through the center of the object. Therefore, the center remains in the same place throughout the process.

Next, since we've done two reflections, the final transformation is orientation-preserving. It turns out that the only rigid transformations that preserve orientation and fix a point $p$ are rotations around $p$. So our final transformation must be a rotation around the center.

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For an intuitive proof of the above fact: imagine putting a thumbtack through the center of the square. Any transformation you can do to it now must fix the center (it's pinned in place!) and must preserve orientation (to flip the square over, you'd need to remove the tack). And with this tack in place, all you can do is rotate the square.

Answer 4

Here is a "really weird way" to look at it, which, if you wait patiently enough, will be useful later on.

Let's write a rotation $r^k$ as $(k,0)$, and a reflection $r^ks$ as $(k,1)$, where $r$ is a rotation "one $n$-th" of a turn (couterclockwise, for definiteness). So if you have a square, $n = 4$ and $r$ is a $90$ degree rotation, if you have a triangle $n = 3$ and $r$ is a $120$ degree rotation. So for $D_3$, for example, the $240$ degree rotation is $(2,0)$. I'll call $r$ a "click".

We're going to make a group$^{\dagger}$ out of $\Bbb Z_n \times \{0,1\}$ in the following way. If we compose rotations, we "add the clicks":

$(k,0)\ast(k',0) = (k+k'\text{ (mod }n),0)$.

So next we'll set $(0,1)$ as our "basic flip" (about the $x$-axis, let's say, with our first vertex of the $n$-gon at $(1,0)$).

So now, we're going to modify our operation $\ast$ so that it also works with elements of the form $(k,1)$. We will set:

$(k,m) \ast (k',m') = (k+ (-1)^mk'\text{ (mod }n),m+m'\text{ (mod }2))$.

It should be clear that this agrees with our previous definition, when $m = m' = 0$.

So what does this mean, geometrically? We can think of this as something $(k',m') $ does after whatever $(k,m)$ does to our original position of the $n$-gon. What the rotations do is clear, they just move the $n$-gon around in $n$-ths of a circle. But what does $(k,1)$ "mean"?

Well, according to our definition above, we have:

$(k,0)\ast (0,1) = (k + (-1)^00 \text{ (mod }n),0+1\text{ (mod }2))$

$= (k + 0\text{ (mod }n), 1\text{ (mod }2)) = (k,1)$.

So $(k,1)$ is a rotation, followed by a (horizontal) flip. So, if we have our first "action" as $(k,1)$, when we follow it by $(k',m')$, we have to reverse the sign of $k'$, because "flipping" changes our counter-clockwise rotation to clockwise rotation. In notation:

$(k,1)\ast(k',m') = (k - k'\text{ (mod }n),1+m'\text{ (mod }2))$

If $m' = 0$, this becomes:

$(k,1)\ast(k',0) = (k - k'(\text{ mod }n),1)$, which is still a reflection (note the $1$ in the second coordinate).

And $(k,0)\ast (k',1) = (k,0)\ast((k',0)\ast(0,1)) = ((k,0)\ast(k',0))\ast(0,1)) = (k+k'\text{ (mod }n),1)$.

(You'll have to take my word for now $\ast$ is associative-you can try to prove it, but it's a bit arduous).

And, at long last, the "answer" to your question:

$(k,1)\ast(k',1) = (k-k'\text{ (mod }n),1+1\text{ (mod }2)) = (k-k'\text{ (mod }n),0)$, which is a rotation (because, just like a light switch, two flips cancel each other out).

So you can think of $(k,m)$ as tracking two different states: a rotational state, and a flipped state. The presence of the $(-1)^m$ term in $\ast$ is to capture how flipping affects rotation.

$ ^{\dagger}$ Note: we haven't "shown" this actually forms a group. You'd have to show $\ast$ is associative, that $(0,0)$ is the identity, and that:

$(k,0)^{-1} = (n-k\text{ (mod }n),0)$

$(k,1)^{-1} = (k,1)$

I've also taken certain liberties writing the congruence class of an integer as that integer, to avoid a lot of extra brackets, and stuff.

Answer 5

A non-identity rotation leaves only one point fixed-the center of rotation. A reflection leaves only the axis of rotation fixed, while a reflection followed by a different reflection leaves only one point fixed-the intersection of the two axes of reflection , so it must be a rotation since only a rotation leaves a point fixed.