Exploring Reflections, Rotations, and Similitude Mappings

[mathmari]

Hey!

I want to check whether the following statements are correct:

1. Each reflection along a line and each reflection along a point is odd.
   
2. For the lines $g,h$ it holds that $\sigma_g\circ\sigma_h=\sigma_h\circ\sigma_g$ iff $g=h$.
   
3. For a rotation $\delta\neq id$ a translation $\tau$, $\delta\circ\tau$ is always a rotation.
   
4. For rotation $\delta\neq id$ and a reflection $\sigma$, $\delta\circ\sigma$ is never a reflection.
   
5. There are the reflections $\sigma_1, \ldots , \sigma_{2017}$ with $\sigma_{2017}\circ\ldots \circ\sigma_2\circ\sigma_1=id$.
   
6. A similitude mapping with exactly one fixed point is a scaling.
   
7. Similitude mappings $\neq id$ with more than one fixed point are reflections.
   
8. Affine mappings $\neq id$ with more than one fixed point are reflections.
   
9. The composition of two rotations with rotation angle $a$ and $b$ is a rotation iff $a+b=k\cdot 2\pi, k\in \mathbb{Z}$.
   
10. The composition of two rotations with rotation angle $a$ and $b$ with $a+b=\pi$ is a reflection along a point.
    
11. If $ABC$ is a triangle, then the center of the incircle is a fixed point of $\sigma_{AB}\circ\sigma_{BC}$.
    
12. For each line $g$ and each similitude mapping $\kappa$ it holds that $\kappa\circ\sigma\circ\kappa^{-1}=\sigma_{\kappa(g)}$.

For some of them I have an idea, for others I have some questions about definitions:

1. By odd do we mean that the reflected object is opposite from the original?
   
2. I think that it holds, but I cannot justify it.
   
3. The tranlation is $\tau (x,y)=(x+h,y+b)$ and the rotation by $\theta$ about the origin is $\delta(\vec x)=M\vec x,$ where $$M=\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}.$$
   
   We have that $$\delta \circ \tau(x,y)=M\tau(x,y)=\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}\binom{x+h}{y+b}=\binom{(x+h)\cos \theta+(y+b)\sin \theta}{-(x+h)\sin \theta+(y+b)\cos \theta}$$
   
   What do we get from here?
4. I think that it can be, it depends on the angle of the rotation.
   
5. If we had two reflections, it would be possible that the composition is the identity, right? Since in this case we have an odd number of compositions, do we have to check if $3$ compositions of reflections is the identity?
   
6. A similitude mapping with exactly one fixed point is a scaling.
   
   At a scaling the scaled object is parallel to the original, or not?
   
   What eactly does it mean a similitude mapping with exactly one fixed point ? That all the similar objectos have exactly one intersection point? (Wondering)
7. 
   
8. 
   
9. 
   
10. I think it is correct, since the points will be then collinear, or not?
    
11. 
    

Unfortnunately, for the others I don't really have an idea. (Wondering)

 

[I like Serena]

1. Each reflection along a line and each reflection along a point is odd.
   
2. For the lines $g,h$ it holds that $\sigma_g\circ\sigma_h=\sigma_h\circ\sigma_g$ iff $g=h$.
   
3. For a rotation $\delta\neq id$ a translation $\tau$, $\delta\circ\tau$ is always a rotation.
   
4. For rotation $\delta\neq id$ and a reflection $\sigma$, $\delta\circ\sigma$ is never a reflection.

1. By odd do we mean that the reflected object is opposite from the original?
   
2. I think that it holds, but I cannot justify it.
   
3. The tranlation is $\tau (x,y)=(x+h,y+b)$ and the rotation by $\theta$ about the origin is $\delta(\vec x)=M\vec x,$ where $$M=\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}.$$
   
   We have that $$\delta \circ \tau(x,y)=M\tau(x,y)=\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}\binom{x+h}{y+b}=\binom{(x+h)\cos \theta+(y+b)\sin \theta}{-(x+h)\sin \theta+(y+b)\cos \theta}$$
   
   What do we get from here?

Hey mathmari! (Smile)

1. By 'odd' we mean an odd permutation.
It takes an even number of odd permutations to get an even permutation. And the identity is even.
Hint: what is the determinant of a reflection?

2. The composition of 2 reflections is a rotation. Suppose g and h have angle phi, which rotation will we get?
Hint: draw a picture.

3. When you say 'rotation' do you mean a rotation around the origin or not? That matters.

4. Consider the group of symmetries of the regular polygon $D_n$. That is known to you isn't it? What happens when we combine a reflection and a rotation?

 

[mathmari]

1. By 'odd' we mean an odd permutation.
It takes an even number of odd permutations to get an even permutation. And the identity is even.
Hint: what is the determinant of a reflection?

The matrix for a reflection is orthogonal with determinant $-1$, right?

But what do we get from that?

When we P around the point Z with rotation angle 180°, then the points P, Z and P' are collinear. This rotation is equivalent to the reflection around the point Z, right?

Since a rotation is a composition of two reflections, we conclude that the reflection around a point is even, right?

2. The composition of 2 reflections is a rotation. Suppose g and h have angle phi, which rotation will we get?
Hint: draw a picture.

I draw a pricture and the angle of rotation seems to be $2\phi$, because of the following:
View attachment 6982

We have that the angle between P and P'' is $a+a+b+b=(a+b)+(a+b)=\phi +\phi =2\phi$.

Is this correct? I saw now in some notes that it holds that $\sigma_g\circ\sigma_h\circ\sigma_g$ even when $g$ is perpendicular to $h$, since then both mappings are reflections along a point, and so the above statement is not correct. I haven't really understood that justification.

3. When you say 'rotation' do you mean a rotation around the origin or not? That matters.

There is no information about that.

I have an idea about that statement:

A translation followed by a rotation of θ is still a rotation by θ. This is because if we take two points a and b which are mapped to a′ and b′, then θ is the angle between the vectors b−a and b′−a′, and even though the second one are translated.

Is this correct? At a suggested solution there is the following justification why the statement is correct:

$\kappa=\sigma\circ\tau$ is an even congruence mapping, so a rotation or a translation. It cannot be a translation since then it would hold that $\sigma=\kappa\circ\tau^{-1}$ is a translation.

Why is $\kappa=\sigma\circ\tau$ an even congruence mapping? Are all the congruence mapping the rotations and the translations?

4. Consider the group of symmetries of the regular polygon $D_n$. That is known to you isn't it? What happens when we combine a reflection and a rotation?

We have that the rotation is a composition of two reflections. Therefore the composition of the rotation and the reflection is the composition of 3 reflections.

Is the composition of three reflections always a reflection?
According to the suggested solution, the statement is wrong, simce if the axis of the reflection passes through the center of rotation, then we have a reflection.
In that case the reflection that we get from the composition, we have the same axis as the initial reflection?

5. There are the reflections $\sigma_1, \ldots , \sigma_{2017}$ with $\sigma_{2017}\circ\ldots \circ\sigma_2\circ\sigma_1=id$.

We have an odd number of permutations. Since the identity is even, the above equality cannot hold.

6. A similitude mapping with exactly one fixed point is a scaling.

That means that one point is unchanged. We have that similitude mapping are compositions of scalings and congruence mappings. Can it not be that the similitude mapping is the composition of scaling and rotation?

7. Similitude mappings $\neq id$ with more than one fixed point are reflections.

According to the suggested solution we have the following:

The statement is correct. Since there are the two fixed points, the scaling factor is $1$, so we have a congruence mapping with two fixed points.

Could you explain this to me? I haven't understood that.

8. Affine mappings $\neq id$ with more than one fixed point are reflections.

Affine mappings are of the form $\vec{x}'=M\vec{x}+\vec{v}$, right? By fixed point do we mean that $\vec{x}$ must be unchanged? Do we maybe take the identity matrix as M and the zero vector as $\vec{v}$ ?

9. The composition of two rotations with rotation angle $a$ and $b$ is a rotation iff $a+b=k\cdot 2\pi, k\in \mathbb{Z}$.

I think that this statement is wrong. Isn't the justification wrong similar to the one of statement 2? From that picture the angles are 2a and 2b. The angle of the composition is then 2a+2b.

So, the composition is always a rotation with the angle the sum of the two angles, right? Using the rotation matrix we have the following:
The one rotation is $\vec{v}\mapsto M(a)\vec{v}$ and the other $\vec{w}\mapsto M(b)\vec{w}$. Their composition is defined as follows:
\begin{align*}M(a)M(b)\vec{w}&=\begin{pmatrix}\cos a & -\sin a \\ \sin a & \cos a \end{pmatrix}\begin{pmatrix}\cos b & -\sin b \\ \sin b & \cos b \end{pmatrix} \\ & =\begin{pmatrix}\cos a\cos b-\sin a \sin b& -\cos a\sin b-\sin a\cos b \\ \sin a\cos b+\cos a\sin b& -\sin a\sin b+\cos a\cos b \end{pmatrix} \\ & =\begin{pmatrix}\cos (a+b)& -\sin(a+b) \\ \sin (a+b)& \cos (a+b)\end{pmatrix}\end{align*}

So we get again a rotation without any constraint about the angle, right?

10. The composition of two rotations with rotation angle $a$ and $b$ with $a+b=\pi$ is a reflection along a point.

When we P around the point Z with rotation angle 180°, then the points P, Z and P' are collinear. This rotation is equivalent to the reflection around the point Z. And we have that the composition of two rotations is a rotation with angle the sum of th two angles. Therefore, the statement is true.

Is this correct? According to the suggested solution we have the following:
$D(a)D(b)=D(\pi)=-I$
So, we have a mapping of the form $\vec{p}\mapsto -\vec{p}+v$.
This is a totation with $180^{\circ}$, so a reflection along a point. The center is $\vec{Z}=-\vec{Z}+v \Rightarrow 2\vec{Z}=\vec{v}$.
($\vec{P}\mapsto -I(\vec{p}-\vec{Z})+\vec{Z}=-\vec{p}+\vec{Z}+\vec{Z}$.

I haven't really understood that. Why do we have a mapping of the form $\vec{p}\mapsto -\vec{p}+v$ ? And why is that the center?

11. If $ABC$ is a triangle, then the center of the incircle is a fixed point of $\sigma_{AB}\circ\sigma_{BC}$.

$\sigma_{AB}\circ\sigma_{BC}$ is the composition of two reflections of lines that have an intersection point. Isn't that point the fixed point? Or are the also other fixed points?

 

[I like Serena]

1. By 'odd' we mean an odd permutation.
It takes an even number of odd permutations to get an even permutation. And the identity is even.
Hint: what is the determinant of a reflection?

The matrix for a reflection is orthogonal with determinant $-1$, right?

But what do we get from that?

When we P around the point Z with rotation angle 180°, then the points P, Z and P' are collinear. This rotation is equivalent to the reflection around the point Z, right?

Since a rotation is a composition of two reflections, we conclude that the reflection around a point is even, right?

Yes. And the determinant of a rotation and identity is $+1$.
So we can map any rotation and/or reflection to either +1 or -1 (the sign function).
-1 is odd, +1 is even.
To get to identity we need an even number of odd transformation.
Consequently the determinant of the composition is $(-1)^{\text{even number}} = +1$, which proves that a reflection is not even and therefore odd.

2. The composition of 2 reflections is a rotation. Suppose g and h have angle phi, which rotation will we get?
Hint: draw a picture.

I draw a pricture and the angle of rotation seems to be $2\phi$, because of the following:
View attachment 6982

We have that the angle between P and P'' is $a+a+b+b=(a+b)+(a+b)=\phi +\phi =2\phi$.

Is this correct?

Good!
If we apply $\sigma_h$ after $\sigma_g$ we rotate over $+2\phi$.
And if we apply $\sigma_g$ after $\sigma_h$ we rotate over $-2\phi$.
When will those rotations be the same? (Wondering)

3. When you say 'rotation' do you mean a rotation around the origin or not? That matters.

There is no information about that.

I have an idea about that statement:

A translation followed by a rotation of θ is still a rotation by θ. This is because if we take two points a and b which are mapped to a′ and b′, then θ is the angle between the vectors b−a and b′−a′, and even though the second one are translated.

Is this correct?

At a suggested solution there is the following justification why the statement is correct:

$\kappa=\sigma\circ\tau$ is an even congruence mapping, so a rotation or a translation. It cannot be a translation since then it would hold that $\sigma=\kappa\circ\tau^{-1}$ is a translation.

Why is $\kappa=\sigma\circ\tau$ an even congruence mapping? Are all the congruence mapping the rotations and the translations?

Now that we're talking about congruence mappings that basically confirms that we're not necessarily rotating around the origin.
In 2D the congruence mappings (the isometry group in 2 dimensions) are the translations, rotations, reflections, and glide reflections.

Reflections and glide reflections are odd (determinant -1), while translations and rotations and even.
Therefore a combination of a translation and a rotation is even, meaning it must be a translation or a rotation as well.

Your argument is not sufficient, since even though the angle is the same, we don't know yet if there is a unique center of the rotation. (Worried)

4. Consider the group of symmetries of the regular polygon $D_n$. That is known to you isn't it? What happens when we combine a reflection and a rotation?

We have that the rotation is a composition of two reflections. Therefore the composition of the rotation and the reflection is the composition of 3 reflections.

Is the composition of three reflections always a reflection?

According to the suggested solution, the statement is wrong, simce if the axis of the reflection passes through the center of rotation, then we have a reflection.
In that case the reflection that we get from the composition, we have the same axis as the initial reflection?

Yes, 3 reflections (or glide reflections) make a reflection (or a glide reflection).
Note that a general rotation (not around the origin) and a reflection typically make a glide reflection.

In the case that both reflection and rotation have the origin as a fixed point, then their composition will be a reflection (with the origin as fixed point), which indeeds contradicts the statement.I'm going to stop here if you don't mind. (Whew)
It's a bit too much otherwise - not too mention that this thread becomes hard to follow.
Perhaps we can put those other items in different threads? (Wondering)

 

[mathmari]

Yes. And the determinant of a rotation and identity is $+1$.
So we can map any rotation and/or reflection to either +1 or -1 (the sign function).
-1 is odd, +1 is even.
To get to identity we need an even number of odd transformation.
Consequently the determinant of the composition is $(-1)^{\text{even number}} = +1$, which proves that a reflection is not even and therefore odd.

Ah ok!

Good!
If we apply $\sigma_h$ after $\sigma_g$ we rotate over $+2\phi$.
And if we apply $\sigma_g$ after $\sigma_h$ we rotate over $-2\phi$.
When will those rotations be the same? (Wondering)

It must hold $+2\phi=-2\phi \Rightarrow \phi =0$, or not? (Wondering)

Now that we're talking about congruence mappings that basically confirms that we're not necessarily rotating around the origin.
In 2D the congruence mappings (the isometry group in 2 dimensions) are the translations, rotations, reflections, and glide reflections.

Reflections and glide reflections are odd (determinant -1), while translations and rotations and even.
Therefore a combination of a translation and a rotation is even, meaning it must be a translation or a rotation as well.

Your argument is not sufficient, since even though the angle is the same, we don't know yet if there is a unique center of the rotation. (Worried)

Ah ok I see!

Yes, 3 reflections (or glide reflections) make a reflection (or a glide reflection).
Note that a general rotation (not around the origin) and a reflection typically make a glide reflection.

In the case that both reflection and rotation have the origin as a fixed point, then their composition will be a reflection (with the origin as fixed point), which indeeds contradicts the statement.

A glide reflection is a composition of a reflection and a translation, right?

So, only when the line along which we reflect passes through the origin, and the rotation has the center at the origin, the compoosition is a reflection. Otherwise we get a glide reflection. Have I understood it correctly? (Wondering)

I'm going to stop here if you don't mind. (Whew)
It's a bit too much otherwise - not too mention that this thread becomes hard to follow.
Perhaps we can put those other items in different threads? (Wondering)

Ok, I will make a new thread.

 

[I like Serena]

It must hold $+2\phi=-2\phi \Rightarrow \phi =0$, or not?

Shouldn't that be $+2\phi=-2\phi + 2\pi k$? (Wondering)

A glide reflection is a composition of a reflection and a translation, right?

So, only when the line along which we reflect passes through the origin, and the rotation has the center at the origin, the compoosition is a reflection. Otherwise we get a glide reflection. Have I understood it correctly? (Wondering)

Actually, a reflection in a line that is not through the origin is also a reflection.
This is what a glide reflection looks like (reflection followed by translation):
View attachment 6983

 

[mathmari]

Shouldn't that be $+2\phi=-2\phi + 2\pi k$? (Wondering)

Why is it like that? (Wondering)

Actually, a reflection in a line that is not through the origin is also a reflection.

So, the difference is whether the rotation is at the origin or not? Or how can we know if we have reflection or glide reflection? (Wondering)

 

[I like Serena]

Why is it like that? (Wondering)

Isn't the rotation over $\frac\pi 6$ the same as a rotation over $\frac\pi 6 + 2\pi$ or a rotation over $\frac\pi 6 + 4\pi$?
The points will be mapped to the same points won't they? (Wondering)

So, the difference is whether the rotation is at the origin or not? Or how can we know if we have reflection or glide reflection? (Wondering)

The fixed points of a reflection form a line.
A translation and a glide reflection have no fixed points. (Thinking)

 

[mathmari]

Isn't the rotation over $\frac\pi 6$ the same as a rotation over $\frac\pi 6 + 2\pi$ or a rotation over $\frac\pi 6 + 4\pi$?
The points will be mapped to the same points won't they? (Wondering)

Ahh yes (Blush)

The fixed points of a reflection form a line.
A translation and a glide reflection have no fixed points. (Thinking)

So, if there are fixed points we have reflections and if not there are glide reflections, right?

We have that a rotation is a composition of two reflections. So, we have three reflections. If the lines of the reflections have an intersection point, this will be the fixed point, and then we have reflection. If the lines are parallel, we have glide reflection.

Is this correct? (Wondering)

 

[I like Serena]

Ahh yes (Blush)

So for which $\phi$ will the rotations be the same? (Wondering)

So, if there are fixed points we have reflections and if not there are glide reflections, right?

What? Wait! (Wait)
Don't rotations and the identity also have fixed points?

We have that a rotation is a composition of two reflections. So, we have three reflections. If the lines of the reflections have an intersection point, this will be the fixed point, and then we have reflection. If the lines are parallel, we have glide reflection.

Huh?

If the lines of 2 reflections are parallel, their composition will again be a reflection, wouldn't it?

 

[mathmari]

So for which $\phi$ will the rotations be the same? (Wondering)

So, we get that \begin{equation*}+2\phi=-2\phi + 2\pi k \Rightarrow 4\phi= 2\pi k \Rightarrow \phi = \frac{k\pi}{2}\end{equation*} So, either the lines are perpendicular when $k$ is odd or parallel then $k$ is even, right?

What? Wait! (Wait)
Don't rotations and the identity also have fixed points?

Huh?

If the lines of 2 reflections are parallel, their composition will again be a reflection, wouldn't it?

I got stuck... If the lines of 2 reflections are parallel, their composition will again be a reflection. Then if we have three lines of 3 reflections, then the two of them make a reflection, and so we have again two lines of reflection, and we again a reflection?

But when do we get a glide reflection? At whch case? (Wondering)

 

[I like Serena]

So, we get that \begin{equation*}+2\phi=-2\phi + 2\pi k \Rightarrow 4\phi= 2\pi k \Rightarrow \phi = \frac{k\pi}{2}\end{equation*} So, either the lines are perpendicular when $k$ is odd or parallel then $k$ is even, right?

Yep.
So statement 2 is false.

I got stuck... If the lines of 2 reflections are parallel, their composition will again be a reflection. Then if we have three lines of 3 reflections, then the two of them make a reflection, and so we have again two lines of reflection, and we again a reflection?

But when do we get a glide reflection? At which case? (Wondering)

My mistake. 2 parallel reflections make a translation.
A reflection and a translation make a glide reflection.

 

[mathmari]

My mistake. 2 parallel reflections make a translation.
A reflection and a translation make a glide reflection.

So, if we have the composition of 3 reflections, and at least two of the lines are parallel, then we have a reflection and a translation, so a glide reflection, right?

Does the third line must also be parallel or can it also have an intersection point with the other two? (Wondering)

 

[I like Serena]

So, if we have the composition of 3 reflections, and at least two of the lines are parallel, then we have a reflection and a translation, so a glide reflection, right?

Does the third line must also be parallel or can it also have an intersection point with the other two?

Only (in general) if 2 consecutive reflections have parallel lines. So the first or the last can have a line intersecting the other two. (Thinking)

 

[mathmari]

Only (in general) if 2 consecutive reflections have parallel lines. So the first or the last can have a line intersecting the other two. (Thinking)

We have the the reflection $\sigma$ along the line through the points $(4,1)$ and $(5,-3)$ and the rotation $\delta$ with rotation angle $\frac{\pi}{3}$ about the point $(-1,-1)$.

I want to check if $\sigma\circ\delta$ and $\delta\circ\sigma$ are reflections or glide reflections.

We have that a rotation is a composition of two reflections.

Therefore both $\sigma\circ\delta$ and $\delta\circ\sigma$ are a product of three reflections.

To check if we have a reflection or a glide reflection, we have to check if the reflection lines are parallel or not.

If they are not parallel does the one line passes through the fixed point of the rotation, i.e. the center of the rotation?

 

[I like Serena]

We have the the reflection $\sigma$ along the line through the points $(4,1)$ and $(5,-3)$ and the rotation $\delta$ with rotation angle $\frac{\pi}{3}$ about the point $(-1,-1)$.

I want to check if $\sigma\circ\delta$ and $\delta\circ\sigma$ are reflections or glide reflections.

We have that a rotation is a composition of two reflections.

Therefore both $\sigma\circ\delta$ and $\delta\circ\sigma$ are a product of three reflections.

To check if we have a reflection or a glide reflection, we have to check if the reflection lines are parallel or not.

If they are not parallel does the one line passes through the fixed point of the rotation, i.e. the center of the rotation?

We can choose to divide the rotation into any 2 reflections we want under certain conditions. (Thinking)

Let's write $\delta = \sigma_2 \circ \sigma_1$.
This will be a rotation by $\frac{\pi}{3}$ about the point $(-1,-1)$ if the lines of $\sigma_1$ and $\sigma_2$ intersect at $(-1,-1)$, and if the angle between the lines is $\frac{\pi}{6}$ (in the right direction).

Then $\sigma\circ\delta$ = $\sigma\circ\sigma_2 \circ \sigma_1$.
Let's pick the lines of $\sigma$ and $\sigma_2$ to be parallel, then $\tau=\sigma\circ\sigma_2$ is a translation that is at an angle with the line of $\sigma_1$.
And thus $\sigma\circ\delta$ is a glide reflection. (Nerd)

 

[mathmari]

Let's write $\delta = \sigma_2 \circ \sigma_1$.
This will be a rotation by $\frac{\pi}{3}$ about the point $(-1,-1)$ if the lines of $\sigma_1$ and $\sigma_2$ intersect at $(-1,-1)$, and if the angle between the lines is $\frac{\pi}{6}$ (in the right direction).

Then $\sigma\circ\delta$ = $\sigma\circ\sigma_2 \circ \sigma_1$.
Let's pick the lines of $\sigma$ and $\sigma_2$ to be parallel, then $\tau=\sigma\circ\sigma_2$ is a translation that is at an angle with the line of $\sigma_1$.
And thus $\sigma\circ\delta$ is a glide reflection. (Nerd)

Do we consider that the lines of $\sigma$ and $\sigma_2$ are parallel or do we know that they are? (Wondering)

 

[I like Serena]

Do we consider that the lines of $\sigma$ and $\sigma_2$ are parallel or do we know that they are? (Wondering)

We have the freedom to choose the line of $\sigma_2$, and we choose it to be parallel to $\sigma$.
Consequently the line of $\sigma_1$ is fixed at an angle of $\frac\pi 6$ with $\sigma_2$.

 

[mathmari]

We have the freedom to choose the line of $\sigma_2$, and we choose it to be parallel to $\sigma$.
Consequently the line of $\sigma_1$ is fixed at an angle of $\frac\pi 6$ with $\sigma_2$.

I understand!

Let $\sigma$ be a reflection and $\lambda$ a glide reflection. Since a glide reflection is a composition of a reflection and a translation, the composition $\sigma\circ\lambda$ is a composition of two reflections and one translation.
If the two reflection lines are parallel, then the composition of the two reflections is a translation, and so $\sigma\circ\lambda$ is a composition of two translations, which is a point reflection.
If the two reflection lines are not parallel, then the composition of the two reflections is a reflection, and so $\sigma\circ\lambda$ is a composition of a reflection and a translation, which is a glide reflection.

Is everything correct? (Wondering)

 

[I like Serena]

I understand!

Let $\sigma$ be a reflection and $\lambda$ a glide reflection. Since a glide reflection is a composition of a reflection and a translation, the composition $\sigma\circ\lambda$ is a composition of two reflections and one translation.
If the two reflection lines are parallel, then the composition of the two reflections is a translation, and so $\sigma\circ\lambda$ is a composition of two translations, which is a point reflection.

Don't two translations make a translation?
Note that a point reflection in 2D is actually a rotation by 180 degrees.

If the two reflection lines are not parallel, then the composition of the two reflections is a reflection, and so $\sigma\circ\lambda$ is a composition of a reflection and a translation, which is a glide reflection.

Don't 2 angled reflections make a rotation? (Wondering)

 

[mathmari]

Don't two translations make a translation?
Note that a point reflection in 2D is actually a rotation by 180 degrees.
Don't 2 angled reflections make a rotation? (Wondering)

Ahh ok (Nerd)

Therefore we have the following:

We have that the composition $\sigma\circ\lambda$ is a composition of two reflections and one translation.

If the two reflection lines are parallel, then the composition of the two reflections is a translation, and so $\sigma\circ\lambda$ is a composition of two translations, which is a translation.

If the two reflection lines are not parallel, then the composition of the two reflections is a rotation, and so $\sigma\circ\lambda$ is a composition of a rotation and a translation, which is a rotation.

Is this correct? (Wondering)

 

[I like Serena]

Ahh ok

Therefore we have the following:

We have that the composition $\sigma\circ\lambda$ is a composition of two reflections and one translation.

If the two reflection lines are parallel, then the composition of the two reflections is a translation, and so $\sigma\circ\lambda$ is a composition of two translations, which is a translation.

If the two reflection lines are not parallel, then the composition of the two reflections is a rotation, and so $\sigma\circ\lambda$ is a composition of a rotation and a translation, which is a rotation.

Is this correct?

Yep! All correct. (Happy)

 

[mathmari]

Yep! All correct. (Happy)

Ok! Thank you so much! (Mmm)