Group theory -- show H is a subgroup of O(2)

[Kara386]

Homework Statement

Let $R(\theta) = \left( \begin{array}{cc} \cos(\theta) & -\sin(\theta)\\ \sin(\theta)& \cos(\theta)\\ \end{array} \right) \in O(2)$ represent a rotation through angle $\theta$, and

$X(\theta) = \left( \begin{array}{cc} \cos(\theta) & \sin(\theta)\\ \sin(\theta)& -\cos(\theta)\\ \end{array} \right) \in O(2)$ represent reflection around $\frac{\theta}{2}$. Let m be a positive integer and H be the set such that $H = \{R(\frac{2q\pi}{m}), X(\frac{2q\pi}{m}) | q = 0, 1, 2..., m-1\}$.

Calculate $R(\theta)R(\phi)$, $R(\theta)X(\phi)$ and $X(\theta)X(\phi)$, express answers in terms of R and X. Show H forms a subgroup of $O(2)$.

Homework Equations

The Attempt at a Solution

$R(\theta) R(\phi) = \left( \begin{array}{cc} \cos(\theta+\phi) & -\sin(\theta+\phi)\\ \sin(\theta+\phi)& \cos(\theta+\phi)\\ \end{array} \right) = R(\theta + \phi)$

$R(\theta) X(\phi) = \left( \begin{array}{cc} \cos(\theta+\phi) & \sin(\theta+\phi)\\ \sin(\theta+\phi)& -\cos(\theta+\phi)\\ \end{array} \right) = X(\theta + \phi)$

$R(\theta) X(\phi) = R(\theta - \phi)$

I'm not really sure of the significance of these calculations, or if there is one. Does this have a geometric interpretation? I've been told that the set $O(2)$ is somehow related to the permutation of 3 points but haven't been able to find out why.
To show it's a subgroup of O(2) I have to show $R \times X \in O(2)$ where $\times$ is matrix multiplication and $a^{-1} \in H$. Do I need to use the argument $\frac{2q\pi}{m}$ in these calculations? So for example let $a = R(\frac{2q\pi}{m})$. Any help is very much appreciated, thank you! :)

I've gone ahead and attempted the problem with the argument above, and set a = R, b = X. I've already calculated RX to be $= R(\theta - \phi)$ above: $= \left( \begin{array}{cc} \cos(0) & \sin(0)\\ \sin(0)& -\cos(0)\\ \end{array} \right) = \left( \begin{array}{cc} 1& 0\\ 0 & 1\\ \end{array} \right)$
I wasn't really expecting to get the identity matrix, I suppose I show it's part of O(2) by showing $A^T A = I$ and of course, it is a real 2x2 matrix.

 

[Orodruin]

To prove that it is a subgroup, you have to show that it is a subset of O(2) that satisfies all the group axioms. That it is a subset of O(2) should be clear so you need to show that the group axioms are satisfied.

 

[Kara386]

To prove that it is a subgroup, you have to show that it is a subset of O(2) that satisfies all the group axioms. That it is a subset of O(2) should be clear so you need to show that the group axioms are satisfied.

By multiplying the two elements and showing they are part of $O(2)$ I'm not demonstrating closure then? I should actually show they are part of $H$? Matrix multiplication is associative, the identity is $\left( \begin{array}{cc} 1& 0\\ 0 & 1\\ \end{array} \right)$ and since $RX = I$ the implication is that $R = X^{-1}$ and $X = R^{-1}$, so the inverses both exist and are part of H. Does this demonstrate that the subset is a subgroup? Or group (not sure of the difference actually)? It certainly satisfies group axioms, so that makes it a subgroup?

 

[fresh_42]

By multiplying the two elements and showing they are part of $O(2)$ I'm not demonstrating closure then?

This is only the closure of $O(2)$, not of $H$.

I should actually show they are part of $H$?

Yes.

Matrix multiplication is associative, the identity is $\left( \begin{array}{cc} 1& 0\\ 0 & 1\\ \end{array} \right)$ and since $RX = I$ the implication is that $R = X^{-1}$ and $X = R^{-1}$, so the inverses both exist and are part of H. Does this demonstrate that the subset is a subgroup?

Yes. But you also have to show $R(\theta) \cdot X(\phi) \in H\, , \,R(\theta) \cdot R(\phi) \in H\; , \;X(\theta) \cdot X(\phi) \in H$ and $X(\phi) \cdot R(\theta) \in H$ for all $\theta, \phi$.

Or group (not sure of the difference actually)? It certainly satisfies group axioms, so that makes it a subgroup?

For a subgroup, you need to show $a\cdot b^{-1} \in H$ for all $a\, , \,b \in H$. Associativity is inherited. (Of course you can as well show $a\cdot b \in H\, , \,a^{-1} \in H\; , \;1 \in H$ separately but $a\cdot b^{-1} \in H$ does it all in one step.) However, it has to be shown for all $a,b \in H$, which means all pairs $(a,b) \in \{R(\theta),X(\phi)\}^2$ in this case.

I'm not sure about the connection to permutations you mentioned. (There is one for its tangent space but I don't know about the group itself. On the other hand it's full of symmetries as you have already calculated above.)

 

[Orodruin]

I should actually show they are part of HHH?

Yes. If the subset is not closed under the multiplication, it is not a group of its own.

By multiplying the two elements and showing they are part of $O(2)$ I'm not demonstrating closure then? I should actually show they are part of $H$? Matrix multiplication is associative, the identity is $\left( \begin{array}{cc} 1& 0\\ 0 & 1\\ \end{array} \right)$ and since $RX = I$ the implication is that $R = X^{-1}$ and $X = R^{-1}$, so the inverses both exist and are part of H. Does this demonstrate that the subset is a subgroup? Or group (not sure of the difference actually)? It certainly satisfies group axioms, so that makes it a subgroup?

What are your R and X here? The inverse of a rotation is not a reflection.

 

[Kara386]

Yes. If the subset is not closed under the multiplication, it is not a group of its own.What are your R and X here? The inverse of a rotation is not a reflection.

Couldn't rotation through $\theta$ undo a reflection through $\frac{\theta}{2}$ in a few cases? I'm not at all sure how this whole thing relates to geometry. What exactly is being rotated or reflected? Or do these matrices just result in rotation or reflection when applied to other matrices or vectors?

 

[fresh_42]

Look in a mirror. Can you reverse the change of left and right by a rotation? Reflections change the orientation, rotations don't. Or in terms of the matrix group here: one has determinant $-1$, the other on $1$.

 

[Orodruin]

Couldn't rotation through $\theta$ undo a reflection through $\frac{\theta}{2}$ in a few cases? I'm not at all sure how this whole thing relates to geometry. What exactly is being rotated or reflected? Or do these matrices just result in rotation or reflection when applied to other matrices or vectors?

Try taking the determinant of $RX$ ... The inverse of a rotation by $\theta$ is a rotation by $-\theta$ (or equivalently, by $2\pi-\theta$).

 

[Kara386]

So $RX$ shouldn't actually be equal to $I$ then? I've probably made a mistake, I'll check. I suppose I was thinking of highly symmetric things like circles, which aren't really changed by these things, when I say rotation could undo reflection, but only in some few cases.

 

[Kara386]

This is only the closure of $O(2)$, not of $H$.
For a subgroup, you need to show $a\cdot b^{-1} \in H$ for all $a\, , \,b \in H$. Associativity is inherited. (Of course you can as well show $a\cdot b \in H\, , \,a^{-1} \in H\; , \;1 \in H$ separately but $a\cdot b^{-1} \in H$ does it all in one step.) However, it has to be shown for all $a,b \in H$, which means all pairs $(a,b) \in \{R(\theta),X(\phi)\}^2$ in this case.

But aren't both theta and phi defined in H as being $\frac{2q\pi}{m}$? I'd shown that $R(\theta)X(\phi) = R(\theta - \phi)$, so $R(\frac{2q\pi}{m})X(\frac{2q\pi}{m}) = R(\frac{2q\pi}{m}- \frac{2q\pi}{m}) = R(0)$, which I thought would be
$R(0) = \left( \begin{array}{cc} 1& 0\\ 0 & 1\\ \end{array} \right)$
I'm also not completely sure what that last bit of notation means: $(a,b) \in \{R(\theta),X(\phi)\}^2$
Why is this bracket squared?

 

[fresh_42]

But aren't both theta and phi defined in H as being $\frac{2q\pi}{m}$?

So? This doesn't change my argument, only the amount of possible angles.

I'd shown that $R(\theta)X(\phi) = R(\theta - \phi)$, so $R(\frac{2q\pi}{m})X(\frac{2q\pi}{m}) = R(\frac{2q\pi}{m}- \frac{2q\pi}{m}) = R(0)$, which I thought would be
$R(0) = \left( \begin{array}{cc} 1& 0\\ 0 & 1\\ \end{array} \right)$

There might be a mistake in here. Check the determinants. And, yes, an equation like this would do the job. But you still have to cover the possibility of different angles $\frac{2q\pi}{m}$ and $\frac{2p\pi}{m}$ in $R(\frac{2q\pi}{m})$ and $X(\frac{2p\pi}{m})$. And what happens with $R \cdot R' \, , \, X \cdot X' \, , \, X\cdot R\,$? The latter because $H$ is eventually not Abelian. (You haven't checked this.)

I'm also not completely sure what that last bit of notation means: $(a,b) \in \{R(\theta),X(\phi)\}^2$
Why is this bracket squared?

It means $(a,b) \in \{R(\theta),X(\phi)\} \times \{R(\theta),X(\phi)\} = \{R(\theta),X(\phi)\}^2$ or $a,b \in \{R(\theta),X(\phi)\}$ or $(a \in \{R(\theta),X(\phi)\}$ and $b \in \{R(\theta),X(\phi)\})$. These are significantly more cases as only $RX$.

 

[Kara386]

So? This doesn't change my argument, only the amount of possible angles.

Ah. I hadn't realized the angles could be different, I assumed it would always be $\theta = \phi = \frac{2q\pi}{m}$, but I see why that's not the case. A shame, it made life that much easier. So what I've worked out is only for the case of $\theta = \phi$. Back to the drawing board!

I'll try that again and see where I get to. Your time and help is much appreciated, thank you! :)