Question on conjugation closure of subgroups

[mnb96]

Hello,

I consider the groups of rotations $R=SO(2)$ and the group $T$ of translations on the 2D Cartesian plane.
Let's define Ω as the group Ω=RT.
Thus Ω is essentially SE(2), the special Euclidean group.

It is known that R and T are respectively 1-dimensional and 2-dimensional Lie groups diffeomorphic to the unit circle and the 2D plane.

My question is:
If I consider now $\left\langle R \right\rangle ^\Omega$, the conjugate closure of R with respect to Ω, what is the "structure" of such a group? Is it still a Lie group? if so, what is the manifold associated with it?

 

[fzero]

Hello,

I consider the groups of rotations $R=SO(2)$ and the group $T$ of translations on the 2D Cartesian plane.
Let's define Ω as the group Ω=RT.
Thus Ω is essentially SE(2), the special Euclidean group.

It is known that R and T are respectively 1-dimensional and 2-dimensional Lie groups diffeomorphic to the unit circle and the 2D plane.

My question is:
If I consider now $\left\langle R \right\rangle ^\Omega$, the conjugate closure of R with respect to Ω, what is the "structure" of such a group? Is it still a Lie group? if so, what is the manifold associated with it?

Since $\Omega$ is abelian, it seems straightforward that $\left\langle R \right\rangle ^\Omega =R$. If we were to consider a higher-dimensional Euclidean group, it seems that this continues to hold, since the result follows from $T$ being a normal subgroup of $\Omega$.

A slightly more interesting example would be to consider a rank $m$ subgroup $R_m$ of a sufficiently large $R_n$ and study $\left\langle R_m \right\rangle ^{\Omega_n}$. It seems easy to convince yourself that this is just $R_n$ itself, though a formal proof would probably be illuminating.

 

[mnb96]

Since $\Omega$ is abelian...

Why is Ω supposed to be abelian?
For sure T and R are abelian groups, but I don't think Ω=RT is abelian, because R is not a normal subgroup of Ω. In general rt≠tr. Or am I missing something?

 

[fzero]

Why is Ω supposed to be abelian?
For sure T and R are abelian groups, but I don't think Ω=RT is abelian, because R is not a normal subgroup of Ω. In general rt≠tr. Or am I missing something?

No, you're correct. So let's work through it explicitly, like I should have done in the first place. A general element of $\Omega$ is

$$ g(a,b,\theta) = t_1 (a) t_2(b) r(\theta). $$

We are to consider $g^{-1} r g$. A particular combination would be $t_1(a)^{-1}r(\phi )t_1(a) = t_1 (a(\cos\phi-1)) t_2(a\sin\phi)$. I expect that considering the general case would lead to $\langle R\rangle^\Omega = \Omega$.

Perhaps my more complicated example would lead to $\langle R_m\rangle^{\Omega_n} = \Omega_m$ when properly analyzed.

 

[mnb96]

Hi fzero,

thanks for your help. I think you are right when you suggest that $\Omega = \left\langle R^\Omega \right\rangle$, because $\left\langle R^\Omega \right\rangle$ is supposed to be a group anyway, and if it's neither T nor R, then I don't know what else it can be, besides Ω itself.

The thing that bothers me is that, I can't prove that $T \subseteq \left\langle R^\Omega \right\rangle$ from the definition of conjugate closure.

---EDIT:---
Maybe I got it: the conjugate closure contains also elements of the kind: $(t^{-1}rt)\,(\tau^{-1}r^{-1}\tau)$ and since T is normal we have that $r(t\tau^{-1})r^{-1}=t' \in T$, thus the above quantity reduces to $tt'\tau$ which is an element of T.