# Question on conjugation closure of subgroups

• mnb96
In summary: Thus T is a subgroup of the conjugate closure, and from the definition of conjugate closure it follows that the conjugate closure is contained in \left\langle R,T \right\rangle, which by definition is the structure of Ω.In summary, the group Ω=RT is essentially the special Euclidean group SE(2), and it is known that R and T are both Lie groups diffeomorphic to the unit circle and the 2D plane respectively. Considering the conjugate closure of R with respect to Ω, it can be shown that this group is still a Lie group and is equal to Ω itself. This is because the conjugate closure includes elements of the form (t^{-1}rt)\,
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?

mnb96 said:
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.

fzero said:
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?

mnb96 said:
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.

1 person
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.

Last edited:

## 1. What is conjugation closure of subgroups?

The conjugation closure of a subgroup is the smallest subgroup that contains all possible conjugates of the original subgroup. This means that if we conjugate any element in the original subgroup by any element in the larger group, the result will still be in the conjugation closure subgroup.

## 2. How is conjugation closure different from normal closure?

Conjugation closure and normal closure are two different ways of extending a subgroup. Normal closure is the smallest normal subgroup that contains the original subgroup, while conjugation closure is the smallest subgroup that contains all possible conjugates of the original subgroup. So, while normal closure focuses on preserving the normality of subgroups, conjugation closure focuses on preserving the conjugacy of subgroups.

## 3. What are some applications of conjugation closure of subgroups?

Conjugation closure of subgroups is useful in various areas of mathematics, including group theory, topology, and algebraic geometry. It is also relevant in understanding the structure of symmetry groups in physics and chemistry.

## 4. Can a subgroup be its own conjugation closure?

Yes, a subgroup can be its own conjugation closure. This is true when the subgroup is already a normal subgroup or when the subgroup is contained in the center of the larger group. In these cases, the conjugation closure is the same as the normal closure, which is equal to the original subgroup.

## 5. How is conjugation closure related to the concept of normalizer?

The normalizer of a subgroup is the largest subgroup of the larger group in which the subgroup is normal. It is closely related to conjugation closure, as the normalizer of a subgroup is always contained in the conjugation closure of that subgroup. In fact, the normalizer is the largest normal subgroup contained in the conjugation closure of the original subgroup.

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