Graduate Difference Between Subgroup & Closed Subgroup of a Group

[LagrangeEuler]

What is difference between subgroup and closed subgroup of the group? It is confusing to me because every group is closed.

In a book Lie groups, Lie algebras and representations by Brian C. Hall is written

"The condition that $G$ is closed subgroup, as opposed to merely a subgroup, should be regarded as technicality, in the most of interesting subgroups of $GL(n,\mathbb{C})$ have this property."

 

[member 587159]

- Subgroup = subgroup in the group theoretic sense.
- Closed subgroup = subgroup in the group theoretic sense and closed in the topological sense.

I don't know if the author means "Lie subgroup" with "subgroup" though.

 

[fresh_42]

We are speaking about topological groups here, i.e. they are topological spaces. Now a subgroup is a subset, in this case a subset of a topological space. Hence this set can be open, closed, or neither. E.g. $GL(n)\subseteq \mathbb{M}(n)$ is not closed.

Edit: ... but neither a subgroup.

 

[Infrared]

$GL(n)\subseteq \mathbb{M}(n)$ is not closed.

$GL(n)$ is not a subgroup of $M(n)$ (think about which group operations you're using).

An example of a non-closed subgroup of $GL(2,\mathbb{R})$ is the subgroup of rotations by rational multiples of $\pi$.

 

[fresh_42]

$GL(n)$ is not a subgroup of $M(n)$ (think about which group operations you're using).

An example of a non-closed subgroup of $GL(2,\mathbb{R})$ is the subgroup of rotations by rational multiples of $\pi$.

Thanks, yes. I only thought about the topology and forgot the group. Silly me.

 

[LagrangeEuler]

$GL(n)$ is not a subgroup of $M(n)$ (think about which group operations you're using).

An example of a non-closed subgroup of $GL(2,\mathbb{R})$ is the subgroup of rotations by rational multiples of $\pi$.

Sorry, could you please explain that. Why this subgroup is not closed?

 

[fresh_42]

Because you can define a sequence of rotations by rational multiples of $\pi$ which converge to a rotation to an irrational multiple of $\pi$: $R_\varphi \circ R_\psi = R_{\varphi +\psi}\,.$ You can add rational numbers to get an irrational number as limit.

 

[Infrared]

Sorry, could you please explain that. Why this subgroup is not closed?

If $X$ is a topological space, a subspace $F$ is said to be closed if the complement $X\setminus F$ is open.

Let $S$ be the subgroup I defined. Let $A$ be a rotation by a irrational multiple of $\pi$. Then $U:=X\setminus S$ is not open because even though $A\in U$, every neighborhood of $A$ contains rotations by rational multiples of $\pi$ (since irrational numbers can be approximated to arbitrary accuracy by rational numbers).

 

[fresh_42]

If $X$ is a topological space, a subspace $F$ is said to be closed if the complement $X\setminus F$ is open.

Let $S$ be the subgroup I defined. Let $A$ be a rotation by a irrational multiple of $\pi$. Then $U:=X\setminus S$ is not open because even though $A\in U$, every neighborhood of $A$ contains rotations by rational multiples of $\pi$ (since irrational numbers can be approximated to arbitrary accuracy by rational numbers).

A little remark: Open doesn't imply not closed. Some connection components which occur often in those groups as covering are both, open and closed.

 

[Infrared]

I did not say "open implies not closed". I said that its complement of $S$ is not open, so by definition $S$ is not closed.

 

[fresh_42]

I did not say "open implies not closed". I said that its complement of $S$ is not open, so by definition $S$ is not closed.

I know, I just mentioned it as the OP explicitly has asked for not closed.

 

[LagrangeEuler]

Because you can define a sequence of rotations by rational multiples of $\pi$ which converge to a rotation to an irrational multiple of $\pi$: $R_\varphi \circ R_\psi = R_{\varphi +\psi}\,.$ You can add rational numbers to get an irrational number as limit.

I understand the point but I am not able to see that. Why if you rotate for rational multiply of $\pi$, you will get at one point rotation for irrational multiply of $\pi$?

 

[martinbn]

This is probably not related, but just to point out that in a topological group an open subgroup is always closed as well.

 

[fresh_42]

I understand the point but I am not able to see that. Why if you rotate for rational multiply of $\pi$, you will get at one point rotation for irrational multiply of $\pi$?

No, that is not the argument. The reasoning goes:

There are rotations $R_{q_n \pi}$ with $q_n\in \mathbb{Q}$ such that $\lim_{n \to \infty} \prod_{n=1}^\infty R_{q_n \pi}= R_{\sqrt{2}\,\pi}$. That is, there is a sequence of rotations in the subgroup which converges to a point outside the subgroup, hence it cannot be closed as we have a found limit point outside.

This is probably not related, but just to point out that in a topological group an open subgroup is always closed as well.

What do you mean here? Why should $SO(2, \pi \mathbb{Q}) \subseteq SO(2,\mathbb{R})$ be closed (see example above)?

 

[Infrared]

The claim was that open subgroups are also closed, $SO(2,\pi\mathbb{Q})$ is not open. For a proof, if $H\subset G$ is an open subgroup, and $g\in G\setminus H$, then $gH$ is an open neighborhood of $g$ disjoint from from $H$.