# Difference Between Subgroup & Closed Subgroup of a Group

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• LagrangeEuler

#### 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."

## Answers and Replies

- 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.

LagrangeEuler
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.

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##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##.

member 587159
##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.

##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?

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.

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).

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.

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 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.

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##?

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

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)?

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##.

fresh_42