# Closure of an abelian subgroup

• Jamma
In summary, the conversation discusses showing that the closure of an abelian subgroup in a topological group is also an abelian subgroup. The speaker mentions that they have already shown it is a subgroup, and now just need to prove it is abelian. They mention using the fact that for a metric space, the elements of the closure can be written as limits of sequences, and therefore can be multiplied using continuity and commuted. They question whether this can be extended to a general topological group, and mention that Hausdorff and second countability are needed for this proof. The other person confirms that convergence of nets or filters can also be used, but emphasizes the importance of Hausdorff.
Jamma
Ok, this is a really easy question, so I apologise in advance.

Let A be an abelian subgroup of a topological group. I want to show that cl(A) is also.

Now I've shown that cl(A) is a subgroup, that is fairly easy. So I just need to show it is abelian.

For a metric space, it is easy, since we can just write for any two elements h_1 h_2 in cl(H) as

h_1=lim_{i -> infty} (h_1(i))
h_2=lim_{i -> infty} (h_2(i))

where h_1(i) and h_2(i) are sequences converging too h_1 and h_2 respectively which are in H. Then to find h_1.h_2 we can just multiply the elements of these limits since multiplication is continuous and then we can commute them to deduce that h_1.h_2=h_2.h_1

Can I do something similar for a general topological group, which is not a metric space? h_1 and h_2 are limit points, if they are not then they are in H anyway and will commute.

Maybe I just need my space to be Hausdorff, so then limits are unique?

Oh, sorry, its going to be second (which implies first) countability that I will need. That will allow me to write the points as limits. Sorry!

Yes, you can use convergence of nets (or filters) to do the proof. But, as you note, Hausdorff is required. Indeed, the result is, in general, false in non-Hausdorff topological groups.

## 1. What is an abelian subgroup?

An abelian subgroup is a subset of a group that is itself a group, and where the group operation is commutative. This means that the order of elements does not affect the result of the operation.

## 2. How is closure defined for an abelian subgroup?

Closure for an abelian subgroup is when the result of the group operation on any two elements within the subgroup is also an element of the subgroup. In other words, the subgroup is closed under the group operation.

## 3. Why is closure important for an abelian subgroup?

Closure is important for an abelian subgroup because it ensures that the subset is a valid group and that the group operation is well-defined. It also allows for the use of important mathematical concepts and theorems, such as Lagrange's theorem.

## 4. How do you prove closure for an abelian subgroup?

To prove closure for an abelian subgroup, you must show that the result of the group operation on any two elements within the subgroup is also an element of the subgroup. This can be done by using the definition of closure and the properties of the group operation, such as commutativity and associativity.

## 5. Can an abelian subgroup have more than one closure?

No, an abelian subgroup can only have one closure. This is because closure is a fundamental property of a group, and a subgroup must have the same closure as the larger group it is a part of. Additionally, if a subgroup had more than one closure, it would no longer be a valid subgroup.

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