## Topological Groups (Subgroups)

We will not extend our discussion of topological groups beyond the basic definitions and consequences relating to topological features to how “construction” features of the spaces (subspaces/subgroups, direct product/product space, etc.) react. For example, is a subgroup a topological group if given the subspace topology? That in particular will the the focus of the post, the answer being, yes. So, we begin by verifying this

**Theorem:** Let be a topological group and . Then, is a topological group with the subspace topology.

**Proof: **It is clear that is a set with both a topological and group structure and thus it remains to show that and are continuous. It should be remarked that the only reason we know these two mappings map into their specified images is because is a subgroup. If it weren’t there’s no guarantee that .

But, the continuity of these maps are apparent since and and since the restriction of a continuous map to a subspace is always continuous the conclusion follows.

*Remark:* As to not cause confusion I will always refer to the multiplication/inversion maps in the ambient topological groups as and and their subgroup analogues as the “theta variant” and “phi variant” and .

So, the natural question is when is a subgroup open or closed? Thus, we begin with some simple theorems relating to this:

**Theorem:** Let . Then, is open if and only if it contains an interior point.

**Proof:** Suppose that it contains an interior point . Then, there exists some such that . So, given any we have that and the last part gotten since multiplication of a subgroup by any of it’s elements merely permutes them, thus leaving the set essentially unchanged. Thus, it follows that . But, since was arbitrary it follows that is open.

Conversely, if is open then every point is an interior point, and since must contain it follows that is an interior point. The subtle point is that if weren’t a subgroup it very well could have been empty in which case it could still be open but not actually contain an interior point.

And in fact we have the following surprising theorem:

**Theorem:** Let be open, then it’s also closed.

**Proof:** If we’re done, so assume not and let . Then, is a neighborhood of and for if not then for some and so which is a contradiction. It follows that . But, since it was arbitrary it follows that is open and thus is closed.

**Corollary:** Let be a connected topological group, then has no proper open subgroups.

**Corollary:** If you’ll remember earlier we proved that if is a symmetric neighborhood of then is a subgroup. But, since it’s evidently open (it’s the union of open sets) it follows from the previous corollary that if the ambient space is connected that it must be true that .

There is a partial converse too.

**Theorem:** Let be closed and then is open.

**Proof:** This follows easily since for some . But, all of these are closed and since they are disjoint (this is basic group theory) it follows that and the right hand side is minus the finite union of closed sets, thus open. The conclusion follows. .

**Corollary:** Let be closed. Then, if is connected it must be that .

*Remark:* This is a convenient time to mention something really powerful about topological groups relating solely to group theory. Suppose that you wanted (I take a very simple example for convenience) to prove that has infinite index. All you’d need to do is pull a “in and out” technique. You note that you can topologize (going “in”) with the normal topology and this turns into a connected topological group for which is a proper closed subgroup. It follows that . But, this is a purely group theoretic matter and so “going out” and forgetting the topology it still must be true that since the topology has nothing to do with index. This will be a recurring technique one can implement. Namely, one wants to prove something about a group. So, you first see if you can topologize the group into a topological group where this quality must be true and then forget the topology.

We now give another example where the above remark can be used to great effect.

**Theorem:** Let , then .

**Proof:** Clearly . Now, let and let be any neighborhood of . Then, , but since there must exists some . So, but . Thus, since was arbitrary it follows that .

Next, let and let be any neighborhood of . Remember that

where each is open. Thus,

Where each being the product of open sets is open. But, since it follows that for some . But, this means that is a neighborhood of and thus it must contain some point . Similarly, there must be some . So,

But, . Thus, since was arbitrary it follows that is an adherent point for and thus . $\blacksquare$

We now can ask about normal subgroups

**Theorem:** Let then .

**Proof:** We have already shown that . Thus, it remains to show that for every .

So, let and let be any neighborhood of it. Then, , but this is a neighborhood of and so it must contain some point . Thus, . But, since we know that . It follows is an adherent point and thus . By previous comment the conclusion follows.

A nice theorem which is tangentially related is the following:

**Theorem:** Let be the component containing . Then, it is closed and .

**Proof:** Being closed is merely a consequence of the fact that a component in any topological space is closed.

Now, we prove that it’s a subgroup. To do this we need only prove that and .

Notice though that and since is a homeomorphism and it follows that is connected. Thus, since and both are connected it follows that is connected. But, by the maximality of it follows that as desired.

Now, for closure under multiplication we note that . But, each being the continuous image of (under ) is connected. But, since we see that and thus their intersection is non-empty. Thus, is connected and contains . It follows that as was desired.

Lastly, for normality we note (similarly as before that) for any is the continuous image of and thus connected. Also, so that is a connected set containing and thus equal to . It follows that . Since was arbitrary the conclusion follows.

Lastly, we prove that a couple of specific kinds of subgroups of are closed. But, first a theorem.

**Theorem:**Let be continuous where is a topological group. Then, the map is continuous.

**Proof: **One must merely note that

And since both and are continuous the conclusion follows. .

So, with this we are able to prove our first theorem

**Theorem:** Recall from group theory that the centralizer is defined to be . Then, is closed in for every if is Hausdorff.

**Proof:** From the above we know that is continuous. And so

The conclusion follows.

From this we can prove the following theorem.

**Theorem:** Let be a topological group which is Hausdorff. Then, (the center) is closed in .

**Proof:** This follows since

But, this is the intersection of closed sets and thus closed.

That’s it for now. Next time we’ll discuss “spaces of cosets”.

No comments yet.

## Leave a Reply