Abstract Nonsense

Crushing one theorem at a time

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 G be a topological group and H\leqslant G. Then, H is a topological group with the subspace topology.

Proof: It is clear that H is a set with both a topological and group structure and thus it remains to show that \vartheta:H\times H\to H:(h_1,h_2)\mapsto h_1h_2 and \varphi:H\to H:h\mapsto h^{-1} are continuous. It should be remarked that the only reason we know these two mappings map into their specified images is because H is a subgroup. If it weren’t there’s no guarantee that h_1,h_2\in H\implies h_1h_2=\vartheta(h_1,h_2)\in H.

But, the continuity of these maps are apparent since \varphi=\phi\mid_H and \vartheta=\theta\mid_{H\times H} and since the restriction of a continuous map to a subspace is always continuous the conclusion follows. \blacksquare

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

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

Theorem: Let H\leqslant G. Then, H is open if and only if it contains an interior point.

Proof: Suppose that it contains an interior point h. Then, there exists some U such that h\in U\subseteq H. So, given any h'\in H we have that h'h^{-1}\in H and h'h^{-1}h\in h'h^{-1}\subseteq h'h^{-1}H=H 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 h'\in H^{\circ}. But, since h' was arbitrary it follows that H is open.

Conversely, if H is open then every point is an interior point, and since H must contain e it follows that e is an interior point. The subtle point is that if H 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. \blacksquare

And in fact we have the following surprising theorem:

Theorem: Let H\leqslant G be open, then it’s also closed.

Proof: If H=G we’re done, so assume not and let g\in G-H. Then, gH is a neighborhood of g and gH\cap H=\varnothing for if not then gh_1=h_2 for some h_1,h_2\in H and so g=h_2h_1^{-1}\in H which is a contradiction. It follows that g\in \left(G-H\right)^{\circ}. But, since it was arbitrary it follows that G-H is open and thus H is closed. \blacksquare

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

Corollary: If you’ll remember earlier we proved that if V is a symmetric neighborhood of e then \displaystyle \Omega=\bigcup_{n=1}^{\infty}V^n 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 G is connected that it must be true that \Omega=G.

There is a partial converse too.

Theorem: Let H\leqslant G be closed and \left[G:H\right]<\infty then H is open.

Proof: This follows easily since G=H\cup g_1H\cup\cdots\cup g_n H for some g_1,\cdots,g_n\in G. But, all of these are closed and since they are disjoint (this is basic group theory) it follows that H=G-\left(g_1H\cup\cdots g_nH\right) and the right hand side is G minus the finite union of closed sets, thus open. The conclusion follows. \blacksquare.

Corollary: Let H<G be closed. Then, if G is connected it must be that \left[G:H\right]=\infty.

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 \mathbb{Z}\leqslant (\mathbb{R},+) has infinite index. All you’d need to do is pull a “in and out” technique. You note that you can topologize (going “in”) (\mathbb{R},+) with the normal topology and this turns \mathbb{R} into a connected topological group for which \mathbb{Z} is a proper closed subgroup. It follows that \left[\mathbb{R}:\mathbb{Z}\right]=\infty. But, this is a purely group theoretic matter and so “going out” and forgetting the topology it still must be true that \left[\mathbb{R}:\mathbb{Z}\right]=\infty 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 H\leqslant G, then \overline{H}\leqslant G.

Proof: Clearly e\in \overline{H}. Now, let h\in\overline{H} and let N be any neighborhood of h^{-1} . Then, h\in N^{-1}, but since h\in\overline{H} there must exists some h'\in N^{-1}\cap H. So, (h')^{-1}\in N but (h')^{-1}\in H. Thus, since N was arbitrary it follows that h^{-1}\in \overline{H}.

Next, let h_1,h_2\in\overline{H} and let N be any neighborhood of h_1h_2. Remember that

\theta^{-1}(N)=\bigcup_{j\in\mathcal{J}}\left(U_j\times V_j\right)

where each U_j,V_j is open. Thus,

\displaystyle N=\theta(\theta^{-1}(N))=\theta\left(\bigcup_{j\in\mathcal{J}}\left(U_j\times V_j\right)\right)=\bigcup_{j\in\mathcal{J}}\theta\left(U_j\times V_j\right)=\bigcup_{j\in\mathcal{J}}U_jV_j

Where each U_jV_j being the product of open sets is open. But, since \theta(h_1,h_2)=h_1h_2\in N it follows that (h_1,h_2)\in U_{j_0}\times V_{j_0} for some j_0\in\mathcal{J}. But, this means that U_{j_0} is a neighborhood of h_1 and thus it must contain some point h'_1\in U_{j_0}\cap H. Similarly, there must be some h'_2\in V_{j_0}\cap H. So,

\displaystyle h'_1h'_2\in U_{j_0}V_{j_0}\subseteq\bigcup_{j\in\mathcal{J}}U_jV_j=N

But, h'_1h'_2\in H. Thus, since N was arbitrary it follows that h_1h_2 is an adherent point for H and thus H_1h_2\in\overline{H}. $\blacksquare$

We now can ask about normal subgroups

Theorem: Let N\unlhd G then \overline{N}\unlhd G.

Proof: We have already shown that \overline{N}\leqslant G. Thus, it remains to show that g\overline{N}g^{-1}\subseteq \overline{N} for every g\in G.

So, let gng^{-1}\in g\overline{N}g^{-1} and let U be any neighborhood of it. Then, n\in g^{-1}Ug, but this is a neighborhood of n and so it must contain some point n'\in N. Thus, gn'g^{-1}\in U. But, since N\unlhd G we know that gn'g^{-1}\in N. It follows gng^{-1} is an adherent point and thus gng^{-1}\in\overline{N}. By previous comment the conclusion follows. \blacksquare

A nice theorem which is tangentially related is the following:

Theorem: Let C\subseteq G be the component containing C. Then, it is closed and C\unlhd G.

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 C^{-1}\subseteq C and C^2\subseteq C.

Notice though that e\in C^{-1} and since \phi is a homeomorphism and C^{-1}=\phi(C) it follows that C^{-1} is connected. Thus, since C\cap C^{-1}\ne\varnothing and both are connected it follows that C\cup C^{-1} is connected. But, by the maximality of C it follows that C\cup C^{-1}=C\implies C^{-1}\subseteq C as desired.

Now, for closure under multiplication we note that \displaystyle C^2=\bigcup_{c\in C}cC. But, each cC being the continuous image of C (under L_c) is connected. But, since c\in C\implies c^{-1}\in C\implies e\in cC^{-1} we see that \displaystyle e\in\bigcap_{c\in C}cC and thus their intersection is non-empty. Thus, \displaystyle \bigcup_{c\in C}cC=C^2 is connected and contains C. It follows that C^2=C as was desired.

Lastly, for normality we note (similarly as before that) gCg^{-1} for any g\in G is the continuous image of C and thus connected. Also, e\in gCg^{-1} so that C\cup gCg^{-1} is a connected set containing C and thus equal to C. It follows that gCg^{-1}\subseteq C. Since g was arbitrary the conclusion follows. \blacksquare

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

Theorem:Let \varphi,\psi:X\to G be continuous where G is a topological group. Then, the map \varphi\psi:X\to G:x\mapsto\varphi(x)\psi(x) is continuous.

Proof: One must merely note that

\varphi\psi=\theta\circ\left(\varphi\oplus\psi\right):X\to G:x\overset{\varphi\oplus\psi}{\longmapsto}(\varphi(x),\psi(x))\overset{\theta}{\longmapsto}\varphi(x)\psi(x)

And since both \varphi\oplus\psi and \theta are continuous the conclusion follows. \blacksquare.

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

Theorem: Recall from group theory that the centralizer C_g is defined to be C(g)=\left\{x\in G:gx=xg\right\}. Then, C(g) is closed in G for every g\in G if G is Hausdorff.

Proof: From the above we know that \phi C_g:G\to G:x\mapsto x^{-1}gxg^{-1} is continuous. And so

C(g)=\left\{x\in G:xg=gx\right\}=\left\{x\in G:e=x^{-1}gxg^{-1}\right\}=\left(\phi C_g\right)^{-1}(\{e\})

The conclusion follows. \blacksquare

From this we can prove the following theorem.

Theorem: Let G be a topological group which is Hausdorff. Then, \mathcal{Z}(G) (the center) is closed in G.

Proof: This follows since

\displaystyle \mathcal{Z}(G)=\left\{x\in X:xg=gx\text{ }\forall g\in G\right\}=\bigcap_{g\in G}C(g)

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

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

Advertisements

April 29, 2010 - Posted by | Algebra, Group Theory, Topological Groups, Topology, Uncategorized | , , , ,

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: