# Abstract Nonsense

## 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 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 $H_1h_2\in\overline{H}$. $\blacksquare$

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”.