Totally disconnected and normal in connected implies central

This page describes additional conditions under which a subgroup property implication can be reversed, viz a weaker subgroup property, namely Normal subgroup (?), can be made to imply a stronger subgroup property, namely central subgroup
View other subgroup property implication-reversing conditions

Statement

Suppose $G$ is a Connected topological group (?) and $H$ is a normal subgroup of $G$, such that $H$ is a totally disconnected space in the subspace topology (i.e., the connected components of $H$ are one-point subsets). Then, $H$ is a central subgroup of $G$, i.e., $H$ is contained in the Center (?) of $G$.

A case of particular interest is where $H$ is a discrete subgroup of $G$, i.e., a closed subgroup of $G$ that is a discrete space under the subspace topology. This particular case says that any discrete normal subgroup of a connected topological group is central. However, there are many examples of totally disconnected normal subgroups that are not discrete.

Related facts

• Central implies normal
• Normal not implies central
• Normal of least prime order implies central
• Cartan-Brauer-Hua theorem

Proof

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Given: A connected topological group $G$, a normal subgroup $H$ that is totally disconnected in the subspace topology.

To prove: $H$ is contained in the center of $G$.

Proof: The idea is to look at a fixed but arbitrary element in $H$ and a continuously varying element of $G$ that conjugates on this element. Note that the construction used in this proof involves conjugation, but viewed in a different way from usual. In the usual way of thinking about conjugation, the conjugating element is kept fixed and the element being conjugated is varied. Here, the element being conjugated is kept fixed and the conjugating element is varied continuously.

Step no.	Assertion/construction	Given data used	Facts used	Previous steps used	Explanation
1	For $h \in H$, consider the map $\pi_h: G \to H$ given by $g \mapsto ghg^{-1}$. This is a well-defined map.	$H$ is normal in $G$			[SHOW MORE] Since $H$ is normal in $G$, $ghg^{-1} \in H$ for $g \in G, h \in H$.
2	$\pi_h$ is a continuous map from $G$ as a topological space to $H$ endowed with the subspace topology.	$G$ is a topological group.		Step (1)	[SHOW MORE] By the definition of topological group, the map $(a,b,c) \mapsto abc$ is continuous from $G \times G \times G$ to $G$. The map $g \mapsto (g,h,g^{-1})$ is also continuous because the inverse map is continuous. Composing these, we see that the map $g \mapsto ghg^{-1}$ is continuous viewed as a map from $G$ to $G$. The definition of subspace topology guarantees that a continuous map to the whole space $G$ that lands completely inside the subspace $H$ is also a continuous map to $H$ with the subspace topology.
3	The image of $G$ under $\pi_h$ is a connected subset of $H$ (viewed with the subspace topology).	$G$ is connected.	The image of a connected set under a continuous map is connected.	Step (2)	[SHOW MORE] Follows from step (2), the fact that $G$ is connected, and the fact that the image of a connected set under a continuous map is connected.
4	The image of $G$ under $\pi_h$ must be a one-point set.	$H$ is totally disconnected in the subspace topology.		Step (3)	[SHOW MORE] By step (3), the image of $G$ under $\pi_h$ is connected in $H$. Since $H$ is totally disconnected, the only possibility is a one-point subset.
5	The image of $G$ under $\pi_h$ is $\{ h \}$, so $ghg^{-1} = h$ for all $g \in G$, and hence $h$ is in the center of $G$.			Step (4)	[SHOW MORE] By step (4), we know that the image is a one-point subset. Thus, to find this point, it suffices to evaluate $\pi_h$ at any point in $G$. We choose the identity element, getting $\pi_h(e) = ehe^{-1} = h$. Thus, the one-point subset is $\{ h \}$. Hence, $\pi_h(g) = ghg^{-1}$ must equal $h$ for all $g \in G$.
6	$H$ is contained in the center of $G$			Steps (1), (5)	[SHOW MORE] The above was all for arbitrary $h \in H$, hence every $h \in H$ is in the center of $G$. Thus, $H$ is contained in the center.