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
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Suppose is a Connected topological group (?) and is a normal subgroup of , such that is a totally disconnected space in the subspace topology (i.e., the connected components of are one-point subsets). Then, is a central subgroup of , i.e., is contained in the Center (?) of .
A case of particular interest is where is a discrete subgroup of , i.e., a closed subgroup of 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.
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Given: A connected topological group , a normal subgroup that is totally disconnected in the subspace topology.
To prove: is contained in the center of .
Proof: The idea is to look at a fixed but arbitrary element in and a continuously varying element of 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 , consider the map given by . This is a well-defined map.||is normal in||[SHOW MORE]|
|2||is a continuous map from as a topological space to endowed with the subspace topology.||is a topological group.||Step (1)||[SHOW MORE]|
|3||The image of under is a connected subset of (viewed with the subspace topology).||is connected.||The image of a connected set under a continuous map is connected.||Step (2)||[SHOW MORE]|
|4||The image of under must be a one-point set.||is totally disconnected in the subspace topology.||Step (3)||[SHOW MORE]|
|5||The image of under is , so for all , and hence is in the center of .||Step (4)||[SHOW MORE]|
|6||is contained in the center of||Steps (1), (5)||[SHOW MORE]|