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 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.
Related facts
- Central implies normal
- Normal not implies central
- Normal of least prime order implies central
- Cartan-Brauer-Hua theorem
Proof
This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format
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 ![]() ![]() ![]() |
![]() ![]() |
[SHOW MORE] | ||
2 | ![]() ![]() ![]() |
![]() |
Step (1) | [SHOW MORE] | |
3 | The image of ![]() ![]() ![]() |
![]() |
The image of a connected set under a continuous map is connected. | Step (2) | [SHOW MORE] |
4 | The image of ![]() ![]() |
![]() |
Step (3) | [SHOW MORE] | |
5 | The image of ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
Step (4) | [SHOW MORE] | ||
6 | ![]() ![]() |
Steps (1), (5) | [SHOW MORE] |