Intermediate Math Problem of the Week 10/17/2017

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  • Thread starter PF PotW Robot
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In summary, the problem states that if ##X## is a topological group and ##A## is a connected subgroup of ##X##, then ##X## is connected. However, this may not hold for all topological groups, as shown by the example of ##GL(n,\mathbb{R})## where the quotient group ##GL_n/GL_n^+## is not connected.
  • #1
PF PotW Robot
Here is this week's intermediate math problem of the week. We have several members who will check solutions, but we also welcome the community in general to step in. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Occasionally there will be prizes for extraordinary or clever methods. Spoiler tags are optional.

Let ##X## be a topological group; let ##A## be a subgroup of ##X## such that ##A## and ##X/A## are connected. Show that ##X## is connected.

(PotW thanks to our friends at http://www.mathhelpboards.com/)
 
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  • #2
PF PotW Robot said:
Let ##X## be a topological group; let ##A## be a subgroup of ##X## such that ##A## and ##X/A## are connected. Show that ##X## is connected.
Is this true? What about the general linear group ##GL(n,\mathbb{R})## (group of all n x n invertible matrices)? It consists of a subgroup ##GL^+(n,\mathbb{R})## of n x n matrices with positive determinant, and the complement ##GL(n,\mathbb{R})/GL^+(n,\mathbb{R})## of n x n matrices with negative determinant (where the slash denotes the set difference, not the quotient group or coset). Both subsets are connected but ##GL(n,\mathbb{R})## isn't.
 
  • #3
TeethWhitener said:
(where the slash denotes the set difference, not the quotient group or coset)
But the slash in ##X/A## is meant to be a quotient, and ##GL_n/GL_n^+ = \mathbb{Z}_2## which is not connected.
 
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  • #4
Ah ok. Thanks for the clarification.
 

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