Proving that a subgroup is normal

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  • #3
WWGD
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In this PDF, http://www.math.unl.edu/~bharbourne1/M417Spr04/M417Exam2Solns.pdf,

in answering why a subgroup of index 2 is normal, the author says that the only two cosets must be ##A## and ##gA##. Why so? Why there can't be another element ##g'## such that ##G = g' A + g A##?
How do you define index? I thought it was the number of cosets.
 
  • #4
mathwonk
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why is this in topology and analysis?
 
  • #6
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How do you define index? I thought it was the number of cosets.
According to my book, it is indeed the number of cosets which defines the index.
 
  • #7
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According to my book, it is indeed the number of cosets which defines the index.
Yes, and in which coset is the neutral element ##e \in G\,##?
 
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  • #8
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Yes, and in which coset is the neutral element ##e \in G\,##?
ohh you made me realise now what the thing is.

One coset must be ##A## because the neutral ##e## should be in ##G##, and it is in ##A##, because ##A## is a subgroup, (not in ##gA##, because ##gA = (g e, ga_1, ...) = (g, ga_1, ...)##). So one coset being ##A##, the only other possible coset is ##gA## so that the index 2 condition is satisfied.
 
  • #9
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ohh you made me realise now what the thing is.

One coset must be ##A## because the neutral ##e## should be in ##G##, and it is in ##A##, because ##A## is a subgroup, (not in ##gA##, because ##gA = (g e, ga_1, ...) = (g, ga_1, ...)##). So one coset being ##A##, the only other possible coset is ##gA## so that the index 2 condition is satisfied.
Yes, and therefore we have ##G=A \cup gA=A \cup Ag## at the same time and on the level of sets. But this can only be if ##gA=Ag## which means ##A## is normal.
 
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