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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##?

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- Thread starter kent davidge
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- #1

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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##?

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WWGD

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How do you define index? I thought it was the number of cosets.

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##?

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mathwonk

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why is this in topology and analysis?

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fresh_42

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Thanks. Moved.why is this in topology and analysis?

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According to my book, it is indeed the number of cosets which defines the index.How do you define index? I thought it was the number of cosets.

- #7

fresh_42

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Yes, and in which coset is the neutral element ##e \in G\,##?According to my book, it is indeed the number of cosets which defines the index.

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ohh you made me realise now what the thing is.Yes, and in which coset is the neutral element ##e \in G\,##?

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

fresh_42

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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.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.

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