- #1

- 933

- 56

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##?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

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.f

- #1

- 933

- 56

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##?

- #2

- 18,482

- 21,232

- #3

Science Advisor

Gold Member

- 6,481

- 9,260

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##?

- #4

Science Advisor

Homework Helper

- 11,523

- 1,773

why is this in topology and analysis?

- #5

- 18,482

- 21,232

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

- #6

- 933

- 56

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

- 18,482

- 21,232

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.

- #8

- 933

- 56

ohh you made me realize 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

- 18,482

- 21,232

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 realize 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.

Share:

- Replies
- 11

- Views
- 515

- Replies
- 3

- Views
- 630

- Replies
- 0

- Views
- 585

- Replies
- 4

- Views
- 347

- Replies
- 14

- Views
- 2K

- Replies
- 2

- Views
- 985

- Replies
- 27

- Views
- 9K

- Replies
- 1

- Views
- 981

- Replies
- 12

- Views
- 3K

- Replies
- 5

- Views
- 1K