# Proving that a subgroup is normal

• I
• kent davidge
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

#### kent davidge

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##?

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.

mathwonk
why is this in topology and analysis?

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

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.

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\,##?

kent davidge
Yes, and in which coset is the neutral element ##e \in G\,##?
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.

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.
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.

kent davidge