Undergrad Proving that a subgroup is normal

[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 $G = g' A + g A$?

 

[fresh_42]

https://www.physicsforums.com/threa...ernating-group-is-normal.955181/#post-6055529

 

[WWGD]

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?

 

[fresh_42]

why is this in topology and analysis?

Thanks. Moved.

 

[kent davidge]

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.

 

[fresh_42]

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.

 

[fresh_42]

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.