# I 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 $G = g' A + g A$?

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#### WWGD

Gold Member
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

Homework Helper
why is this in topology and analysis?

#### fresh_42

Mentor
2018 Award
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

Mentor
2018 Award
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 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.

#### fresh_42

Mentor
2018 Award
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.

"Proving that a subgroup is normal"

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