This question will be really easy to answer I think, I know that there are other questions on roughly the same topic but honestly I could not find what i was looking for in an answer.
I am a physics student trying to learn group theory, I am studying on Pierre Ramond's book "Group theory – a physicist's survey".
I am just really confused about a pretty elementary fact.
When talking about finite size groups the book introduces the notion of quotient groups and explains their construction. Given a group $G$ and a normal subgroup $\mathcal{H}$ then the cosets $g_i\mathcal{H}$ can be given a group structure using the normality properties of $\mathcal{H}$, .
The book also says that the size of this group of cosets $G/\mathcal{H}$ is $\dfrac{n_G}{n_\mathcal{H}}$ with $n_G$ being the size of $G$ and $n_\mathcal{H}$ being the size of the subgroup $\mathcal{H}$ the book does not give any proof of this statement about the size of the quotient group treating it as a trivial result but the counting just doesn't add up to me.
I mean, in $G$ we have $n_G-n_\mathcal{H}$ elements that are not in $\mathcal{H}$, in fact we could write sumeting like $G = \{g_1, g_2, g_3, …, g_{n_G-n_\mathcal{H}}\} \bigcup \mathcal{H}$. This would imply that the cosets I can build are:
$$
G/\mathcal{H} = \{\mathcal{H}, g_1 \mathcal{H}, g_2 \mathcal{H},…, g_{n_G-n_\mathcal{H}}\mathcal{H}\}
$$
Giving it a size of $n_G – n_\mathcal{H} + 1$.
I know i must be at fault on this thing since the fact that the size of the quotient group is $\dfrac{n_G}{n_\mathcal{H}}$ is stated pretty much everywhere in group theory books but i just can't seem to find where I am wrong about this.
Can you give me a simple proof that the size of $G/\mathcal{H}$ is $\dfrac{n_G}{n_\mathcal{H}}$?
Best Answer
What you're missing is that $gH$ and $kH$ can be the same coset, even if $g$ and $k$ are different elements.
We define an equivalence relation as follows: $g \sim k$ if and only if $g = hk$ for some $h \in H$. It's easy to check that this is indeed an equivalence relation, and that each equivalence class has size $|H|$. You should also try proving that $gH = kH$ if and only if $g \sim k$; this gives you the conclusion.