Size of quotient groups

Question

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 $\dfrac{n_G}{n_\mathcal{H}}$?

Answer 1

While it is true that $G=\{g_1,\ldots, g_{n_G-n_H}\}\cup H$, many of the elements $g_i$ give the same coset. In fact, you should prove that $g_iH=g_jH$ if and only if $g_j^{-1}g_i\in H$. Here's an outline for the proof of the fact that$|G/H|=n_G/n_H$:

First, one shows that all the cosets $gH$ have the same size. In particular, they all have size $|1\cdot H|=|H|=n_H$. Now show that $G$ is the disjoint union of the distinct cosets, so you get $$n_G=|G|=\left|\bigcup g_iH\right|.$$ But there are exactly $|G/H|$ cosets, so the right hand side is $|G/H|\cdot n_H$. Thus $n_G=|G/H|\cdot n_H$ which is what you're looking for.

Answer 2

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.

Answer 3

Two distinct cosets of $H$ in $G$ are disjoint. That means that the size of $G$ is the sum of the sizes of each coset of $H$ in $G$. But all cosets have the same size: that of $H$. To see this just consider the map $gh\mapsto g'h$ form a coset $gH$ to a coset $g'H$ and prove that it is a bijection.