Proof of Sylow's Theorem (Herstein) - why is $no(H) = o(G)$? The theorem is:

(Sylow's theorem): If $p$ is a prime number, and $p^\alpha |o(G)$, then $G$ has a subgroup of order $p^\alpha$.

Right before the proof, the author has established that if $n = p^\alpha m$ (where $p$ is prime), and if $p^r|m$ but $p^{r+1}\nmid m$, then $$p^r | {p^\alpha m\choose p^\alpha} \text{ but } p^{r+1} \nmid {p^\alpha m\choose p^\alpha}$$
This notation is used in the proof too.

I'm posting an image here, and the highlighted part is what bothers me. I'm sorry I didn't type it all up in MathJax, but as you can imagine it is a lot to type!



Question: How do I prove that $no(H) = o(G)$?
It's probably some class equation or such but I am not very comfortable with those yet and I would appreciate any help! Thanks a lot.
Clarification: $no(H)$ means $n$ multiplied with $o(H)$, the order of $H$.
Update (based on comments):Herstein's proof never explicitly introduced group actions, and therefore I'm having trouble connecting it to the orbit-stabilizer theorem. Could someone please spell it out? In terms of group actions, what does the equivalence relation $\sim$ (and equivalence classes thereof) in Herstein's proof correspond to? What does $H$ mean? $H$ looks very similar to what a stabilizer of an element is defined as, but it's definitely not the same concept.
 A: $G$ acts (from right) on $\mathcal{M}$ by right multiplication (in fact, $|Mg|=|M|$ for every $M\in\mathcal{M}$ and $g\in G$). $H$ is $\operatorname{Stab}(M_1)$ and $\{M_1,\dots,M_n\}$ is $\operatorname{Orb}(M_1)$. By the orbit-stabilizer theorem, then, $|\operatorname{Orb}(M_1)||\operatorname{Stab}(M_1)|=|G|$, i.e, in book's notation, "$no(H)=o(G)$".
A: How do I prove that $no(H) = o(G)$?
Here this fact is proved by showing that there is a bijection from $\{Ha;\;a\in G\}$ to $\{M_1,...,M_n\}$.
Details:
Given a set $A$, let us write $|A|=$ cardinality of $A$.
Let $C=\{Ha;\; a\in G\}$ be the set of right cosets of $H$. Then,
$$|C|=\frac{o(G)|}{o(H)}.\tag{1}$$
Note that
$$Ha=Hb\quad\Leftrightarrow\quad ab^{-1}\in H\quad\Leftrightarrow\quad M_1ab^{-1}=M_1\quad\Leftrightarrow\quad M_1a=M_1b.$$
Therefore, the function $f:C\to \mathscr{M}$ given by $f(Ha)=M_1a$ is well-defined and injective. Since $M_1aa^{-1}=M_1$, we have $M_1a\sim M_1$ and thus $f(C)\subset \{M_1,...,M_n\}$. On the other hand, each $M_i$ has the form $M_1a_i$ because $M_i\sim M_1$. So, $M_i=f(Ha_i)$ and thus $\{M_1,...,M_n\}\subset f(C)$. This shows that $f(C)=\{M_1,...,M_n\}$. Since $f$ is one-to-one, we conclude that
$$|C|=|\{M_1,...,M_n\}|.\tag{2}$$
The desired result follows from (1) and (2).
