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

Question

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.

I could be wrong , but is this not a matter of the Orbit-Stabilizer theorem (of Frobenius-Burnside formula) https://en.wikipedia.org/wiki/Burnside%27s_lemma ? — Nicky Hekster, Feb 19 '21 at 14:57
You didn't ask, but this is my least favorite proof of the Sylow theorems. The McKay proof is so much easier for me to comprehend. — Randall, Feb 19 '21 at 14:58
@Randall aren't you mixing up Sylow and Cauchy's Theorem on the existence of an element of order $p$? — Nicky Hekster, Feb 19 '21 at 14:59
Yes it's just the Orbit-Stabilizer Theorem. Herstein was obviously familiar with this, but at the time he wrote the book it had not been formulated as a specific result. — Derek Holt, Feb 19 '21 at 15:07
You also might to take a look at https://math.stackexchange.com/questions/479839/wielandts-proof-of-sylows-theorem since Herstein is following the proof of Helmut Wielandt (see https://mathshistory.st-andrews.ac.uk/Biographies/Wielandt/) of the Sylow Theorems. This is very different from the proof of Sylow himself, since it uses actions of groups on sets. Henceforth the need of the Orbit-Stabilizer Theorem. — Nicky Hekster, Feb 19 '21 at 15:31
What is $no(H)$? This is not standard notation so you should explain it — D_S, Feb 19 '21 at 15:43
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. — stoic-santiago, Feb 19 '21 at 15:44
I guess that the "[...] the argument used in the counting principle in Section 2.11 [...]" is the way to get the result without (explicitly) using group actions. — , Feb 19 '21 at 15:47
@user That "counting principle" is $$o(G) = \sum \frac{o(G)}{o(N(a))}$$ where this sum runs over one element $a$ from each conjugate class. $N(a)$ is the normalizer of $a\in G$. How does this relate to my problem? — stoic-santiago, Feb 19 '21 at 15:50
@strawberry-sunshine, I don't think the "your" $n$ ("$n=p^\alpha m$") is the same as excerpt's "$n$", though. — , Feb 19 '21 at 15:55
$H$ is the stabilizer of the action which is a subgroup of $G$, hence its order divides the order of $G$. — Vajra, Feb 19 '21 at 15:56
The text says, "Define $M_1 \sim M_2$ if there exists $g \in G$ such that $M_1 = M_2 g$". So the action of $G$ is right multiplication on the subsets of size $p^a$. — Ted, Feb 19 '21 at 16:42

score 2 · Accepted Answer · answered Feb 19 '21 at 16:34

2

$G$ 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, "$no(H)=o(G)$".

answered Feb 19 '21 at 16:34

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Thanks a lot! This resolves it. – stoic-santiago Feb 19 '21 at 16:44

score 2 · Answer 2 · answered Oct 12 '21 at 18:45

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).

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

2 Answers2