I'm reading I. N. Herstein's proof of Sylow's third theorem:
Theorem: The number of $p$-Sylow subgroups in $G$, for a given prime, is of the form $1+kp$.
Here is a picture of the proof, for reference.
All but one part of the proof makes sense to me. First, I'm aware that double cosets form an equivalence relation in a group, so for a subgroup $P<G$ we can write
$$
G=\bigsqcup_{g\in\mathcal I}PgP\quad\text{and}\quad|G|=\sum_{g\in\mathcal I}|PgP|,
$$
where $\mathcal I$ is a complete set of double coset representatives for $G$. Now, the proof splits the summation into two separate summations:
$$
|G|=\sum_{g\in\text N(P)}|PgP|+\sum_{g\not\in\text N(P)}|PgP|,
$$
where $\text N(P)$ denotes the normalizer of $P$ in $G$. This part is fine. In the first summation, which extends over $g\in\text N(P)$, we know that each term $PgP=P(Pg)=Pg$. So the first summation can be expressed as
$$
\sum_{g\in\text N(P)}|PgP|=\sum_{g\in\text N(P)}|Pg|.
$$
This part is also clear to me. But now the proof asserts that, since the summation extends over distinct cosets of $P$ in $\text N(P)$, then $\sum_{g\in\text N(P)}|Pg|=|\text N(P)|$. I don't understand two things:
- It's clear that each $Pg$ is a distinct coset, since the original summation was over distinct coset representatives. But why is it that $\sum_{g\in\text N(P)}|Pg|=|\text N(P)|$, necessarily? This might be a dumb question, but it is generally true that if $H<G$, then
$$
|G|=\sum_{g\in\mathcal I}|Hg|\quad\text{entails}\quad|H|=\sum_{g\in \mathcal I\cap H}|Hg|\,?
$$
- Why did we have to rewrite $PgP$ as $Pg$? If the answer to my first question is yes, and the double cosets $PgP$ partition $P$ anyway, could we not have written
$$
|\text N(P)|=\sum_{g\in\text N(P)}|PgP|\,?
$$
Thanks in advance.
Best Answer
You have already recalled that if $H$ is a subgroup of $G$, and we have that $g_1,\ldots,g_k$ is a complete set of coset representatives for $H$ in $G$, then $$G = \mathop{\amalg}\limits_{i=1}^n Hg,\qquad\text{and}\qquad |G|=\sum_{i=1}^n |Hg_i|.$$
So now, consider $P$ as a subgroup of $N(P)$. If $g_1,\ldots,g_n$ are the coset representatives for $P$ in $N(P)$, then we will certainly have that $\sum_{i=1}^n |Pg| = |N(P)|$.
So the only question is whether double coset representatives for $P$ in $N(P)$ coincide with coset representatives of $P$ in $N(P)$. You already noted that if $g\in N(P)$, then the double coset $PgP$ is equal to the right coset $Pg$. Conversely, if $h\in N(P)$, then $Ph = PPh = Phh^{-1}Ph = PhP$. Thus, every right coset of $P$ in $N(P)$ corresponds to one of the double cosets you have. That’s why you can assert the equality.
As to why he rewrites... in part because you are more used to cosets; here it is clear that they are the cosets of $P$ in $N(P)$, and so the sum is $|N(P)|$, as is usually the case for any subgroup and group.
But mostly he rewrote because he wanted to count how many elements there are in $PxP$; while every coset has the same size, the same does not hold for double cosets.
The double coset $HeH$ has cardinality $|H|$, but if $x\notin N(H)$, then in general $HxH$ has different cardinality. E.g., in $S_3$, the double coset $H(123)H$ of $H=\{e,(12)\}$ consists of $e(123)e=(123)$, $e(123)(12)= (13)$, $(12)(123)e = (23)$, and $(12)(123)(12) = (132)$, so $H(123)H$ has four elements. )
So he first calculated how many elements there are in $PxP$ when $x\notin N(P)$; and then he rewrites to easily calculate how many elements there are in $PxP$ when $x\in N(P)$: since in that case $PxP=Px$, then there are exactly $|P|$ elements.