this is not a duplicate .. I'm trying to understand a specific answer of this question. the parts that I don't understand are marked in (1) , (2), (3) and explanations for what i don't understand are in the bottom. thanks
P is a p group and H is a proper subgroup of P . Prove that $N_p(H) \ne H$
I saw many answers but I'm trying to understand a specific one and having difficulties:
my answer is as follows -
Im trying to find a specific $p \in N_p(H)$ such that $p \notin H$
I look at the group $X = \{pH |p \in P \}$
and the operation of ?(1)? on X
$ h \in H$
$ \pi (pH) = hpH$
from the orbit stabiliser i get
$|X| = \sum_I [P:P_{x_i}] + |x_0|$ ?(2)?
for each i,
$p|[P:P_{x_i}]$ because it is an index of a p group ?(3)?
from here i assume p divides $x_0$ and i show an element in $x_0$ which
is in $N_p(H)$ and not not in H.
What I don't understand in this solution is -
(1) who operates on X? is it P or H?
(2) I assumed in (1) that it is H operating on X because they mention $h \in H$, than how come
they use P in the orbit stabiliser theorem?
(3) please help me understand why $p | [P:P_{x_i}]$
any help will be very appreciated ,
thanks
First, this theorem holds (and the proof is identical) with the more general hypothesis:
[Clearly if $G$ is a $p$-group then any subgroup is a $p$-subgroup and the fact that $H$ is proper implies that $p | [G : H]$.]
Let $X = \{gH\, .\, g \in G\}$ be the set of left cosets of $H$, then we consider the action
$$ \begin{align*} H \times X &\to X\\ (h, gH) &\mapsto (hg)H \end{align*} $$
[The answers to (1) is therefore that it is $H$ acting on the set of left cosets by left multiplication.]
Now, by orbit-stabilizer theorem the order of any orbit is a divisor of $H$, but since $H$ is a $p$-subgroup it must be power of $p$. Also, it is known that $X$ is the disjoint union of its orbits, so
$$p | [G : H] = |X| = \sum_{\substack{O_x \in X_G \\ p | |O_x|}} |O_x| + \sum_{\substack{O_x \in X_G \\ |O_x| = 1}} 1$$
What I'm doing is to separate the sum between the orbits whose order is divisible by $p$ and these of order $p^0 = 1$.
Since all the terms of the first summatory are divisible by $p$ it follows that the second summatory, which is essentialy the number of trivial orbits, is divisible by $p$. Since $H$ has a trivial orbit it follows that there are at least other $p-1$ trivial orbits. Let $\{gH\}$ be one of these, being the only element in the orbit means $H g H = g H$, then
$$g^{-1} H g \subseteq g^{-1} H g H = g^{-1} g H = H$$
this proves that $g \in N_G(H)$, where $g \notin H$.
You can also infer that $gH$ is in $N_G(H)$ by remembering that $N_G(H)$ is a group and therefore closed under product.
To conclude, I don't really understand after $\pi (p H) = \ldots$ but here are my guess-answer to (2) and (3):
$P$ (or $G$ in my setting) arises from the orbit stabilizer theorem: $O_x \cong G / G_x$, so $|O_x| = [G : G_x]$, I avoided this passage though.
$p | [G:G_x]$ because $G_x$ is a subgroup of $H$, therefore $[G:G_x] = [G:H]\, [H:G_x]$ and we know that $[G:H]$ is divisible by $p$.
As a bonus, if you continue this proof just a bit you get the first Sylow theorem, as it has been suggested here, upper part of page 3.
P.S. with $X_G$ I mean the orbit space, the quotient by the usual equivalence relation given by the action, the place where the orbits live.