Need help with the follow lemma in Sylow theorems

Question

Lemma Let G be a finite group and let $P$ be a p-subgroup of $G$, i.e. a subgroup of $G$ which is a p-group. Then $$[N_G(P):P]\equiv [G:P] \mod p$$

Proof: Consider the action of $G$ on $X=G/P$ by left multiplication: $a*gP=agP$. We restrict this action to an action of $P$ on $G/P$. For this action we have $(G/P)^P=N_G(P)/P$. In fact, for $gP \in G/P$ one has

$gP \in (G/P)^P \leftrightarrow agP=gP$ for all $a \in P \leftrightarrow g^{-1}ag \in P \text{ How?}$ for all $a \in P$

$\leftrightarrow a \in gPg^{-1}$ for all $a \in P$ $\leftrightarrow P\leq gPg^{-1}$ How? $\leftrightarrow P=gPg^{-1}$ How?

$\leftrightarrow g \in N_G(P) \leftrightarrow gP \in N_G(P)/P$ HOW?

Now, Corollary 11.12 implies this result

Let $G$ be a p group and and let $X$ be a finite G set. Then

$$|X^G| \equiv |X| \text{ }\mod p$$

Basically I don't understand the whole proof. It would be nice if someone explains the details. There are a lot of things that are obvious to the author which I don't get.

I always have asked myself why $gHg^{-1} \in P$ implies $H \in g^{-1}Pg$

Why do we treat inclusion like equality because it looks like we can multiply on both sides like we do with an equality?

Answer 1

For example, if $agP=gP\Leftrightarrow\exists x,y\in P \ ;agx=gy \Leftrightarrow g^{-1}ag=yx^{-1}\in P$, i.e., $agP=gP\Leftrightarrow g^{-1}ag\in P$.

The other cases are similar.

Answer 2

For $gP \in G/P$ one has

$gP \in (G/P)^P \leftrightarrow agP=gP$ for all $a \in P $

That's the definition of a fixed point under the action of $P$.

$\leftrightarrow g^{-1}ag \in P \text{ How?}$ for all $a \in P$

Multiply both sides by $g^{-1}$.

$\leftrightarrow a \in gPg^{-1}$ for all $a \in P$ $\leftrightarrow > P\leq gPg^{-1}$ How?

Take the inverse automorphism of conjugation by $g$, which is conjugation by $g^{-1}$. This proves every element of $P$ is in $gPg^{-1}$, whence the inclusion.

$\leftrightarrow P=gPg^{-1}$ How?

$P$ and $gPg^{-1}$ have the same number of elements, since the latter is the isomorphic image of the former. As one contains the other, they're actually equal.

$\leftrightarrow g \in N_G(P) \leftrightarrow gP \in N_G(P)/P$ HOW?

The first equivalence corresponds to the definition of the normaliser of $P$ in $G$. The second equivalence corresponds to the definition of the quotient group $N_G(P)/P$.

I always have asked myself why $gHg^{-1} \in P$ implies $H \in > g^{-1}Pg$

Always the inverse automorphism. But you should use $\subset$, not $\in$.