$\newcommand{\fix}{\text{Fix}}$
$\newcommand{\orb}{\text{Orb}}$
$\newcommand{\stab}{\text{Stab}}$
I'm having an issue concluding the proof of Sylow's 1st theorem in the case where we assume $p \nmid |Z(G)|$.
Statement: Let $G$ be a finite group and $p$ a prime dividing $|G|$. Then $G$ has a Sylow p-subgroup.
To do the proof we induct on $|G|$ and consider two cases where $p||Z(G)|$ and $p \nmid |Z(G)|$, I'm having a modular arithmetic issue with the latter.
Proof: Next we consider the case where $p \nmid |Z(G)|$. Let $G$ act on itself by conjugation; i.e $g \ast a = gag^{-1}$ for all $a \in G$. Because $G$ acts on itself the orbits under this action form a partition of $G$ and moreover we know that $|\fix(G)| = |Z(G)|$ under this action. But now since $G$ is partitioned into it's orbits we know that
$$
|G| = \sum_{i=1}^N |\orb(x_i)|
$$
where $x_1,\dots,x_N$ are elements of the orbits partitioning $G$. But more than that we that $|\fix(G)|$ is the number of elements of $X$ with trivial orbits. So the order of $G$ actually becomes
$$
|G| = |\fix(G)| + \sum_{i=1}^k |\orb(x_k)|,
$$
where we have removed the orbits containing only a single element. Using the fact that the fixed set of $G$ is the centre we have that
$$
|G| = |Z(G)| + \sum_{i=1}^k |\orb(x_i)|,
$$
but recall we assumed from the start that $p||G|$ and so this means that $|G| \equiv 0 (\bmod p)$. Because we assumed that $p$ did not divide the order of the centre we have that $|Z(G| \not\equiv 0 (\bmod p)$ but then for this to make sense we must have that the sum of the order of the orbits containing more than one element must be not equivalent to $0$ modulo $p$. But by the Orbit-Stabilizer theorem we know that $|\orb(x_i)| = |G|/|\stab(x_i)|$, which implies that
$$
\sum_{i=1}^k \frac{|G|}{|\stab(x_i)|} \not\equiv 0 (\bmod p).
$$
So for some $1 \leq i \leq k$ we are guaranteed that we have $|G|/|\stab(x_i)| \not\equiv 0 (\bmod p)$. But this means that every power of $p$ that appears in $|G|$ must also appear in the order of the stabilizer.
The next line then states that $p^r||\stab(x_i)|$ and hence $G$ has a $p$-Sylow subgroup of order $p^r$ finishing the induction. I have two issues.
- Where do we get that $p^r$ divides the order of the stabilizer? I assume it's evident, but I can only see that if the ratio is not $0$ modulo p then $|G|/|\stab(x_i)| = k+np$ for some $1 \leq k < p$ and $n \in \mathbb{Z}$ but this doesn't seem to help any. Is it because as a subgroup the order of the stabilizer divides the order of $G$? But then what does that give us? Is it that somehow that $p^r||G|$ and $|G|/|\stab(x_i)| \not\equiv 0 (\bmod p)$ means we must have $p^r||\stab(x_i)|$?
- How do we know for sure that $\stab(x_i)$ is in fact a proper subgroup of $G$? Because if it was secretly all of $G$ we wouldn't be able to use the inductive hypothesis and conclude it contained a group of order $p^r$ right?
Thanks in advance for the help.
Best Answer
For the first, think about the prime decomposition of $|G|$, every $p$ appearing in the decomposition must be cancelled with a $p$ from $\text{Stab}(x_i)$, or else you have $\frac{|G|}{\text{Stab}(x_i)}=p \cdot(\text{some other integer})$.
For the second the stabiliser cannot be the whole group, as this implies $\text{Orb}(x_i)$ has only one element, and we have already dealt with such orbits.