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Fix: Typos and changes to the Sylow proof
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jfab20 committed Jan 9, 2025
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19 changes: 9 additions & 10 deletions tex/H113/sylow.tex
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Expand Up @@ -142,7 +142,7 @@ \subsection{Step 1: Prove that a Sylow $p$-subgroup exists}

Here is the proof.
\begin{itemize}
\ii Let $G$ act on $X$ by $g \cdot X \defeq \left\{ gx \mid x \in X \right\}$.
\ii Let $G$ act on $X$ by $g \cdot S \defeq \left\{ gs \mid s \in S \right\}$.
\ii Take an orbit $\OO$ with size not divisible by $p$.
(This is possible because of our deep number theoretic fact.
Since $\left\lvert X \right\rvert$ is nonzero mod $p$ and the orbits partition $X$,
Expand All @@ -159,10 +159,10 @@ \subsection{Step 1: Prove that a Sylow $p$-subgroup exists}


\subsection{Step 2: Any two Sylow $p$-subgroups are conjugate}
Let $P$ be a Sylow $p$-subgroup (which exists by the previous step).
We now prove that for any $p$-group $Q$, $Q \subseteq gPg\inv$.
Note that if $Q$ is also a Sylow $p$-subgroup, then $Q = gPg\inv$ for size reasons;
this implies that any two Sylow subgroups are indeed conjugate.
If $P$ is a Sylow $p$-subgroup and $Q$ is a $p$-group, we prove $Q
\subseteq gPg\inv$. Note that if $Q$ is also a Sylow $p$-subgroup, then
$Q = gPg\inv$ for size reasons; this implies that any two Sylow
subgroups are indeed conjugate.

\textbf{Let $Q$ act on the set of left cosets of $P$ by left multiplication.}
Note that
Expand All @@ -176,15 +176,14 @@ \subsection{Step 2: Any two Sylow $p$-subgroups are conjugate}


\subsection{Step 3: Showing $n_p \equiv 1 \pmod p$}
Let $\mathcal S$ denote the set of all the Sylow $p$-subgroups.
Let $P \in \mathcal S$ be arbitrary.
Let $\mathcal S$ denote the set of all the Sylow $p$-subgroups. By our
first step, there exists some $P \in \mathcal S$.
\begin{ques}
Why does $\left\lvert \mathcal S \right\rvert$ equal $n_p$?
(In other words, are you awake?)
\end{ques}
Now we can proceed with the proof.
Let $P$ act on $\mathcal S$ by conjugation.
Then:
Now we can proceed with the proof. Let $P$ act on $\mathcal S$ by
conjugation. Then:
\begin{itemize}
\ii Because $P$ is a $p$-group, $n_p \pmod p$ is the number of fixed points
of this action.
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