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Section 1.21 Sylow theorems: conclusion of proof
Subsection 1.21.1 Second action: \(H\) on \(G/K\)
We now move on to a proof of the the
Sylow 2 . That proof makes use of the left multiplication action of
\(G\) on the coset space
\(G/K\text{,}\) where
\(K\in \Syl_p(G)\text{.}\) The proof will make use of the following proposition, which is interesting in its own right.
Proposition 1.21.1 . Fix point of \(p\) -group action.
Let
\(G\) be a
\(p\) -group. If
\(G\) acts on a finite set
\(A\) satisfying
\(p\nmid \abs{A}\text{,}\) then there is an element
\(a\in A\) with
\(G_a=G\text{:}\) i.e.,
\(A\) contains an element that is fixed by all elements of
\(G\text{.}\)
Proof.
Decompose \(A\) into its distict orbits \(O_{a_1}, O_{a_2}, \dots, O_{a_r}\text{.}\) Since \(\abs{O_{a_i}}=[G\colon G_{a_i}]\text{,}\) we see that either \(\abs{O_{a_i}}\) is divisible by \(p\text{,}\) or \(\abs{O_{a_i}}=1\text{,}\) in which case \(G_{a_i}=G\text{.}\) But we cannot have \(p\mid \abs{O_{a_i}}\) for all \(i\text{,}\) since then we’d have
\begin{align*}
\abs{A} \amp =\sum_{i=1}^r\abs{O_{a_i}}\\
\amp \equiv 0 \pmod p\text{,}
\end{align*}
contradicting the fact that \(p\nmid \abs{A}\text{.}\) Thus we have \(G_{a_i}=G\) for some \(i\text{,}\) as desired.
Proof of Sylow 2. Let
\(p\) be a prime divisor of
\(\abs{G}\text{,}\) let
\(K\in Syl_p(G)\text{,}\) let
\(H\) be a
\(p\) -subgroup of
\(G\text{,}\) and consider the action of
\(H\) on
\(G/K\text{.}\) Since
\(\abs{G/K}=m\text{,}\) where
\(p\nmid m\text{,}\) it follows that one of the cosets
\(gK\) of
\(G/K\) is fixed under left multiplication by
\(H\text{:}\) i.e.,
\(h\cdot gK=gK\) for all
\(h\in H\text{.}\) But by
Theorem 1.17.3 it follows that
\(H\leq G_{gK}=gKg^{-1}\text{,}\) as desired.
Subsection 1.21.2 Third action: \(G\) on \(\Syl_p(G)\)
