GroupActionsAndSylowTheorem

Notation. Let $G$ be a group acting on a set $X$.

• For $x\in X$, $Gx=\{gx\mid g\in G\}$ and $G_x=\{g\in G\mid gx=x\}$.
• $X^G=\{x\in X\mid gx=x$ for all $g\in G\}$.

Facts.

1. For $x\in X$, $|Gx|=[G:G_x]$.
2. If $X$ is finite, then $|X|=|X^G|+\sum |Gx_i|$.
3. If $G$ is a $p$-group, then $p\mid [G:Gx_i]$ for all $i$. Hence $|X|\equiv |X^G|\pmod p$.
4. Let $G$ be a finite group acting on $X=G$ by conjugations. Then $X^G=Z(G)$, the center of $G$. The class equation says $$|G|=|Z(G)|+\sum |\hbox{conjugacy classes with size} > 1|.$$
5. If $G$ is a finite $p$-group, then $|Z(G)|\equiv 0\pmod p$, and so $|Z(G)|\geq p$.

Sylow Subgroups. Let $G$ be a finite group, $p$ a prime, and $n\geq 0$, such that $p^n\,|\,|G|$ but $p^{n+1}\,\not|\, |G|$. A subgroup of $G$ is a Sylow $p$-subgroup if is has order $p^n$.

Theorem.

1. There is a Sylow $p$-subgroup in $G$.
2. Let $P$ and $Q$ be Sylow $p$-subgroups of $G$. Then there is a $g\in G$ such that $Q=P^g=gPg^{-1}$.
3. Let $m=|\{P\leq G\mid P, |P|=p^n\}$. Then $m\mid |G|$ and $m\equiv 1\pmod p$.

Proof.

1. By induction on $|G|$. If $|G|=1$, then the result holds trivially. Assume that $|G|>1$ and that the result has been proved for any group with smaller orders. There are two cases. (1) There is some proper subgroup $H$ of $G$ with $p$ not dividing ${[G:H]}$. Then, by induction, $H$ has a Sylow $p$-subgroup of order $p^n$, which is also a Sylow $p$-subgroup of $G$. (2) For any proper subgroup $H$ of $G$, $p$ divides ${[G:H]}$. In this case, we have $|G|\equiv |Z(G)|\pmod p$. Since $|G|\equiv 0\pmod p$, we have $Z(G)$ is not trivial. Let $a$ be a non-identity element in $Z(G)$ with order $p$, and let $A$ be the subgroup generated by $a$. Then $G/A$ is of order less than $|G|$, and $p^{n-1}$, but not $p^n$, divides $G/A$. Thus, there is a subgroup $\bar B$ of $G/A$ with order $p^{n-1}$. The preimage of $\bar B$ in $G$ is then a desired subgroup of order $p^n$.
2. Let $X=\{gP\mid g\in G$, and let $Q$ act on $X$ by $h\cdot (gP)=hgP$ for all $h\in Q$ and $g\in G$. Since $|X|=[G:P]$ is not divisible by $p$. Thus $|X^Q|\equiv |X|\not\equiv 0\pmod p$, and so $X^Q\not=\varnothing$. Let $gP\in X^Q$. Then $a\cdot(gP)=gP$ for all $a\in Q$. This gives $g^{-1}ag\in P$ for all $a\in Q$. Therefore, $Q=gPg^{-1}$.
3. From the above, we immediately see that $G$ has a unique Sylow $p$-subgroup if and only if every Sylow $p$-subgroup is normal. Consequently, if a Sylow $p$-subgroup $Q$ is in the normalizer of $P$, then $Q=P$ since they are both normal Sylow $p$-subgroups of the group $N(P)$.
   Now, let $X=\{P\mid P$ is a Sylow $p$-subgroup of $G\}$. Let $Q$ acts on $X$ by conjugation. Then $|X|\equiv |X^Q|\pmod p$. But $X^Q=\{P\in X\mid P^q=P$ for all $q\in Q\}$, which is $\{P\in X\mid Q\leq N(P)\}=\{P\in X\mid Q=P\}$. Therefore, $|X^Q|=1$, and so $|X|\equiv1\pmod p$.
   Now, let $G$ act on $X$ by conjugation. Then $X$ is a single orbit of $Q$. Then $|X|=|G\cdot Q|=[G:G_Q]$ which divides $|G|$.