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Problem 1 (Sylow Theorem). Let $G$ be a finite group of order $p^{r} m,$ where $p$ is prime, $r$ is a positive integer, and $p$ does not divide $m$. Then
(1) $G$ has at least one Sylow $p$ -subgroup, and every $p$ -subgroup of $G$ is contained in a Sylow $p$ -subgroup.
(2) Let $n_{p}$ be the number of Sylow $p$ -subgroups of $G .$ Then $n_{p} \equiv 1 \bmod p$ and $n_{p}$ divides $m$
(3) All Sylow $p$ -subgroups are conjugate. Thus if we define an equivalence relation on subgroups by $H \sim K$ iff $H=g K g^{-1}$ for some $g \in G,$ then the Sylow $p$ -subgroups comprise a single equivalence class. [Note that the conjugate of a Sylow $p$ -subgroup is also a Sylow p-subgroup, since it has the same number of elements $p^{r}$.
Proof . (1) Let $G$ act on subsets of $G$ of size $p^{r}$ by left multiplication. The number of such subsets is $\left(\begin{array}{c}p^{r} m \\ p^{r}\end{array}\right),$ which is not divisible by $p$  Consequently, since orbits partition the set acted on by the group, there is at least one subset $S$ whose orbit size is not divisible by $p$. If $P$ is the stabilizer of $S$, then the size of the orbit is $[G: P]=|G| /|P|=p^{r} m /|P| .$ For this to fail to be divisible by $p,$ we must have $p^{r}|| P \mid,$ and therefore $p^{r} \leq|P| .$ But for any fixed $x \in S,$ the map of $P$ into $S$ given by $g \rightarrow g x$ is injective. (The map is indeed into $S$ because $g$ belongs to the stabilizer of $S$, so that $g S=$ S.) Thus $|P| \leq|S|=p^{r}$. We conclude that $|P|=p^{r}$, hence $P$ is a Sylow $p$ -subgroup.

So far, we have shown that a Sylow $p$ -subgroup $P$ exists, but not that every $p$ -subgroup is contained in a Sylow $p$ -subgroup. We will return to this in the course of proving (2) and (3).

(2) and (3) Let $X$ be the set of all Sylow p-subgroups of $G$. Then $|X|=n_{p}$ and $P$ acts on $X$ by conjugation, i.e., $g \bullet Q=g Q g^{-1}, g \in P .$  the size of any orbit divides $|P|=p^{r}$, hence is a power of $p$. Suppose that there is an orbit of size 1 , that is, a Sylow $p$ -subgroup $Q \in X$ such that $g Q g^{-1}=Q,$ and therefore $g Q=Q g,$ for every $g \in P .$ (There is at least one such subgroup, namely P.) Then $P Q=Q P,$ so $P Q=\langle P, Q,\rangle,$ the subgroup generated by $P$ and $Q .$ Since $|P|=|Q|=p^{r}$ it follows from (5.2.4) that $|P Q|$ is a power of $p,$ say $p^{c}$. We must have $c \leq r$ because $P Q$ is a subgroup of $G$ (hence $|P Q|$ divides $|G|$ ). Thus
$$p^{r}=|P| \leq|P Q| \leq p^{r}, \text { so }|P|=|P Q|=p^{r}$$
But $P$ is a subset of $P Q,$ and since all sets are finite, we conclude that $P=P Q,$ and therefore $Q \subseteq P$. Since both $P$ and $Q$ are of size $p^{r}$, we have $P=Q$. Thus there is only one orbit of size $1,$ namely $\{P\} .$ Since $all other orbit sizes are of the form$p^{c}$where$c \geq 1,$it follows that$n_{p} \equiv 1 \bmod p$. Now let$R$be a$p$-subgroup of$G,$and let$R$act by multiplication on$Y,$the set of left cosets of$P$. Since$|Y|=[G: P]=|G| /|P|=p^{r} m / p^{r}=m, p$does not divide$|Y|$Therefore some orbit size is not divisible by$p$. , every orbit size divides$|R|$hence is a power of$p$. . We are not going around in circles only depend on the existence of Sylow subgroups, which we have already established.) Thus there must be an orbit of size$1,$say$\{g P\}$with$g \in G .$If$h \in R$then$h g P=g P,$that is,$g^{-1} h g \in P,$or equally well,$h \in g P g^{-1}$. Consequently,$R$is contained in a conjugate of$P .$If$R$is a Sylow$p$-subgroup to begin with, then$R$is a conjugate of$P$, completing the proof of (1) and (3). To finish (2), we must show that$n_{p}$divides$m$. Let$G$act on subgroups by conjugation. The orbit of$P$has size$n_{p}$by$(3),$so$n_{p}$divides$|G|=p^{r} m .$But$p$cannot be a prime factor of$n_{p},$since$n_{p} \equiv 1 \bmod p .$It follows that$n_{p}$must divide$m$. Problem 2. Under the hypothesis of the Sylow theorems, show that$G$has a subgroup of index$n_{p}$Proof . Let$G$act on subgroups by conjugation. If$P$is a Sylow$p$-subgroup, then the stabilizer of$P$is$N_{G}(P)$, so the index of$N_{G}(P)$is$n_{p}$. Problem 3. Let$P$be a Sylow$p$-subgroup of the finite group$G,$and let$H$be a subgroup of$G$that contains the normalizer$N_{G}(P)$. (a) If$g \in N_{G}(H),$show that$P$and$g P g^{-1}$are Sylow$p$-subgroups of$H,$hence they are conjugate in$H$(b) Show that$N_{G}(H)=H$. Proof . (a) By definition of normalizer, we have$g P g^{-1} \leq g N_{G}(P) g^{-1} \leq g H g^{-1}=H$. Thus$P$and$g P g^{-1}$are subgroups of$H,$and since they are$p$-subgroups of maximum possible size, they are Sylow$p$-subgroups of$H$. (b) Since$H$is always a subgroup of its normalizer, let$g \in N_{G}(H)$. By (a),$P$and$g P g^{-1}$are conjugate in$H,$so for some$h \in H$we have$g P g^{-1}=h P h^{-1}$. Thus$\left(h^{-1} g\right) P\left(h^{-1} g\right)^{-1}=P,$so$h^{-1} g \in N_{G}(P) \leq H .$But then$g \in H,$and the result follows. Problem 4. The Sylow theorems are about subgroups whose order is a power of a prime$p$. Here is a result about subgroups of index$p$. Let$H$be a subgroup of the finite group$G,$and assume that$[G: H]=p .$Let$N$be a normal subgroup of$G$such that$N \leq H$and$[G: N]$divides$p !$. Show that$[H: N]$divides$(p-1) !$Proof .$[G: N]=[G: H][H: N]=p[H: N],$and since$[G: N]$divides$p !=p(p-1) !\$, the result follows.