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以下是UCLA的一次关于Sylow Theorems的 assignment.更多的经典案例请参阅以往案例，关于abstract algebra的更多的以往案例可以参阅相关文章。abstract algebra代写请认准UpriviateTA.
(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 $p^{r}$.
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 $m$.
(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 $N_{G}(H)=H$.
(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.
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