Abstract algebra不算是一门简单的学科,这门学科在国内叫做抽象代数,经常有很多学生在学linear algebra或者analysis(advance calculus)的时候觉得并不困难,但是却觉得Abstract algebra很难,这是因为没有找到正确的方法学习Abstract algebra,UpriviateTA有一系列非常擅长Abstract algebra的老师,可以确保您在Abstract algebra取得满意的成绩。

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.

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.
abstract algebra代写请认准UpriviateTA

abstract algebra代写请认准UpriviateTA. UpriviateTA为您的留学生涯保驾护航。