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# 数学代写|抽象代数代写Abstract Algebra代考|MATH4200 Applications of Sylow Theorems

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## 数学代写|抽象代数代写Abstract Algebra代考|Applications of Sylow Theorems

Theorem $21.1$ gives us a complete description of all finite fields. The following theorem gives us a complete description of all the subfields of a finite field. Notice the close analogy between this theorem and Theorem 4.3, which describes all the subgroups of a finite cyclic group.
Theorem 21.11 Subfields of a Finite Field
For each divisor $m$ of $n, \operatorname{GF}\left(p^n\right)$ has a unique subfield of order $p^m$. Moreover, these are the only subfields of $\mathrm{GF}\left(p^n\right)$.

PROOF To show the existence portion of the theorem, suppose that $m$ divides $n$. Then, since
$$p^n-1=\left(p^m-1\right)\left(p^{n-m}+p^{n-2 m}+\cdots+p^m+1\right),$$
we see that $p^m-1$ divides $p^n-1$. For simplicity, write $p^n-1=\left(p^m-1\right) t$. Let $K=\left{x \in \operatorname{GF}\left(p^n\right) \mid x^{p^m}=x\right}$. We leave it as an easy exercise for the reader to show that $K$ is a subfield of $\operatorname{GF}\left(p^n\right)$ (Exercise 37). Since the polynomial $x^{p^m}-x$ has at most $p^m$ zeros in $\operatorname{GF}\left(p^n\right)$, we have $|K| \leq p^m$. Let $\langle a\rangle=\operatorname{GF}\left(p^n\right)^*$. Then $\left|a^t\right|=p^m-1$, and since $\left(a^t\right)^{p^{m-1}}=1$ , it follows that $a^t \in K$. So, $K$ is a subfield of $\operatorname{GF}\left(p^n\right)$ of order $p^m$.

The uniqueness portion of the theorem follows from the observation that if $\operatorname{GF}\left(p^n\right)$ had two distinct subfields of order $p^m$, then the polynomial $x^{p^m}-x$ would have more than $p^m$ zeros in $\operatorname{GF}\left(p^n\right)$. This contradicts Theorem 16.3.

## 数学代写|抽象代数代写Abstract Algebra代考|Cyclic Groups of Order $p q$

If $G$ is a group of order $p q$, where $p$ and $q$ are primes, $p<q$, and $p$ does not divide $q-1$, then $G$ is cyclic. In particular, $G$ is isomorphic to $Z_{p q}$.

PROOF Let $H$ be a Sylow $p$-subgroup of $G$ and let $K$ be a Sylow $q$-subgroup of $G$. Sylow’s Third Theorem states that the number of Sylow $p$-subgroups of $G$ is of the form $1+k p$ and divides $q$. So $1+k p=1$ or $1+k p=q$. Since $p$ does not divide $q-1$, we have that $k=0$ and therefore $H$ is the only Sylow $p$-subgroup of $G$.

Similarly, there is only one Sylow $q$-subgroup of $G$ (see Exercise 25). Thus, by the corollary to Theorem 23.5, $H$ and $K$ are normal subgroups of $G$. Moreover, from Theorem $7.2$ and Lagrange, we have $G=H K$ and $H \cap K={e}$. This tells us that $G=H \times K$. Finally, by Theorem $8.2, G \approx Z_p \oplus Z_q \approx Z_{p q}$.

Theorem $23.6$ demonstrates the power of the Sylow theorems in classifying the finite groups whose orders have small numbers of prime factors. Similar results exist for groups of orders $p^2 q, p^2 q^2, p^3$, and $p^4$, where $p$ and $q$ are prime.

For your amusement, Figure $23.2$ lists the number of nonisomorphic groups with order at most 100. Note in particular the large number of groups of order 64. Also observe that, generally speaking, it is not the size of the group that gives rise to a large number of groups of that size but the number of prime factors involved. In all, there are 1047 nonisomorphic groups with 100 or fewer elements. Contrast this with the fact that there are 49,487,365,422 groups of order $1024=2^{10}$. The number of groups of any order less than 2048 is given at http://oeis.org/A000001/b000001.txt.

## 数学代写|抽象代数代写Abstract Algebra代考|Applications of Sylow Theorems

$$p^n-1=\left(p^m-1\right)\left(p^{n-m}+p^{n-2 m}+\cdots+p^m+1\right),$$

\left 缺少或无法识别的分隔符 $\quad$. 我们把它作为一个简单的练习留给读者来证明 $K$ 是一个子 字段 $G F\left(p^n\right)$ (练习37) 。由于多项式 $x^{p^m}-x$ 至多有 $p^m$ 归霝 $\mathrm{GF}\left(p^n\right)$ ，我们有 $|K| \leq p^m$. 让
$\langle a\rangle=\operatorname{GF}\left(p^n\right)^*$. 然后 $\left|a^t\right|=p^m-1$ ，并且因为 $\left(a^t\right)^{p^{m-1}}=1$ ，它逽循 $a^t \in K$. 所以， $K$ 是一个子字段 $\operatorname{GF}\left(p^n\right)$ 秩序 $p^m$.

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