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数学代写|抽象代数代写Abstract Algebra代考|MATH411 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

A few numerical examples will make the Sylow theorems come to life.
I EXAMPLE 3 Say $\mathrm{G}$ is a group of order 40. What do the Sylow theorems tell us about G? A great deal! Since 1 is the only divisor of 40 that is congruent to 1 modulo 5 , we know that $\mathrm{G}$ has exactly one subgroup of order 5 , and therefore it is normal. Similarly, $\mathrm{G}$ has either one or five subgroups of order 8 . If there is only one subgroup of order 8 , it is normal. If there are five subgroups of order 8 , none is normal and all five can be obtained by starting with any particular one, say $\mathrm{H}$, and computing $x H x^{-1}$ for various $x$ ‘s. Finally, if we let $\mathrm{K}$ denote the normal subgroup of order 5 and let $\mathrm{H}$ denote any subgroup of order 8, then $G=H K$. (See Example 5 in Chapter 9.) If $\mathrm{H}$ happens to be normal, we can say even more: $G=H \times K$.

I EXAMPLE 4 Consider a group of order 30. By Sylow’s Third Theorem, it must have either one or six subgroups of order 5 and one or 10 subgroups of order 3 . However, $G$ cannot have both six subgroups of order 5 and 10 subgroups of order 3 (for then G would have more than 30 elements). Thus, the subgroup of order 3 is unique or the subgroup of order 5 is unique (or both are unique) and therefore is normal in G. It follows, then, that the product of a subgroup of order 3 and one of order 5 is a group of order 15 that is both cyclic (Exercise 35) and normal (Exercise 9 in Chapter 9) in G. [This, in turn, implies that both the subgroup of order 3 and the subgroup of order 5 are normal in $G$ (Exercise 59 in Chapter 9).] So, if we let $y$ be a generator of the cyclic subgroup of order 15 and let $x$ be an element of order 2 (the existence of which is guaranteed by Cauchy’s Theorem), we see that $G=\left{x^i y^j \mid 0 \leq i \leq 1,0 \leq j \leq 14\right}$.
I EXAMPLE 5 We show that any group $\mathrm{G}$ of order 72 must have a proper, nontrivial normal subgroup. Our arguments are a preview of those in Chapter 24. By Sylow’s Third Theorem, the number of Sylow 3-subgroups of $\mathrm{G}$ is equal to $1 \bmod 3$ and divides 8 . Thus, the number is 1 or 4 . If there is only one, then it is normal by the corollary of Sylow’s Third Theorem. Otherwise, let $H$ and $H^{\prime}$ be two distinct Sylow 3-subgroups. By Theorem 7.2, we have that $\left|H H^{\prime}\right|=|H|\left|H^{\prime}\right| /\left|H \cap H^{\prime}\right|=81 /\left|H \cap H^{\prime}\right|$. Since $|G|=72$ and $\left|H \cap H^{\prime}\right|$ is a subgroup of $\mathrm{H}$ and $H^{\prime}$, we know that $\left|H \cap H^{\prime}\right|=3$. By the corollary to Theorem 23.2, $N\left(H \cap H^{\prime}\right)$ contains both $\mathrm{H}$ and $H^{\prime}$. Thus, $\left|N\left(H \cap H^{\prime}\right)\right|$ divides 72 , is divisible by 9 , and has at least $\left|H H^{\prime}\right|=27$ elements. This leaves only 36 or 72 for $\left|N\left(H \cap H^{\prime}\right)\right|$. In the first case, we have from Exercise 9 of Chapter 9 that $N\left(H \cap H^{\prime}\right)$ is normal in $G$. In the second case, we have by definition that $H \cap H^{\prime}$ is normal in $G$.

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

1 示例 4 考虑一个 30 阶群。根据 Sylow 第三定理，它必须有一个或六个 5 阶子群和一个或 10 个 3 阶子群。然而， $G$ 不能同时有 6 个 5 阶子群和 10 个 3 阶子群 (因为那时 $G$ 将有超过 30 个元溸) 。因此，3 阶子群是唯一的或 5 阶子群是唯一的（或两者都是 唯一的），因此在 $G$ 中是正规的。因此，3 阶子群与 5 阶子群之一的乘积是 $G$ 中的一个 15 阶群，它既是徣环的（练习 35) 又是 正规的（第 9 章的练习 9) 。[这又意味着 3 阶的子群和 5 阶的子群都是正规的 $G$ (第 9 章练习 59) ] 所以，如果我们让 $y$ 是 15 阶循环子群的生成器并令 $x$ 是 2 阶元㨞（其存在由柯西定理保证)，我们看到 $\backslash$ lef $t$ 缺少或无法识别的分隔符
1 示例 5 我们证明任何组G72 阶的子群必须有一个适当的、非平凡的正规子群。我们的论点是第 24 章中那些论点的预演。根据 Sylow 第三定理，sylow 3-子群的数量G等于 $1 \mathrm{mod} 3$ 并除以 8。因此，数字是 1 或 4。如果只有一个，则根据 Sylow 第三定 理的推论，它是正常的。否则，让 $H$ 和 $H^{\prime}$ 是两个不同的 Sylow 3-子群。根据定理 7.2，我们有
$\left|H H^{\prime}\right|=|H|\left|H^{\prime}\right| /\left|H \cap H^{\prime}\right|=81 /\left|H \cap H^{\prime}\right|$. 自从 $|G|=72$ 和 $\left|H \cap H^{\prime}\right|$ 是一个子群 $\mathrm{H}$ 和 $H^{\prime}$ ， 我们知道 $\left|H \cap H^{\prime}\right|=3$. 根 据定理 $23.2$ 的推论， $N\left(H \cap H^{\prime}\right)$ 包含两者H和 $H^{\prime}$. 因此， $\left|N\left(H \cap H^{\prime}\right)\right|$ 被 72 整除，能被 9 整除，并且至少有 $\left|H H^{\prime}\right|=27$ 元 䍱。这只剩下 36 或 $72\left|N\left(H \cap H^{\prime}\right)\right|$. 在第一种情况下，我们从第 9 章的练习 9 中得到 $N\left(H \cap H^{\prime}\right)$ 是正常的 $G$. 在第二种情况 下，根据定义我们有 $H \cap H^{\prime}$ 是正常的 $G$.

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