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数学代写|现代代数代写Modern Algebra代考|Sylow p-Subgroup

If $G$ is a finite abelian group and $p$ is a prime such that $p$ divides $|G|$, then $G_p$ is a Sylow $p$-subgroup.

Proof Assume that $G$ is a finite abelian group such that $p^m$ divides $|G|$ but $p^{m+1}$ does not divide $|G|$. Then $|G|=p^m k$, where $p$ and $k$ are relatively prime. We need to prove that $G_p$ has order $p^m$.

We first argue that $\left|G_p\right|$ is a power of $p$. If $\left|G_p\right|$ had a prime factor $q$ different from $p$, then $G_p$ would have to contain an element of order $q$, according to Cauchy’s Theorem. This would contradict the very definition of $G_p$, so we conclude that $\left|G_p\right|$ is a power of $p$. Let $\left|G_p\right|=p^t$.

Suppose now that $\left|G_p\right|<p^m$-that is, that $t<m$. Then the quotient group $G / G_p$ has order $p^m k / p^t=p^{m-t} k$, which is divisible by $p$. Hence $G / G_p$ contains an element $a+G_p$ of order $p$, by Theorem 4.41 . Then
$$
G_p=p\left(a+G_p\right)=p a+G_p,
$$
and this implies that $p a \in G_p$. Thus $p a$ has order that is a power of $p$. This implies that $a$ has order a power of $p$, and therefore $a \in G_p$; that is, $a+G_p=G_p$. This is a contradiction to the fact that $a+G_p$ has order $p$. Therefore, $\left|G_p\right|=p^m$, and $G_p$ is a Sylow $p$-subgroup of $G$.

The next theorem shows the true significance of the Sylow $p$-subgroups in the structure of abelian groups.

Let $G$ be an abelian group of order $n=p_1^{m_1} p_2^{m_2} \cdots p_r^{m_r}$ where the $p_i$ are distinct primes and each $m_i$ is a positive integer. Then
$$
G=G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}
$$
where $G_{p_i}$ is the Sylow $p_i$-subgroup of $G$ that corresponds to the prime $p_i$.
Proof Assume the hypothesis of the theorem. For each prime $p_i, G_{p_i}$ is a Sylow $p$-subgroup of $G$ by Theorem 4.42. Suppose an element $a_1 \in G_{p_1}$ is also in the subgroup generated by $G_{p_2}, G_{p_3}, \ldots, G_{p_r}$. Then
$$
a_1=a_2+a_3+\cdots+a_r
$$
where $a_i \in G_{p_i}$. Since $G_{p_i}$ has order $p_i^{m_i}, p_i^{m_i} a_i=0$ for $i=2, \ldots, r$. Hence
$$
p_2^{m_2} p_3^{m_3} \cdots p_r^{m_r} a_1=0 \text {. }
$$
Since the order of any $a_1 \in G_{p_1}$ is a power of $p_1$, and $p_1$ is relatively prime to $p_2^{m_2} p_3^{m_3} \cdots p_r^{m_r}$, this requires that $a_1=0$. A similar argument shows that the intersection of any $G_{p_i}$ with the subgroup generated by the remaining subgroups
$$
G_{p_1}, G_{p_2}, \ldots, G_{p_{i-1}}, G_{p_{i+1}}, \ldots, G_{p_r}
$$
is the identity subgroup ${0}$. Hence the sum
$$
G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}
$$
is direct and has order equal to the product of the orders $p_i^{m_i}$ :
$$
\left|G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}\right|=p_1^{m_1} p_2^{m_2} \cdots p_r^{m_r}=|G| .
$$
Therefore,
$$
G=G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}
$$

数学代写|现代代数代写Modern Algebra代考|Direct Sum of Cyclic Groups

Any finitely generated abelian group $G$ (and therefore any finite abelian group) is a direct sum of cyclic groups.

Proof The proof is by induction on the rank of $G$. If $G$ has rank 1 , then $G$ is cyclic and the theorem is true.

Assume that the theorem is true for any group of rank $k-1$, and let $G$ be a group of rank $k$. We consider two cases.

Case 1 Suppose there exists a minimal generating set $\left{a_1, a_2, \ldots, a_k\right}$ for $G$ such that any relation of the form
$$
z_1 a_1+z_2 a_2+\cdots+z_k a_k=0
$$
with $z_i \in \mathbf{Z}$ implies that $z_1 a_1=z_2 a_2=\cdots=z_k a_k=0$. Then
$$
G=\left\langle a_1\right\rangle+\left\langle a_2\right\rangle+\cdots+\left\langle a_k\right\rangle,
$$
and the theorem is true for this case.

Case 2 Suppose that Case 1 does not hold. That is, for any minimal generating set $\left{a_1, a_2, \ldots, a_k\right}$ of $G$, there exists a relation of the form
$$
z_1 a_1+z_2 a_2+\cdots+z_k a_k=0
$$
with $z_i \in \mathbf{Z}$ such that some of the $z_i a_i \neq 0$. Among all the minimal generating sets and all the relations of this form, there exists a smallest positive integer $\bar{z}_i$ that occurs as a coefficient in one of these relations. Suppose this $\bar{z}_i$ occurs in a relation with the generating set $\left{b_1, b_2, \ldots, b_k\right}$. If necessary, the elements in $\left{b_1, b_2, \ldots, b_k\right}$ can be rearranged so that this smallest positive coefficient occurs as $\bar{z}_1$ with $b_1$ in
$$
\bar{z}_1 b_1+\bar{z}_2 b_2+\cdots+\bar{z}_k b_k=0 .
$$
Now let $s_1, s_2, \ldots, s_k$ be any set of integers that occur as coefficients in a relation of the form
$$
s_1 b_1+s_2 b_2+\cdots+s_k b_k=0
$$
with these generators $b_i$. We shall show that $\bar{z}_1$ divides $s_1$. By the Division Algorithm, $s_1=\bar{z}_1 q_1+r_1$, where $0 \leq r_1<\bar{z}_1$. Multiplying equation (1) by $q_1$ and subtracting the result from equation (2), we have
$$
r_1 b_1+\left(s_2-\bar{z}_2 q_1\right) b_2+\cdots+\left(s_k-\bar{z}_k q_1\right) b_k=0 .
$$

现代代数代考

数学代写|现代代数代写Modern Algebra代考|Sylow p-Subgroup

如果$G$是一个有限阿贝尔群，并且$p$是一个质数，使得$p$可以除$|G|$，那么$G_p$是一个Sylow $p$ -子群。

证明假设$G$是一个有限阿贝尔群，使得$p^m$能除$|G|$，但$p^{m+1}$不能除$|G|$。然后是$|G|=p^m k$，其中$p$和$k$是相对主要的。我们需要证明$G_p$有阶$p^m$。

我们首先论证$\left|G_p\right|$是$p$的一种力量。如果$\left|G_p\right|$有一个不同于$p$的质因数$q$，那么根据柯西定理，$G_p$必须包含一个顺序为$q$的元素。这将与$G_p$的定义相矛盾，因此我们得出结论:$\left|G_p\right|$是$p$的幂。让$\left|G_p\right|=p^t$。

假设现在是$\left|G_p\right|<p^m$，也就是$t<m$。那么商群$G / G_p$的阶为$p^m k / p^t=p^{m-t} k$，能被$p$整除。因此，根据定理4.41,$G / G_p$包含一个阶为$p$的元素$a+G_p$。然后
$$
G_p=p\left(a+G_p\right)=p a+G_p,
$$
这意味着$p a \in G_p$。因此$p a$的秩序是$p$的力量。这意味着$a$有$p$的次幂，因此$a \in G_p$;也就是$a+G_p=G_p$。这与$a+G_p$有秩序这一事实相矛盾$p$。因此，$\left|G_p\right|=p^m$和$G_p$是$G$的Sylow $p$ -子组。

下一个定理说明了Sylow $p$ -子群在阿贝尔群结构中的真正意义。

设$G$是一个阶为$n=p_1^{m_1} p_2^{m_2} \cdots p_r^{m_r}$的阿贝尔群，其中$p_i$是不同的素数，每个$m_i$都是正整数。然后
$$
G=G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}
$$
其中$G_{p_i}$是$G$的Sylow $p_i$子组，对应素数$p_i$。
假设定理的假设。根据定理4.42，对于每个质数$p_i, G_{p_i}$都是$G$的一个Sylow $p$ -子群。假设一个元素$a_1 \in G_{p_1}$也在$G_{p_2}, G_{p_3}, \ldots, G_{p_r}$生成的子组中。然后
$$
a_1=a_2+a_3+\cdots+a_r
$$
在哪里$a_i \in G_{p_i}$。因为$G_{p_i}$为$i=2, \ldots, r$订购了$p_i^{m_i}, p_i^{m_i} a_i=0$。因此
$$
p_2^{m_2} p_3^{m_3} \cdots p_r^{m_r} a_1=0 \text {. }
$$
因为任何$a_1 \in G_{p_1}$的阶是$p_1$的幂次，而$p_1$相对于$p_2^{m_2} p_3^{m_3} \cdots p_r^{m_r}$是质数，这就要求$a_1=0$。类似的论证表明，任何$G_{p_i}$与由其余子组生成的子组的交集
$$
G_{p_1}, G_{p_2}, \ldots, G_{p_{i-1}}, G_{p_{i+1}}, \ldots, G_{p_r}
$$
是身份子组${0}$。因此求和
$$
G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}
$$
是直接的，阶数等于阶数的乘积$p_i^{m_i}$:
$$
\left|G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}\right|=p_1^{m_1} p_2^{m_2} \cdots p_r^{m_r}=|G| .
$$
因此，
$$
G=G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}
$$

数学代写|现代代数代写Modern Algebra代考|Direct Sum of Cyclic Groups

任何有限生成的阿贝尔群$G$(因此任何有限阿贝尔群)都是循环群的直接和。

证明是通过归纳法对$G$的秩进行的。如果$G$的秩为1，则$G$是循环的，定理成立。

假设该定理对任何秩为$k-1$的组成立，设$G$为秩为$k$的组。我们考虑两种情况。

假设存在一个最小发电集$\left{a_1, a_2, \ldots, a_k\right}$对于$G$，使得任何形式的关系
$$
z_1 a_1+z_2 a_2+\cdots+z_k a_k=0
$$
用$z_i \in \mathbf{Z}$表示$z_1 a_1=z_2 a_2=\cdots=z_k a_k=0$。然后
$$
G=\left\langle a_1\right\rangle+\left\langle a_2\right\rangle+\cdots+\left\langle a_k\right\rangle,
$$
这个定理对这个例子是成立的。

假设情况1不成立。即对于$G$的任何最小发电机组$\left{a_1, a_2, \ldots, a_k\right}$，都存在如下形式的关系
$$
z_1 a_1+z_2 a_2+\cdots+z_k a_k=0
$$
有了$z_i \in \mathbf{Z}$这样的一些$z_i a_i \neq 0$。在所有这种形式的最小发电机组和所有这种形式的关系中，存在一个最小的正整数$\bar{z}_i$，它作为系数出现在其中一个关系中。假设这个$\bar{z}_i$发生在与发电集$\left{b_1, b_2, \ldots, b_k\right}$的关系中。如果有必要，可以重新排列$\left{b_1, b_2, \ldots, b_k\right}$中的元素，使最小的正系数变为$\bar{z}_1$和$b_1$ in
$$
\bar{z}_1 b_1+\bar{z}_2 b_2+\cdots+\bar{z}_k b_k=0 .
$$
现在设$s_1, s_2, \ldots, s_k$为任意整数的集合它们作为系数出现在如下形式的关系中
$$
s_1 b_1+s_2 b_2+\cdots+s_k b_k=0
$$
有了这些生成器$b_i$。我们将证明$\bar{z}_1$除$s_1$。按除法算法，$s_1=\bar{z}_1 q_1+r_1$，其中$0 \leq r_1<\bar{z}_1$。将式(1)乘以$q_1$，再减去式(2)的结果，我们得到
$$
r_1 b_1+\left(s_2-\bar{z}_2 q_1\right) b_2+\cdots+\left(s_k-\bar{z}_k q_1\right) b_k=0 .
$$

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