如果你也在 怎样代写现代代数Modern Algebra 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。现代代数Modern Algebra就像数学的其他分支一样——只有从最基本的思想和例子中仔细地推导才能掌握。但这需要时间，而且有些目标在你实现之前是不明确的。

现代代数Modern Algebra这门学科的思想和方法几乎渗透到现代数学的每一个部分。此外，没有一门学科更适合培养处理抽象概念的能力，即理解和处理问题或学科的基本要素。这包括阅读数学的能力，提出正确的问题，解决问题，运用演绎推理，以及写出正确、切中要害、清晰的数学。

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数学代写|现代代数代写Modern Algebra代考|Rational number reconstruction

The integer analog of rational function reconstruction is, given integers $m>g \geq 0$ and $k \in{1, \ldots, m}$, to compute a rational number $r / t \in \mathbb{Q}$, with $r, t \in \mathbb{Z}$, such that
$$
\operatorname{gcd}(t, m)=1 \text { and } r t^{-1} \equiv g \bmod m, \quad|r|<k, \quad 0 \leq t \leq \frac{m}{k},
$$
where $t^{-1}$ is the inverse of $t$ modulo $m$. As in the polynomial case, we will see that the related problem
$$
r \equiv t g \bmod m, \quad|r|<k, \quad 0 \leq t \leq \frac{m}{k},
$$
is always solvable, while (24) need not have a solution. The uniqueness statements are a bit weaker than in the polynomial case, however. The following lemma is the integer analog of the Uniqueness Lemma 5.15.

LeMmA 5.25. Let $f, g \in \mathbb{N}$ and $r, s, t \in \mathbb{Z}$ with $r=s f+t g$, and suppose that
$$
|r|<k \text { and } 0<t \leq \frac{f}{k} \text { for some } k \in{1, \ldots, f}
$$
We let $r_i, s_i, t_i \in \mathbb{Z}$ for $0 \leq i \leq \ell+1$ be the results of the traditional Extended Euclidean Algorithm for $f, g$, with $r_i \geq 0$ for all $i$. Moreover, we define $j \in$ ${1, \ldots, \ell+1}$ by
$$
r_j<k \leq r_{j-1}
$$
and if $j \leq \ell$, we choose $q \in \mathbb{N}{\geq 1}$ such that $$ r{j-1}-q r_j<k \leq r_{j-1}-(q-1) r_j
$$
and let $q=0$ if $j=\ell+1$. Then there exists a nonzero $\alpha \in \mathbb{Z}$ such that
$$
\text { either }(r, s, t)=\left(\alpha r_j, \alpha s_j, \alpha t_j\right) \text { or }(r, s, t)=\left(\alpha r_j^, \alpha s_j^, \alpha t_j^\right), $$ where $r_j^=r_{j-1}-q r_j, s_j^=s_{j-1}-q s_j$, and $t_j^=t_{j-1}-q t_j$.

数学代写|现代代数代写Modern Algebra代考|Partial fraction decomposition

Let $F$ be a field, $f_1, \ldots, f_r \in F[x]$ nonconstant monic and pairwise coprime polynomials, $e_1, \ldots, e_r \in \mathbb{N}$ positive integers, and $f=f_1^{e_1} \cdots f_r^{e_r}$. (We will see in Part III how to factor polynomials over finite fields and over $\mathbb{Q}$ into irreducible factors, but here we do not assume irreducibility of the $f_i$.) For another polynomial $g \in F[x]$ of degree less than $n=\operatorname{deg} f$, the partial fraction decomposition of the rational function $g / f \in F(x)$ with respect to the given factorization of the denominator $f$ is
$$
\frac{g}{f}=\frac{g_{1,1}}{f_1}+\cdots+\frac{g_{1, e_1}}{f_1^{e_1}}+\cdots+\frac{g_{r, 1}}{f_r}+\cdots+\frac{g_{r, e_r}}{f_r^{e_r}},
$$
with $g_{i j} \in F[x]$ of smaller degree than $f_i$, for all $i, j$. If all $f_i$ are linear polynomials, then the $g_{i j}$ are just constants.

EXAMPLE 5.28. Let $F=\mathbb{Q}, f=x^4-x^2$, and $g=x^3+4 x^2-x-2$. The partial fraction decomposition of $g / f$ with respect to the factorization $f=x^2(x-1)(x+1)$ of $f$ into linear polynomials is
$$
\frac{x^3+4 x^2-x-2}{x^4-x^2}=\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x-1}+\frac{-1}{x+1} .
$$
The following questions pose themselves: Does a decomposition as in (31) always exist uniquely, and how can we compute it? The next lemma is a first step towards an answer.

Lemma 5.29. There exist unique polynomials $c_i \in F[x]$ with $\operatorname{deg} c_i<e_i \operatorname{deg} f_i$ for all $i$ such that
$$
\frac{g}{f}=\frac{c_1}{f_1^{e_1}}+\cdots+\frac{c_r}{f_r^{e_r}} .
$$
Proof. We multiply both sides in (33) by $f$ and obtain the linear equation
$$
g=c_1 \prod_{j \neq 1} f_j^{e_j}+\cdots+c_r \prod_{j \neq r} f_j^{e_j}
$$
with “unknowns” $c_1, \ldots, c_r$. (We have already seen in Section 4.5 how to find polynomial solutions of such equations.) For any $i \leq r$, each summand with the possible exception of the $i$ th one is divisible by $f_i^{e_i}$, whence $g \equiv c_i \prod_{j \neq i} f_j^{e_j} \bmod f_i^{e_i}$. Now each $f_j$ is coprime to $f_i$ and hence invertible modulo $f_i^{e_i}$, and we obtain
$$
c_i \equiv g \prod_{j \neq i} f_j^{-e_j} \bmod f_i^{e_i},
$$
which together with $\operatorname{deg} c_i<\operatorname{deg} f_i^{e_i}$ uniquely determines $c_i$.

现代代数代考

数学代写|现代代数代写Modern Algebra代考|Rational number reconstruction

有理数函数重构的整数类比是，给定整数$m>g \geq 0$和$k \in{1, \ldots, m}$，用$r, t \in \mathbb{Z}$计算有理数$r / t \in \mathbb{Q}$，使得
$$
\operatorname{gcd}(t, m)=1 \text { and } r t^{-1} \equiv g \bmod m, \quad|r|<k, \quad 0 \leq t \leq \frac{m}{k},
$$
其中$t^{-1}$是$t$模$m$的倒数。在多项式的情况下，我们会看到相关的问题
$$
r \equiv t g \bmod m, \quad|r|<k, \quad 0 \leq t \leq \frac{m}{k},
$$
是永远可解的，而(24)不必有解。然而，唯一性语句比多项式的情况弱一些。下面的引理是唯一性引理5.15的整数类比。

引理5.25。让$f, g \in \mathbb{N}$和$r, s, t \in \mathbb{Z}$等于$r=s f+t g$，假设
$$
|r|<k \text { and } 0<t \leq \frac{f}{k} \text { for some } k \in{1, \ldots, f}
$$
我们设$r_i, s_i, t_i \in \mathbb{Z}$为$0 \leq i \leq \ell+1$的传统扩展欧几里得算法对$f, g$的结果，$r_i \geq 0$为所有$i$。此外，我们通过定义$j \in$${1, \ldots, \ell+1}$
$$
r_j<k \leq r_{j-1}
$$
如果$j \leq \ell$，我们选择$q \in \mathbb{N}{\geq 1}$使得$$ r{j-1}-q r_j<k \leq r_{j-1}-(q-1) r_j
$$
让$q=0$ if $j=\ell+1$。那么存在一个非零$\alpha \in \mathbb{Z}$，使得
$$
\text { either }(r, s, t)=\left(\alpha r_j, \alpha s_j, \alpha t_j\right) \text { or }(r, s, t)=\left(\alpha r_j^, \alpha s_j^, \alpha t_j^\right), $$，其中$r_j^=r_{j-1}-q r_j, s_j^=s_{j-1}-q s_j$和$t_j^=t_{j-1}-q t_j$。

数学代写|现代代数代写Modern Algebra代考|Partial fraction decomposition

设$F$为一个域，$f_1, \ldots, f_r \in F[x]$为非常一元多项式和对素数多项式，$e_1, \ldots, e_r \in \mathbb{N}$为正整数，$f=f_1^{e_1} \cdots f_r^{e_r}$为正整数。(我们将在第三部分看到如何将有限域和$\mathbb{Q}$上的多项式分解为不可约因子，但在这里我们不假设$f_i$不可约。)对于另一个次小于$n=\operatorname{deg} f$的多项式$g \in F[x]$，有理函数$g / f \in F(x)$相对于给定的分母$f$的因式分解的部分分式分解为
$$
\frac{g}{f}=\frac{g_{1,1}}{f_1}+\cdots+\frac{g_{1, e_1}}{f_1^{e_1}}+\cdots+\frac{g_{r, 1}}{f_r}+\cdots+\frac{g_{r, e_r}}{f_r^{e_r}},
$$
对于所有$i, j$, $g_{i j} \in F[x]$的度数小于$f_i$。如果所有的$f_i$都是线性多项式，那么$g_{i j}$就是常数。

例5.28。设$F=\mathbb{Q}, f=x^4-x^2$和$g=x^3+4 x^2-x-2$。$g / f$对于$f$的因式分解$f=x^2(x-1)(x+1)$的部分分式分解为线性多项式为
$$
\frac{x^3+4 x^2-x-2}{x^4-x^2}=\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x-1}+\frac{-1}{x+1} .
$$
下面的问题提出了:(31)中的分解是否总是唯一存在，我们如何计算它?下一个引理是找到答案的第一步。

引理5.29。对于所有$i$存在唯一的多项式$c_i \in F[x]$和$\operatorname{deg} c_i<e_i \operatorname{deg} f_i$，使得
$$
\frac{g}{f}=\frac{c_1}{f_1^{e_1}}+\cdots+\frac{c_r}{f_r^{e_r}} .
$$
证明。我们在(33)两边乘以$f$得到线性方程
$$
g=c_1 \prod_{j \neq 1} f_j^{e_j}+\cdots+c_r \prod_{j \neq r} f_j^{e_j}
$$
与“未知”$c_1, \ldots, c_r$。(我们已经在第4.5节看到了如何找到这些方程的多项式解。)对于任何$i \leq r$，除$i$外，每个求和都可以被$f_i^{e_i}$整除，因此$g \equiv c_i \prod_{j \neq i} f_j^{e_j} \bmod f_i^{e_i}$。现在每个$f_j$都是$f_i$的互素数因此对$f_i^{e_i}$取可逆模，我们得到
$$
c_i \equiv g \prod_{j \neq i} f_j^{-e_j} \bmod f_i^{e_i},
$$
它和$\operatorname{deg} c_i<\operatorname{deg} f_i^{e_i}$一起决定了$c_i$。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

如果你也在 怎样代写现代代数Modern Algebra 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。现代代数Modern Algebra就像数学的其他分支一样——只有从最基本的思想和例子中仔细地推导才能掌握。但这需要时间，而且有些目标在你实现之前是不明确的。

现代代数Modern Algebra这门学科的思想和方法几乎渗透到现代数学的每一个部分。此外，没有一门学科更适合培养处理抽象概念的能力，即理解和处理问题或学科的基本要素。这包括阅读数学的能力，提出正确的问题，解决问题，运用演绎推理，以及写出正确、切中要害、清晰的数学。

statistics-lab™ 为您的留学生涯保驾护航 在代写现代代数Modern Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写现代代数Modern Algebra代写方面经验极为丰富，各种代写现代代数Modern Algebra相关的作业也就用不着说。

数学代写|现代代数代写Modern Algebra代考|Quotient Group

If $H$ is a normal subgroup of $G$, the group $G / H$ that consists of the cosets of $H$ in $G$ is called the quotient group or factor group of $G$ by $H$.

If the group $G$ is abelian, then so is the quotient group $G / H$. Let $a$ and $b$ be elements of $G$, then
$$
\begin{aligned}
a H b H & =a b H & & \text { since } H \text { is normal } \
& =b a H & & \text { since } G \text { is abelian } \
& =b H a H & & \text { since } H \text { is normal }
\end{aligned}
$$
and $G / H$ is abelian.
Suppose the group $G$ has finite order $n$ and the normal subgroup $H$ has order $m$. Then by Lagrange’s Theorem, we have
$$
|G|=|H| \cdot|G / H|
$$
or
$$
n=m \cdot|G / H|,
$$
and the order of the quotient group is $|G / H|=n / m$.
Example 1 Let $G$ be the octic group as given in Example 3 of Section 4.5:
$$
D_4=\left{e, \alpha, \alpha^2, \alpha^3, \beta, \gamma, \Delta, \theta\right} .
$$
It can be readily verified that $H=\left{e, \gamma, \theta, \alpha^2\right}$ is a normal subgroup of $D_4$. The distinct cosets of $\mathrm{H}$ in $\mathrm{D}_4$ are
$$
H=e H=\gamma H=\theta H=\alpha^2 H=\left{e, \gamma, \theta, \alpha^2\right}
$$
and
$$
\alpha H=\alpha^3 H=\beta H=\Delta H=\left{\alpha, \alpha^3, \beta, \Delta\right} .
$$
Thus $D_4 / H={H, \alpha H}$, and a multiplication table for $D_4 / H$ is as follows.

数学代写|现代代数代写Modern Algebra代考|Quotient Group => Homomorphic Image

Let $G$ be a group, and let $H$ be a normal subgroup of $G$. The mapping $\phi: G \rightarrow G / H$ defined by
$$
\phi(a)=a H
$$
is an epimorphism from $G$ to $G / H$.
Proof The rule $\phi(a)=a H$ clearly defines a mapping from $G$ to $G / H$. For any $a$ and $b$ in $G$,
$$
\begin{aligned}
\phi(a) \cdot \phi(b) & =(a H)(b H) \
& =a b H \quad \text { since } H \text { is normal in } G \
& =\phi(a b) .
\end{aligned}
$$
Thus $\phi$ is a homomorphism. Every element of $G / H$ is a coset of $H$ in $G$ that has the form $a H$ for some $a$ in $G$. For any such $a$, we have $\phi(a)=a H$. Therefore, $\phi$ is an epimorphism.
Example 2 Consider the octic group
$$
D_4=\left{e, \alpha, \alpha^2, \alpha^3, \beta, \gamma, \Delta, \theta\right}
$$
and its normal subgroup
$$
H=\left{e, \gamma, \theta, \alpha^2\right} .
$$
We saw in Example 1 that $D_4 / H={H, \alpha H}$. Theorem 4.25 assures us that the mapping $\phi: D_4 \rightarrow D_4 / H$ defined by
$$
\phi(a)=a H
$$
is an epimorphism. The values of $\phi$ are given in this case by
$$
\begin{gathered}
\phi(e)=\phi(\gamma)=\phi(\theta)=\phi\left(\alpha^2\right)=H \
\phi(\alpha)=\phi\left(\alpha^3\right)=\phi(\beta)=\phi(\Delta)=\alpha H .
\end{gathered}
$$

现代代数代考

数学代写|现代代数代写Modern Algebra代考|Quotient Group

如果$H$是$G$的正常子组，则由$G$中$H$的余集组成的组$G / H$被$H$称为$G$的商组或因子组。

如果群$G$是阿贝尔，那么商群$G / H$也是阿贝尔。那么，让$a$和$b$成为$G$的元素
$$
\begin{aligned}
a H b H & =a b H & & \text { since } H \text { is normal } \
& =b a H & & \text { since } G \text { is abelian } \
& =b H a H & & \text { since } H \text { is normal }
\end{aligned}
$$
$G / H$是阿贝尔的。
假设群$G$有有限阶$n$，正规子群$H$有阶$m$。根据拉格朗日定理，我们有
$$
|G|=|H| \cdot|G / H|
$$
或
$$
n=m \cdot|G / H|,
$$
商群的阶是$|G / H|=n / m$。
设$G$为第4.5节例3中给出的octic组:
$$
D_4=\left{e, \alpha, \alpha^2, \alpha^3, \beta, \gamma, \Delta, \theta\right} .
$$
可以很容易地验证$H=\left{e, \gamma, \theta, \alpha^2\right}$是$D_4$的正常子组。$\mathrm{D}_4$中$\mathrm{H}$的不同的集是
$$
H=e H=\gamma H=\theta H=\alpha^2 H=\left{e, \gamma, \theta, \alpha^2\right}
$$
和
$$
\alpha H=\alpha^3 H=\beta H=\Delta H=\left{\alpha, \alpha^3, \beta, \Delta\right} .
$$
因此是$D_4 / H={H, \alpha H}$，下面是$D_4 / H$的乘法表。

数学代写|现代代数代写Modern Algebra代考|Quotient Group => Homomorphic Image

设$G$为一个组，设$H$为$G$的正常子组。定义的映射$\phi: G \rightarrow G / H$
$$
\phi(a)=a H
$$
是从$G$到$G / H$的表属关系。
证明规则$\phi(a)=a H$明确定义了$G$到$G / H$的映射关系。有关$G$中的$a$和$b$，
$$
\begin{aligned}
\phi(a) \cdot \phi(b) & =(a H)(b H) \
& =a b H \quad \text { since } H \text { is normal in } G \
& =\phi(a b) .
\end{aligned}
$$
因此$\phi$是一个同态。$G / H$的每个元素都是$G$中的$H$的协集，对于$G$中的某些$a$具有$a H$的形式。对于任何这样的$a$，我们有$\phi(a)=a H$。因此，$\phi$是一个外胚。
例2考虑octic组
$$
D_4=\left{e, \alpha, \alpha^2, \alpha^3, \beta, \gamma, \Delta, \theta\right}
$$
和它的正规子群
$$
H=\left{e, \gamma, \theta, \alpha^2\right} .
$$
我们在例1中看到$D_4 / H={H, \alpha H}$。定理4.25保证由。定义的映射$\phi: D_4 \rightarrow D_4 / H$
$$
\phi(a)=a H
$$
是一个外属词。在本例中，$\phi$的值由
$$
\begin{gathered}
\phi(e)=\phi(\gamma)=\phi(\theta)=\phi\left(\alpha^2\right)=H \
\phi(\alpha)=\phi\left(\alpha^3\right)=\phi(\beta)=\phi(\Delta)=\alpha H .
\end{gathered}
$$

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如果你也在 怎样代写现代代数Modern Algebra 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。现代代数Modern Algebra就像数学的其他分支一样——只有从最基本的思想和例子中仔细地推导才能掌握。但这需要时间，而且有些目标在你实现之前是不明确的。

现代代数Modern Algebra这门学科的思想和方法几乎渗透到现代数学的每一个部分。此外，没有一门学科更适合培养处理抽象概念的能力，即理解和处理问题或学科的基本要素。这包括阅读数学的能力，提出正确的问题，解决问题，运用演绎推理，以及写出正确、切中要害、清晰的数学。

statistics-lab™ 为您的留学生涯保驾护航 在代写现代代数Modern Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写现代代数Modern Algebra代写方面经验极为丰富，各种代写现代代数Modern Algebra相关的作业也就用不着说。

数学代写|现代代数代写Modern Algebra代考|Left Coset Partition

Let $H$ be a subgroup of the group $G$. The distinct left cosets of $H$ in $G$ form a partition of $G$; that is, they separate the elements of $G$ into mutually disjoint subsets.

Proof It is sufficient to show that any two left cosets of $H$ that are not disjoint must be the same left coset.

Suppose $a H$ and $b H$ have at least one element in common-say, $z \in a H \cap b H$. Then $z=a h_1$ for some $h_1 \in H$, and $z=b h_2$ for some $h_2 \in H$. This means that $a h_1=b h_2$ and $a=b h_2 h_1^{-1}$. We have that $h_2 h_1^{-1}$ is in $H$ since $H$ is a subgroup, so $a=b h_3$ where $h_3=h_2 h_1^{-1} \in H$. Now, for every $h \in H$,
$$
\begin{aligned}
a h & =b h_3 h \
& =b h_4
\end{aligned}
$$
where $h_4=h_3 \cdot h$ is in $H$. That is, $a h \in b H$ for all $h \in H$. This proves that $a H \subseteq b H$. A similar argument shows that $b H \subseteq a H$, and thus $a H=b H$.

The distinct right cosets of a subgroup $H$ of a group $G$ also form a partition of $G$. That is, Lemma 4.13 can be restated in terms of right cosets (see Exercise 13).
Example 4 Consider again the subgroup
$$
K={(1),(1,2)}
$$
of
$$
G=S_3={(1),(1,2,3),(1,3,2),(1,2),(1,3),(2,3)} .
$$
In Example 3 of this section, we saw that
$$
(1,2,3) K={(1,2,3),(1,3)} .
$$
Since $(1,3)$ is in this left coset, it follows from Lemma 4.13 that
$$
(1,3) K=(1,2,3) K={(1,2,3),(1,3)} .
$$
Straightforward computations show that
$$
(1) K=(1,2) K={(1),(1,2)}=K
$$
and
$$
(2,3) K=(1,3,2) K={(1,3,2),(2,3)} .
$$
Thus the distinct left cosets of $K$ in $G$ are given by
$$
K,(1,2,3) K,(1,3,2) K
$$
and a partition of $G$ is
$$
G=K \cup(1,2,3) K \cup(1,3,2) K \text {. }
$$

数学代写|现代代数代写Modern Algebra代考|Normal Subgroup

Let $H$ be a subgroup of $G$. Then $H$ is a normal (or invariant) subgroup of $G$ if $x H=H x$ for all $x \in G$.

Note that the condition $x H=H x$ is an equality of sets, and it does not require that $x h=h x$ for all $h$ in $H$.
Example 1 Let
$$
H=A_3={(1),(1,2,3),(1,3,2)}=\langle(1,2,3)\rangle
$$
and
$$
G=S_3={(1),(1,2,3),(1,3,2),(1,2),(1,3),(2,3)} .
$$
For $x=(1,2)$ we have
$$
\begin{aligned}
x H & ={(1,2)(1),(1,2)(1,2,3),(1,2)(1,3,2)} \
& ={(1,2),(2,3),(1,3)}
\end{aligned}
$$
and
$$
\begin{aligned}
H x & ={(1)(1,2),(1,2,3)(1,2),(1,3,2)(1,2)} \
& ={(1,2),(1,3),(2,3)} .
\end{aligned}
$$
We have $x H=H x$, but $x h \neq h x$ when $h=(1,2,3) \in H$. Similar computations show that
$$
\begin{aligned}
(1) H=(1,2,3) H=(1,3,2) H & ={(1),(1,2,3),(1,3,2)}=H \
H(1)=H(1,2,3)=H(1,3,2) & ={(1),(1,2,3),(1,3,2)}=H \
(1,2) H=(1,3) H=(2,3) H & ={(1,2),(1,3),(2,3)} \
H(1,2)=H(1,3)=H(2,3) & ={(1,2),(1,3),(2,3)} .
\end{aligned}
$$
Thus $H$ is a normal subgroup of $G$. Additionally, we note that $G$ can be expressed as
$$
G=H \cup(1,2) H .
$$

现代代数代考

数学代写|现代代数代写Modern Algebra代考|Left Coset Partition

设$H$为组$G$的子组。$G$中$H$的不同左余集形成了$G$的一个分区;也就是说，它们将$G$的元素分离为互不相交的子集。

证明$H$的任意两个不相交的左余集必定是相同的左余集。

假设$a H$和$b H$至少有一个共同的元素，比如$z \in a H \cap b H$。然后$z=a h_1$表示一些$h_1 \in H$, $z=b h_2$表示一些$h_2 \in H$。这意味着$a h_1=b h_2$和$a=b h_2 h_1^{-1}$。我们知道$h_2 h_1^{-1}$在$H$中，因为$H$是子组，所以$a=b h_3$在$h_3=h_2 h_1^{-1} \in H$中。对于每个$h \in H$，
$$
\begin{aligned}
a h & =b h_3 h \
& =b h_4
\end{aligned}
$$
$h_4=h_3 \cdot h$在$H$中。也就是说，所有$h \in H$都是$a h \in b H$。这证明了$a H \subseteq b H$。类似的论点表明$b H \subseteq a H$，因此$a H=b H$。

群$G$的子群$H$的不同的右余集也形成了$G$的一个分区。也就是说，引理4.13可以用右集来重述(参见练习13)。
再次考虑子组
$$
K={(1),(1,2)}
$$
的
$$
G=S_3={(1),(1,2,3),(1,3,2),(1,2),(1,3),(2,3)} .
$$
在本节的示例3中，我们看到了这一点
$$
(1,2,3) K={(1,2,3),(1,3)} .
$$
因为$(1,3)$在这个左余集，从引理4.13可以得出
$$
(1,3) K=(1,2,3) K={(1,2,3),(1,3)} .
$$
简单的计算表明了这一点
$$
(1) K=(1,2) K={(1),(1,2)}=K
$$
和
$$
(2,3) K=(1,3,2) K={(1,3,2),(2,3)} .
$$
因此，$G$中$K$的不同左余集由式给出
$$
K,(1,2,3) K,(1,3,2) K
$$
$G$的分区是
$$
G=K \cup(1,2,3) K \cup(1,3,2) K \text {. }
$$

数学代写|现代代数代写Modern Algebra代考|Normal Subgroup

设$H$为$G$的子组。那么$H$是$G$的正常(或不变)子组，如果$x H=H x$适用于所有$x \in G$。

注意，条件$x H=H x$是集合的等式，它不要求$H$中的所有$h$都是$x h=h x$。
例1
$$
H=A_3={(1),(1,2,3),(1,3,2)}=\langle(1,2,3)\rangle
$$
和
$$
G=S_3={(1),(1,2,3),(1,3,2),(1,2),(1,3),(2,3)} .
$$
对于$x=(1,2)$我们有
$$
\begin{aligned}
x H & ={(1,2)(1),(1,2)(1,2,3),(1,2)(1,3,2)} \
& ={(1,2),(2,3),(1,3)}
\end{aligned}
$$
和
$$
\begin{aligned}
H x & ={(1)(1,2),(1,2,3)(1,2),(1,3,2)(1,2)} \
& ={(1,2),(1,3),(2,3)} .
\end{aligned}
$$
我们有$x H=H x$，但$x h \neq h x$当$h=(1,2,3) \in H$。类似的计算表明
$$
\begin{aligned}
(1) H=(1,2,3) H=(1,3,2) H & ={(1),(1,2,3),(1,3,2)}=H \
H(1)=H(1,2,3)=H(1,3,2) & ={(1),(1,2,3),(1,3,2)}=H \
(1,2) H=(1,3) H=(2,3) H & ={(1,2),(1,3),(2,3)} \
H(1,2)=H(1,3)=H(2,3) & ={(1,2),(1,3),(2,3)} .
\end{aligned}
$$
因此$H$是$G$的正常子组。另外，我们注意到$G$可以表示为
$$
G=H \cup(1,2) H .
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。