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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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