指数为2的子群为正规子群

hbghlyj

Abel’s Theorem in Problems & Solutions(大字号版 | 小字号版)
Problem-105 Let $n$ be the order of a group $G$, $m$ the order of subgroup $H$ and $m =\frac n2$. Prove that $H$ is a normal subgroup of the group $G$.

Czhang271828

正规子群 $H\lhd G$ 的等价定义:

1. 对任意 $g\in G$, $g^{-1}H g=H$ ($H$ 在 $G$ 中的共轭子群仅有自身).
2. $H$ 的左陪集均为右陪集 (等价地, 右陪集均为左陪集).
3. 存在同态 $\varphi:G\to G'$ 使得 $\ker \varphi =H$.

这里用 $2$, 考察单位元的位置, 显然.

可以参考*近世代数导引* (冯克勤, 章璞著).

hbghlyj

明白了.谢谢.今天又遇见这个题了:

这份2014年讲义的命题207(第55页)

hbghlyj

又见于Groups and Symmetry(Theorem 15.4)
If the index of H in G is equal to 2, then H is a normal subgroup of G and the quotient group G/H is isomorphic to $\mathbb Z_2$

$G\cong\mathbb Z_2\rtimes H$

hbghlyj

回复 2# Czhang271828
我来补一个链接

近世代数引论 (冯克勤, 章璞著)

Czhang271828

回复 5# hbghlyj

找书当然去 z-lib 啊.

补充一个近世代数 300 题
.

hbghlyj

上次我上传了kuing的题集到library.lol时,发现第3个镜像节点zlibrary马上就有了,我感觉这两个站点之间是互通有无的

Czhang271828

回复 7# hbghlyj

可能节点有讲究? 我是直连香港某高校的

abababa

写一个详细的证明：设$(G,\circ)$是群而$N\le G$且$[G:N]=2$。
若$b\in N$，则显然有$bN=Nb$(1)。
若$b\in G$且$b\not\in N$，则由于$[G: N]=2$，因此$G$被分为两个左陪集，其中一个为$eN=N$，另一个为$bN$，由左陪集分解知$N\cup bN=G$且$N\cap bN=\varnothing$，但同时$G$也被分为两个右陪集$N$和$Nb$，由右陪集分解知$N\cup Nb=G$且$N\cap Nb=\varnothing$。

于是$N\cup Nb=G=N \cup bN$，且$N\cap Nb=N\cap bN=\varnothing$，所以$Nb=bN$(2)。

根据(1)(2)就知道，对任意的$b\in G$都有$bN=Nb$，根据定义$N$就是$G$的正规子群。

hbghlyj

Subgroup of Index 2 is Normal
1965: Seth Warner: Modern Algebra: Chapter II: New Structures from Old: §11: Quotient Structures: Exercise 11.9
1966: Richard A. Dean: Elements of Abstract Algebra: §1.10: Theorem 23
1971: Allan Clark: Elements of Abstract Algebra: Chapter 2: Conjugacy, Normal Subgroups, and Quotient Groups: §46γ
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra: §49.2 Normal subgroups
1996: John F. Humphreys: A Course in Group Theory: Chapter 7: Normal subgroups and quotient groups: Example 7.6

In general, we have
     Subgroup of Index Least Prime Divisor is Normal

Theorem
Let $G$ be a finite group of order $n>1$
Let $p$ be the least prime divisor of $n$
Let $H$ be a subgroup of index $p$
Then $H$ is normal.

hbghlyj

数论的例子: The quadratic residues are a subgroup $G$ of index 2 of $(\mathbb Z/p\mathbb Z)^×$, and $(\mathbb Z/p\mathbb Z)^×/G=\{\pm1\}$.
Lemma 4.2.
(i) The product of two quadratic residues is a quadratic residue;
(ii) The product of a quadratic residue and a quadratic nonresidue is a quadratic nonresidue;
(iii) The product of two quadratic nonresidues is a quadratic residue.

There is a (unique!) map $ψ : (\mathbb Z/p\mathbb Z)^× → \{−1, 1\}$ such that $ψ(ab) = ψ(a)ψ(b)$ and such that $a$ is a quadratic residue if and only if $ψ(a) = 1$.
(Legendre symbol). $ψ(a)=(\frac ap)$
(Euler’s criterion). $ψ$ is the power map $a\mapsto a^{p-1\over2}$
证明
$p$的quadratic residue都为$x^{p-1\over2}-1$的根. 只需证明$x^{p-1\over2}-1$没有其它的根:
因为$x^{p-1}-1$有$(p-1)$个根[是整个$(\mathbb Z/p\mathbb Z)^×$]
且$x^{p-1}-1=(x^{p-1\over2}-1)(x^{p-1\over2}+1)$ [每个是$p-1\over2$次,故有$\le{p-1\over2}$个根,因$(\mathbb Z/p\mathbb Z)^×$为整域]
所以$x^{p-1\over2}-1$与$x^{p-1\over2}+1$各有$p-1\over2$个根. $_\square$

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提取一个结论: 在整域上, 多项式$f$有$\deg f$个根, 且$f=gh$, 则$g$有$\deg g$个根, $h$有$\deg h$个根.

hbghlyj

Free group n contains subgroup of index 2 @spin的回答

For any subgroup $H$ of $G$ and elements $a$ and $b$ of $G$ the following statements hold.

• If $a \in H$ and $b \in H$, then $ab \in H$
• If $a \in H$ and $b \not\in H$, then $ab \not\in H$
• If $a \not\in H$ and $b \in H$, then $ab \not\in H$

Hence it is natural to ask when $a \not\in H$ and $b \not\in H$ implies $ab \in H$, ie. when a subgroup is a "parity subgroup", as in $G = \mathbb{Z}$ and $H = 2\mathbb{Z}$ or in $G = S_n$ and $H = A_n$. Suppose that $H$ is a proper subgroup since the case $H = G$ is not interesting. Then the following statements are equivalent:

1. $[G:H] = 2$
2. For all elements $a \not\in H$ and $b \not\in H$ of $G$, we have $ab \in H$.
3. There exists a homomorphism $\phi: G \rightarrow \{1, -1\}$ with $\operatorname{Ker}(\phi) = H$.