Basic of Group Theory

半群：封闭性，结合律

monoid(幺半群):半群含单位元

循环群

a group of order n is cyclic if
and only if it contains an element of order n.

if p is prime, the
group $((\mathbb{Z}/p\mathbb{Z})^*,\cdot )$ is cyclic.

$\mathrm{Aut}(\mathbb{Z}/n\mathbb{Z})\cong ((\mathbb{Z}/n\mathbb{Z})^*,\cdot )$

The group $(\mathbb{Z}/n\mathbb{Z})^\times$ is cyclic if and only if n is $1,2,4,p^k,2p^k$ , where p is an odd prime and k > 0

Multiplicative group of integers modulo n;

群同态

Let ϕ : G → H be a group homomorphism, and let g ∈ G be an
element of finite order. Then |ϕ(g)| divides |g|.

群同构

Number of groups of order n

子群

子群的像，原像是子群

$\mathrm{ker},\mathrm{im}$是子群

$$N\le Z(G)\Rightarrow N\unlhd G$$

$A\le C\Rightarrow AB\cap C=A(B\cap C)$

$$\{kH|k\in K\}=\frac{KH}{H}$$

Sylow 子群

A subgroup of a finite group is termed a normal Sylow subgroup if it satisfies the following equivalent conditions:

• It is a Sylow subgroup, and is normal in the whole group.
  • It is a Sylow subgroup, and is subnormal in + the whole group.
  • It is a Sylow subgroup, and is characteristic in the whole group.
    It is a Sylow subgroup, and is fully characteristic in the whole group.

正规闭包

$H\le G$，则$H$的正规闭包

$$H^G=\langle H^x|x\in G \rangle=\{h^g|h\in H,g\in H\}=\bigcap_{K\ge H}K$$

即，包含$H$的最小的$G$的正规子群

子群的核

$H\le G$,则$H$在$G$中的核

$$H_G=\bigcap_{g\in G}H^g$$

积，包含在$H$中的极大正规子群
它$G$在$H$右陪集上的作用

$$P(x):Hg\longmapsto Hgx$$

的kernel

单群

order of finite simple groups

List_of_finite_simple_group

可解群

$H\unlhd G,$则$G’\le H \Leftrightarrow G/H$交换

Lagerange’s theorem

The order |g| of any element g of a finite group G is a divisor of
|G|

If |G| is a prime integer p, then necessarily $G\cong \mathbb{Z}/n\mathbb{Z}$

群构造

semidirect product

1. Proof:

   (a)
   It is sufficent to proof $H\cap N=1$

   Assume $x\in H\cap N$,then $|x|\big||H|$ and $|x|\big||N|$,but $(|H|,|N|)=1$,therefore $|x|=1\Rightarrow x=1\Rightarrow H\cap N=1$

   (b)Since $|g(N)|=|N|$,
   we will proof that the subgroup of order $|N|$ in is unique. Assume $H$ is a subgroup of order $|N|$,let $\overline{H}$ be the image of $H$ in $G/N$,that is,$\overline{H}=HN/N\simeq H/H\cap N$,therefore,$|\overline{H}|\big| |H|$ and $|\overline{H}|\big| |G/N|$,but $(|H|,|G/N|)=1$ ,we must have $\overline{H}=1$,that is $HN=N$,since $N\subseteq HN$,hence $H=N$.

Remark

According to the proof ,if $N$ is a normal group of $G$,and $(|N|,|G/N|)=1$,then $N$ is the characteristic subgroup of $G$,and such normal group is unique.And If $G$ has a group $H$ of order $|G/N|$,then the complement subgroup for $N$ exists,that is $G=HN$,and $G$ is the semidirect product of $N$ by $H$.If $|G/N|=p^r$ is a power of prime ,the Sylow theorem tells us such order of subgroup must exist, in such situation ,we call $N$ the normal p-complement of $G$,and we call G is p-nilpotent.If we have learned the Schur-Zassenhaus theorem,we will know that the $|G/N|$ order of subgroup always exists,or ,$N$ always has complement,and any two complemnt subgroup of $N$ are conjugated.

The conjugation action