抽象代数学习笔记（群论）

Group Theory

2.26

Definition. A group is an ordered pair $(G,*)$, such that:

Exists an identity $e$, s.t. $e*a=a*e=a$ for $\forall a\in G$
$\forall a\in G$, $\exists a^{-1}\in G$, s.t. $a*a^{-1}=a^{-1}*a=e$
$\forall a,b,c\in G$, $a*(b*c)=(a*b)*c$

$(G,*)$ is called Abelian (or commutative) if $a*b=b*a$ for $\forall a,b\in G$

Examples:

$\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}$ are all groups under "$+$" ($e=0$, $a^{-1}=-a$)
$GL_n(\mathbb{R})$ (matrix of $n\times n$ with rank $=n$) is a group under matrix multiplication ($e=I$)
$\mathbb{Z}/n\mathbb{Z}$ (integers modulo $n$) is a group under "$+$"

Properties of groups: Let $(G,*)$ be a group, then

$e$ is unique.
$a^{-1}$ is unique $\forall a\in G$
$(a^{-1})^{-1}=a$, $\forall a\in G$
$(ab)^{-1}=(b^{-1})(a^{-1})$, pay attention to the order because a group may not be commutative in general.
$a_1*a_2*\cdots*a_n$ is independent of how the expression is bracketed.

For a group $G$ and $x\in G$, define the order of $x$ as the smallest positive integer $n$ s.t. $x^n=1$, denoted as $|x|$. If such $n$ does not exist, define $|x|$ as $\infty$.

Cancellation law: $\forall a,u,v\in G$, $au=av\Rightarrow u=v, vu=va\Rightarrow u=v$

Dihedral groups:

r: rotation clockwise $\dfrac{2\pi}{n}$ rotation.
s: reflection along the dished line.

Then:

$1,r,r^2,\cdots,r^{n-1}$ are all distinct and $r^n=1$
$s\ne r^j$ for $\forall j\in\mathbb{Z}$
$sr^i\ne sr^j$ for all $0\le i<j\le n-1$
$rs=sr^{-1},r^js=sr^{-j}$

Let $D_{2n}=\{1,,r,r^2,\cdots,r^{n-1},s,sr,sr^2,\cdots,sr^{n-1}\}$

Generator: A subset $S$ of a group $G$ s.t. $\forall g\in G$, $g$ can be written as a product of elements of $S$ and their inverses.

Relations: Equations satisfied by generators.

Presentations: $G=\lang S|R_1,R_2,\cdots,R_m\rang$, where $R_1,R_2,\cdots,R_m$ are the relations.

Example: $D_{2n}=\lang r,s|r^n=s^2=1,rs=sr^{-1}\rang$

3.4

Symmetric groups: Let $\Omega=\{1,2,3,\cdots,n\}$, $S_n$ denotes all permutations of $\Omega$. Obviously, $S_n$ forms a group under the permutation "$\circ$", and $|S_n|=n!$.

Cycle decomposition: $(a_1,a_2,\cdots,a_m)$ means $a_1\to a_2,a_2\to a_3,\cdots,a_{m-1}\to a_m,a_m\to a_1$.

The Quaternion group $Q_8=\{\pm 1, \pm i, \pm j, \pm k\}$ with identity $1$. Their product is defined as:

$i·i=j·j=k·k=-1$
$i·j=k,j·i=-k,,j·k=i,k·j=-i,k·i=j,i·k=-j$

$|Q_8|=8$ and $Q_8$ is non-Abelian.

Homomorphisms & Isomorphisms

Let $(G,*)$ and $(H,\circ)$ be groups. If there exists a map $\varphi:G\to H$ such that $\forall x,y\in G$, $\varphi(x*y)=\varphi(x)\circ\varphi(y)$, then $\varphi$ is called a Homomorphism.

If in addition $\varphi$ is bijective, then $\varphi$ is an Isomorphism, written as $G\cong H$. Namely, $G$ and $H$ are isomorphic if there exists a bijection between them which preserves the group operations.

Examples:

$G\cong G$, $\cong$ is an equivalence relation.
$(\mathbb{R},+)\to(\mathbb{R}^+,\times)$ defined by $\exp(x)=e^x$ is Isomorphism.
The symmetric groups $S_{\Delta}\cong S_{\Omega}$ if and only if $|\Delta|=|\Omega|$.

Necessary conditions for Isomorphisms:

If $G\cong H$, then

$|G|=|H|$
$G$ is Abelian if and only if $H$ is Abelian
$\forall x\in G, |x|=|\varphi(x)|$

Thus $S_3$ and $Z/6Z$ are not isomorphic, since $S_3$ is non-Abelian, but $Z/6Z$ is. And $(\mathbb{R}-\{0\},\times)$ and $(\mathbb{R},+)$ are not isomorphic as well.

Group actions

Definition. A group action of $(G,\times)$ on a set $A$ is a map from $G\times A$ to $A$ (written as $g·a$ for $g\in G,a\in A$), obeying:

$g_1·(g_2·a)=(g_1\times g_2)·a$, $\forall g_1,g_2\in G,a\in A$
$1·a=a$ $\forall a\in A$.

Let $G$ act on $A$, define a map $\sigma_g:A\to A$ by $\sigma_g(a)=g·a$, then

$\sigma_g$ is a permutation of $A$
The map from $G$ to $S_A$ defined by $g\to\sigma_g$ is homomorphism.

Examples:

Trivial action: let $g·a=a$ for $\forall g\in G,a\in A$. Let $G$ act on $B$, if $\forall g_1,g_2\in G\Rightarrow\sigma_{g_1}\ne\sigma_{g_2}$, then the action is said to be faithful.

The kernel of action is $\{g\in G|gb=b,\forall b\in B\}$
$A\ne\varnothing$, $S_A$ acts on $A$ by $\sigma_{a}=a$.
Let $G$ be a group, $A=G$, define a group action of $G$ on itself by $g·a=ga$, called the left regular of $G$ on itself.

Subgroups

Definition. let $(G,\times)$ be a group, a non-empty subset $H$ of $G$ is a subgroup of $G$ if $H$ is closed under $\times$ and inverse, written as $H\leq G$.

Examples:

$\mathbb{Z}\subseteq\mathbb{Q}$ and $\mathbb{Q}\subseteq\mathbb{R}$ under "+".
$G\le G$, $1\leq G$.
Let $H=\lang r\rang$, $H\leq D_{2n}$.

A subset $H$ of a group $G$ is a subgroup if and only if:

$H\ne\varnothing$
For $\forall x,y\in H$, $xy^{-1}\in H$

Furthermore, if $H$ is a finite group, then it suffices to check that $H$ is non-empty and closed under multiplication.

Centralizer

Let $G$ be a group, $A$ be any non-empty subset of $G$. $C_G(A)=\{g\in G|gag^{-1}=a\forall a\in A\}$.

Then $C_G(A)\leq G$ since:

$e\in C_G(A)\Rightarrow C_G(A)\ne\varnothing$
$\forall x,y\in C_G(A),\forall a\in A$, $(xy^{-1})a(xy^{-1})^{-1}=x(y^{-1}ay)x^{-1}=xax^{-1}=a$, $xy^{-1}\in C_G(A)$.

Examples:

If $G$ is Abelian, then $C_G(A)=G$ for $\forall A\in G$.
$C_{Q_8}(i)=\{\pm 1,\pm i\}$.

Center of $G$: $Z(G)=\{g\in G|gx=xg\forall x\in G\}$, $Z(G)=C_G(G)$.

Normalizer

Let $G$ be a group, $A$ be any non-empty subset of $G$. $N_G(A)=\{g\in G|gAg^{-1}=A\}$, where $gAg^{-1}=\{gag^{-1}|a\in A\}$.

$N_G(A)\leq G$, $C_G(A)\leq N_G(A)$.

Examples:

If $G$ is Abelian, then $Z(G)=N_G(A)=C_G(A)=G$ for all $A\le G$.
Let $G=D_8$, $A=\{1,r,r^2,r^3\}$, then $C_{D_8}(A)=A$, $N_{D_8}(A)=D_8$, $Z(D_8)=\{1,r^2\}$.

Stabilizers of group actions

Let $G$ acts on $S$ and $s\in S$, then the stabilizer of $s$ in $G$ is $G_s=\{g\in G|gs=s\}$

$G_s\le G$

The kernel of the action $\{g\in G|gs=s,\forall s\in S\}$

Cyclic groups

Definition: A group $H$ is cyclic if $H$ can be generated by a single element, i.e, there exists $x$ such that $H=\lang x\rang$.

The generator may NOT be unique, since $H=\lang x\rang=\lang x^{-1}\rang$.
Cyclic groups are always Abelian.

Examples:

$H=\lang r\rang\le D_{2n}=\lang r,s\rang$.
$\mathbb{Z}=\lang1\rang$.

Propositions:

If $H=\lang x\rang$, then $|H|=|x|$.
Let $G$ be an arbitrary group, $x\in G$, and $m,n\in\mathbb{Z}$. If $x^n=1$ and $x^m=1$, then $x^d=1$, where $d=\gcd(m,n)$. In particular, if $x^m=1$, then $|x|\mid m$.
Any two cyclic groups of the same order are isomorphic.
Let $G$ be a group, $x\in G$, and $a\in\mathbb{Z}-\{0\}$, then
- If $|x|=\infty$, then $|x^a|=\infty$.
- If $|x|=n<\infty$, then $|x^a|=\dfrac{n}{\gcd(n,a)}$.
- If $|x|=n<\infty$, and $a$ is a positive integer dividing $n$, then $|x^a|=\dfrac{n}{a}$.
Let $H=\lang x\rang$,
- Assume $|x|=\infty$, then $H=\lang x^a\rang$ if and only if $a=\pm 1$.
- Assume $|x|=n<\infty$, then $H=\lang x^a\rang$ if and only if $\gcd(a,n)=1$. (the number of generators of $H$ is $\varphi(n)$)

Subgroups generated by subsets

Let $G$ be a group, $A$ is a subset of $G$, then a subgroup of $G$ generated by $A$ is:

$A=\{a_1^{\alpha_1}a_2^{\alpha_2}a_3^{\alpha_3}\cdots a_n^{\alpha_n}|a_i\ne a_{i+1},a_i\in A,\alpha_i\in\mathbb{Z},n\in\mathbb{Z}^+\}$.

3.11

The lattice of subgroups of a group:

We use lattice diagrams to "see" the relationships among subgroups of a finite group $G$

Steps:

Plot all subgroups of $G$ with $1$ at the bottom and $G$ at the top, roughly subgroups of larger order positioned higher.
Draw paths between subgroups. Draw a path from $G_1$ to $G_2$ if $G_2$ is a subgroup of $G_1$ and there is no subgroups properly between $G_1$ and $G_2$.

Limitation to this process:

Don't work for infinite group.
Even for finite group, sometimes the lattices can be very complicated.

The Klein 4 group: $V_4=\{1,a,b,c\}$. Multiplication table:

\	1	a	b	c
1	1	a	b	c
a	a	1	c	b
b	b	c	1	a
c	c	b	a	1

Every group of size $4$ is isomorphism to either $V_4$ or $Z/4Z$.

Proof:

If there exists $a\in G$ such that $|a|=4$, then $G$ is isomorphism to $Z/4Z$.
Otherwise, $|a|\le 3$. If $|a|=3$, then $a^2=b$ or $c$. Assume $a^2=b$, then $a^{-1}=b$. Consider $ac\ne 1,ac\ne c,ac\ne a,ac\ne b$, this is impossible. So $|a|=2$. Which means all elements have order $2$, $G$ is isomorphism to $V_4$.

Quotient groups

Let $\varphi$ be a homomorphism from $G$ to $H$.

Recall the fibers of $\varphi$ are the sets of elements of $G$ projecting to some single element in $H$.

Let $X_a$ and $X_b$ denote the fibers above $a$ and $b$, then we define $X_aX_b=X_{ab}$.

This makes the set of fibers into a group, called quotient group of $G$:

Identity is $X_1$.
The inverse of $X_a$ is $X_{a^{-1}}$.
Associativity: $(X_aX_b)X_c=X_{ab}X_c=X_{abc}=X_aX_{bc}=X_a(X_bX_c)$.

If $\varphi:G\to H$ is a homomorphism, the kernel of $\varphi$: $\ker\varphi=\{g\in G|\varphi(g)=1\}$.

Proposition 1: Let $G$ and $H$ be groups and $\varphi:G\to H$ be a homomorphism, then:

$\varphi(1_G)=1_H$
$\varphi(g^{-1})=\varphi(g)^{-1}$
$\varphi(g^n)=\varphi(g)^n$
$\ker\varphi\le G$
$\text{im}(\varphi)=\varphi(G)\le H$

Proof:

Since $\varphi(1_G)=\varphi(1_G1_G)=\varphi(1_G)\varphi(1_G)$, $\varphi(1_G)=\varphi(1_H)$.

$\varphi(1_G)=\varphi(gg^{-1})\Rightarrow \varphi(g)\varphi(g^{-1})=1_H\Rightarrow\varphi(g^{-1})=\varphi(g)^{-1}$.

Easy proof by induction.

Since $1_G\in\ker\varphi$, the kernel of $\varphi$ is not empty. Let $x,y\in\ker\varphi$, that is $\varphi(x)=\varphi(y)\in 1_H$, then $\varphi(xy^{-1})=\varphi(x)\varphi(y^{-1})=1_H$, thus $xy^{-1}\in\ker\varphi$, $\ker\varphi\le G$.

Since $\varphi(1_G)=1_H\in\varphi(G)$, $\varphi(G)$ is non-empty. Let $x,y\in\varphi(G)$, say $x=\varphi(a),y=\varphi(b)$, then $xy^{-1}=\varphi(a)\varphi(b)^{-1}=\varphi(ab^{-1})\in\varphi(G)$, thus $xy^{-1}\in\varphi(G)$, $\varphi(G)\le H$.

Definition. Let $\varphi:G\to H$ be a homomorphism with kernel $K$. The quotient group or factor group $G/K$ is the whose elements are fibers of $\varphi$ with group operation $X_aX_b=X_{ab}$.

Proposition 2: Let $\varphi:G\to H$ be a homomorphism with kernel $K$. Let $X\in G/K$ be the fiber above $a$, i.e. $X=\varphi^{-1}(a)$, then

$\forall u\in X,X=\{uk|k\in K\}=uK$.
$\forall u\in X,X=\{ku|k\in K\}=Ku$.

Proof:

Let $u\in X$, then $\varphi(u)=a$. we first prove that $uK\subseteq X$, since $\forall k\in K$, $\varphi(uk)=\varphi(u)\varphi(k)=a$, that is $uk\in X$, so $uK\subseteq X$. In addition, for $\forall g\in X$, let $k=u^{-1}g$, then $\varphi(k)=\varphi(u)^{-1}\varphi(g)=1$, thus $k\in\ker\varphi$, $g=uk\in uK$, $X\subseteq uK$. So $uK=X$.

Similar as (1).

Definition. For any $N\le G$ and $g\in G$, define:

Left coset: $gN=\{gn|n\in N\}$
Right coset: $Ng=\{ng|n\in N\}$

An element in the coset is called the representative in the coset.

Theorem 3: Let $G$ be a group and $K$ be the kernel of some homomorphism from $G$ to another group. Then the set of let cosets with operation $uK·vK=(uv)K$ form a group, $G/K$. The same is true for right cosets.

Proof. Let $X,Y\in G/K$ and $Z=XY\in G/K$, so that by proposition 2, $X,Y,Z$ are left cosets of $K$. In addition, $K$ is the kernel of some homomorphism $\varphi:G\to H$, so $X=\varphi^{-1}(a),Y=\varphi^{1}(b),Z=\varphi^{-1}(ab)$ for some $a,b\in H$. Let $u,v$ be arbitrary representatives of $X,Y$ respectively, s.t. $\varphi(u)=a,\varphi(v)=b$ and $X=uK,Y=vK$, so since $\varphi(u)\varphi(v)=ab$, $uv\in\varphi^{-1}(ab)$, so $uv\in Z$. On the other hand, $\forall z\in Z,z=uv$ for some $u\in X,v\in Y$, since $\varphi(u^{-1}z)=\varphi(u^{-1})\varphi(z)=a^{-1}(ab)=b$. Thus $Z=uvK$.

This also proves that the product of $X,Y$ is the coset $uvK$ for any choice of representatives $u\in X,v\in Y$. By proposition 2, $uK=Ku$ for all $u\in G$, thus the statement is true for right cosets.

Proposition 4: Let $N\le G$, then the set of left cosets of $N$ in $G$ forms a partition of $G$. Furthermore, for $\forall u,v\in G$, $uN=vN$ if and only if $v^{-1}u\in N$ and in particular, $uN=vN$ if and only if $u$ and $v$ are representatives of the same coset.

Proof: Since $N\le G$, $1\in N$ and $g=g·1$ for $\forall g\in G$, i.e., $G=\cup_{g\in G}gN$. Moreover, suppose $uN\cap vN\ne\varnothing$, then $\exists x\in uN\cap vN$, write $x$ as $un$ and $vm$ for certain $n,m\in N$, then $u=vmn^{-1}$ and $\forall ut\in uN$, $ut=vmn^{-1}t\in vN$, hence $uN\subseteq vN$. Similarly $vN\subseteq uN$, hence $uN=vN$. By the first part, $uN=vN\Leftrightarrow u\in vN\Leftrightarrow v^{-1}u\in N$, as desired.

Proposition 5: Let $N\le G$:

$uN·vN=(uv)N$ is well-defined if and only if $gng^{-1}\in N$ for $\forall g\in G,n\in N$.
If the above operation is well-defined, then it makes the set of left cosets of $N$ into a group.

Proof:

Assume first that the operation is well-defined, that is $\forall u,v\in G$, if $uu_1\in uN$ and $vv_1\in vN$, then $uvN=u_1v_1N$. Let $\forall g\in G,\forall n\in N$, choose $u=1,u_1=n,v=v_1=g^{-1}$, then $1g^{-1}N=ng^{-1}N$, i.e. $g^{-1}N=ng^{-1}N$, since $1\in N$, $ng^{-1}\in ng^{-1}N=g^{-1}N$, thus $ng^{-1}=g^{-1}n_1$ for certain $n_1\in N$, $gng^{-1}=n_1\in N$, as desired. Conversely, assume $gng^{-1}\in N$ for $\forall g\in G,n\in N$. Let $u,u_1\in uN,v,v_1\in vN$, we may write $u_1=un$ and $v_1=vm$ for some $n,m\in N$, then $u_1v_1=unvm=uvv^{-1}nvm=uv(n_1m)$ where $n_1=v^{-1}nv\in N$, thus $u_1v_1\in uvN$, the operation is well-defined.

The identity is obviously $1N$. The inverse of $gN$ is $g^{-1}N$. The associativity: $(uN)(vNwN)=(uvw)N=(uNvN)(wN)$. Thus, it is a group.

Definition. Let $N\le G$, the $gng^{-1}$ is called the conjugate of $n$ by $g$. The set $gNg^{-1}=\{gng^{-1}|n\in N\}$ is called the conjugate of $N$ by $g$. The element is said to normalize $N$ if $gNg^{-1}=N$. $N$ is called normal if $gNg^{-1}=N$ for $\forall g\in G$. Written as $N\unlhd G$.

Proposition 6: Let $N\le G$, then the following are equivalent:

$N\unlhd G$.
$N_G(N)=G$.
$gN=Ng$ for $\forall g\in G$.
$gNg^{-1}=N$ for $\forall g\in G$.
The operation defined by $(aH)·(bH)=(ab)·H$ is well-defined, and the set of left cosets forms a group.

Proposition 7: A subgroup $N$ of $G$ is normal if and only if it is the kernel of some homomorphism.

Proof:

If $N$ is the kernel of the homomorphism $\varphi$, then by proposition 2, $\forall g\in G$, $gN=Ng$, and $N\unlhd G$.

Conversely, if $N\unlhd G$, let $H=G/N$ and define $\pi:G\to H$ by $\pi(g)=gN$ for $\forall g\in G$. Then $\pi(g_1g_2)=g_1g_2N=g_1Ng_2N=\pi(g_1)\pi(g_2)$, $\pi$ is a homomorphism.

The $\pi$ define above is called the natural projection of $G$ onto $G/N$. If $\bar{H}\le G/N$, the complete preimage of $\bar{H}$ is the preimage of $\bar{H}$ under $\pi$.

Theorem 8 (Lagrange's Theorem). If $G$ is a finite group and $H\le G$, then $|H|$ is a divisor of $|G|$ and the number of cosets of $H$ in $G$ equals $\dfrac{|G|}{|H|}$.

Proof: Let $|H|=n$ and the number of cosets of $H$ in $G$ equals to $k$. By proposition 4 the left cosets of $H$ in $G$ is a partition of $G$. Consider a map $f:H\to gH$ defined by $f(h)=gh$ for $\forall g\in H$. Obviously, $f$ is a surjection. The left cancellation law implies that $f$ is also injective. Thus $f$ is a bijection, $|gH|=|H|=n$. Since $G$ is partitioned into $k$ disjoint cosets each of which has cardinality $n$, $|G|=kn$, $k=\dfrac{|G|}{|H|}$.

Definition: The number of left cosets of $H$ in $G$ is called the index of $H$ in $G$, written as $|G:H|$.

Corollary 9. If $x$ is a finite group, $\forall x\in G$, $|x|\mid |G|$, $x^{|G|}=1$.

Proof: Since $|x|=|\lang x\rang|$ and $\lang x\rang\le G$, $|x|\mid|G|$.

Corollary 10. If $|G|=p$, where $p$ is a prime number, then $G\cong Z_P$.

Proof: Let $x\ne G$, $x\ne 1$, thus $|x|>1$ and since $|x|\mid|G|$, and $|G|$ is prime, we must have $|x|=|G|$, thus $G=\lang x\rang$ is a cyclic group.

Euler's Theorem: Let $a\in\mathbb{Z},n\in\mathbb{Z}^+$, $\varphi(n)$ be the Euler's $\varphi$-function, if $\gcd(a,n)=1$, then $a^{\varphi(n)}\equiv 1\pmod{n}$.

Proof: Consider the multiplicative group $G=(Z/nZ)^{\times}=\{a\in Z/nZ|(a,n)=1\}$ with operation $ab=a\times b\bmod n$. First $G$ is indeed a group since:

Closeness: If $(a,n)=1,(b,n)=1$, then $(ab,n)=1$.

Identity: $1$.

Associativity: $a(bc)=(ab)c$

Inverse: $\forall a\in G$, $(a,n)=1$, so $\exists x,y\in Z/nZ$ such that $ax+ny\equiv 1\pmod{n}$, so $ax\equiv 1\pmod{n}$ and $a^{-1}=x$ ($x\in G$ since $(x,n)=1$).

By the definition of Euler function, $|G|=\varphi(n)$, thus $a^{\varphi(n)}=a^{|G|}=1$ for $a\in G$.

Fermat's Theorem: If $p$ is a prime, then $a^p\equiv a\pmod{p}$ for $\forall a\in\mathbb{Z}$.

3.18

Definition: Let $H,K\le G$, define $HK=\{hk|h\in H,k\in K\}$. $HK$ may not be a subgroup of $G$.

Property 13. $|HK|=\dfrac{|H||K|}{|H\cap K|}$.

Proof: Note that $HK=\cup_{h\in H}hK$, since each coset of $K$ has $|K|$ elements, it suffices to find the number of distinct left cosets of the form $hK$, where $h\in H$. But $h_1K=h_2K\Leftrightarrow h_1h_2^{-1}\in K$, and $H$ is a group, so $h_1Kh_2K\Leftrightarrow h_1(H\cap K)=h_2(H\cap K)$. So the number of distinct cosets of the form $hK$ is equal to the number of distinct cosets $h(H\cap K)$, which is, by Lagrange's Theorem, $\dfrac{|H|}{|H\cap K|}$. Thus, $|HK|=\dfrac{|H||K|}{|H\cap K|}$.

Property 14. If $H,K\le G$, $HK\le G$ if and only if $HK=KH$.

Proof: Assume first that $HK=KH$ and let $a,b\in HK$, so $a=h_1k_1,b=h_2k_2$ for some $h_1,h_2\in H,k_1,k_2\in K$. Thus $b^{-1}=k_2^{-1}h_2^{-1}$, $ab^{-1}=h_1k_1k_2^{-1}h_2^{-1}$. Let $k_3=k_1k_2^{-1}\in K$, $h_3=h_2^{-1}\in H$, then since $HK=KH$, $k_3h_3=h_4k_4$ for some $h_4\in H, k_4\in K$, $ab^{-1}=h_1h_4k_4\in HK$. Also obviously $1\in HK$, so $HK\le G$.

Conversely, assume $HK\le G$, since $K\le HK,H\le HK$, $KH\subseteq HK$ by closeness of $HK$. Moreover, let $hk\in HK$, write $hk=a^{-1}$ for some $a\in HK$. If $a=h_1k_1$, then $hk=k_1^{-1}h_1^{-1}\in KH$, thus $HK\subseteq KH$. So $HK=KH$.

Corollary 15. If $H\le G,K\le G,H\le N_G(K)$, then $HK\le G$. In particular, if $K\unlhd G$, then $HK\le G$ for any $H\subseteq G$.

Proof: Let $h\in H,k\in K$, by assumption, $hkh^{-1}\in K$, hence $hk=(hkh^{-1})h\in KH$, so $HK\subseteq KH$, similarly, $kh=h(h^{-1}kh)\in HK$, so $KH\subseteq HK$. Hence $HK=KH\le G$.

If $K\unlhd G$, then $N_G(K)=G$, $HK\le G$ for any $H\le G$.

Theorem 16. (The first Isomorphism Theorem) If $G\to H$ is a homomorphism of groups, then $\ker\varphi\unlhd G$, and $G/\ker\varphi\cong\varphi(G)$.

Corollary 17. Let $\varphi:G\to H$ be a homomorphism of groups:

$\varphi$ is injective if and only if $\ker\varphi=1$.
$|G:\ker\varphi|=|\varphi(G)|$.

Proof:

If $\varphi$ is injective, then by definition, $\ker\varphi=1$. Conversely, if $\ker\varphi=1$, $G\cong G/\ker\varphi\cong\varphi(G)$, $\varphi$ is injective.

Obvious from $G/\ker\varphi\cong\varphi(G)$.

Theorem 18. (The second Isomorphism Theorem / Diamond Theorem) Let $A,B\le G$, and $A\le N_G(B)$, then $AB\le G$, $A\cap B\unlhd A$, and $AB/B\cong A/(A\cap B)$.

Proof: $A\le N_G(B)\Rightarrow AB\le G$, since $A\le N_G(B),B\le N_G(B)$. Thus $AB\le N_G(B)$, i.e., $B\unlhd AB$. Define $\varphi:A\to AB/B$ by $\varphi(a)=aB$, then $\varphi$ is a homomorphism and by the definition of $AB$, $\varphi$ is surjective. $\ker\varphi=\{a\in A|aB=1B\}=A\cap B$. Thus $A\cap B\unlhd A$, and $AB/B\cong A/(A\cap B)$ by the first Isomorphism Theorem.

Theorem 19. (The third Isomorphism Theorem) Let $H,K\unlhd G$, and $H\le K$, then $K/H\unlhd G/H$ and $(G/H)/(K/H)\cong G/K$.

Proof: Since $H\unlhd G,K\unlhd G$, then $N_K(H)=K$, $H\unlhd K$. So $K/H$ is well-defined. Obvious $K/H\le G/H$ since $K\unlhd G$. Consider $aH\in G/H$, $\forall a\in G$, $bH\in K/H$, $\forall b\in K$, then $(aH)(bH)(a^{-1}H)=(aba^{-1})H$, so $K/H\unlhd G/H$.

Define $\varphi:G/H\to G/K$ as $\varphi(gH)=gK$. To show $\varphi$ is well-defined, suppose $g_1H=g_2H$. Then $g_1=g_2h$ for some $h\in H$, because $H\le K$, $h\in K$, hence $g_1K=g_2K$, thus $\varphi$ is well-defined. Since $g$ is chosen arbitrarily in $G$, $\varphi$ is a surjective homomorphism. Finally, $\ker\varphi=\{gH\in G/H|\varphi(gH)=1K\}=\{gH\in G/H|g\in K\}=K/H$, according to the first isomorphism theorem, $(G/H)/(K/H)\cong G/K$.

Theorem 20. (The fourth Isomorphism Theorem / Lattice Theorem) Let $N\unlhd G$, then there exists a bijection from the set of subgroups $A$ of $G$ which contain $N$ onto the set of subgroups $\bar{A}=A/N$ of $G/N$.

In particular, every subgroup of $\bar{G}=G/N$ is of the form $A/N$ for some $A\le G$ and $N\le A$. This bijection has the following properties: for $\forall A,B\le G$, with $N\le A,N\le B$:

$A\le B$ if and only if $\bar{A}\le \bar{B}$.
If $A\le B$, then $|B:A|=|\bar{B}:\bar{A}|$.
$\bar{\lang A,B\rang}=\lang\bar{A},\bar{B}\rang$.
$\bar{A\cap B}=\bar{A}\cap\bar{B}$.
$A\unlhd G$ if and only if $\bar{A}\unlhd\bar{G}$.

Cauchy's Theorem: If $|G|<\infty$, and $p$ is a prime dividing $|G|$, then $\exists x\in G$, s,t, $|x|=p$, and $\lang x\rang$ has order $p$.

Property 21 (a weaker version of Cauchy's Theorem). If $G$ is a finite Abelian group and $p$ is a prime dividing $G$, then $G$ contains an element of order $p$.

Proof: The proof proceeds by induction on $|G|$.Namely, we assume the result is valid for every group whose order is smaller than $|G|$ and then prove the result for $G$. Since $|G|>1$, $\exists x\in G$ with $x\ne 1$. If $|G|=1$, then $|x|=p$, done. We therefore assume $|G|>p$. Further suppose $p\mid |x|$, and $|x|=pn$ for some $n\in\mathbb{Z}^+$, $|x^n|=p$, done. Thus we may assume $p\nmid |x|$. Let $N=\lang x\rang$. Since $G$ is Abelian, $N\unlhd G$. by Lagrange's Theorem, $|G/N|=\dfrac{|G|}{|N|}<|G|$. Moreover, since $p\nmid |N|$, $p\mid |G/N|$, by induction, it contains an element $\bar{y}=yN$ of order $p$. Since $y\notin N$, but $y^p\in N$, we must have $\lang y^p\rang\ne\lang y\rang$, $|y^p|<|y|$, thus $p\mid|y|$. Let $z=y^{|y|/p}$, $|z|=p$, done.

Definition: A group $G$ is called simple if $|G|>1$ and the only normal subgroups of $G$ are $1$ and $G$.

Example: If $|G|$ is prime, then $|G|$ is simple.

Simple groups cannot be "factored" into pieces in the form $N$ and $G/N$ and thus they play a role analogous to that of primes in $\mathbb{Z}$.

Definition: In a group $G$, a sequence of subgroups

\[1=N_0\le N_1\le N_2\le N_3\cdots\le N_k=G \]

is called a composition series if $N_i\unlhd N_{i+1}$ and $N_{i+1}/N_i$ is simple. And the quotient groups $N_{i+1}/N_{i}$ are called the composition factors of $G$.

Theorem 22. (Jordan-Holder Theorem) Let $G$ be a finite group with $G\ne 1$, then:

$G$ has a composition series
The composition factors are unique. Namely, if $1=N_0\le N_1\le N_2\le\cdots\le N_r=G,1=M_0\le M_1\le M_2\le\cdots\le R_s=G$ are two composition series, then $r=s$ and there exists a permutation $\pi$ satisfying $M_{\pi(i)}/M_{\pi(i)-1}\cong N_i/N_{i-1}$)

Theorem 23 (Feit-Tompson Theorem) If $G$ is a simple group of odd order, then $G\cong Z_p$ for some prime number $p$.

Solvable groups

Definition. A group $G$ is solvable if there is a chain of subgroups

\[1=G_0\unlhd G_1\unlhd G_2\unlhd\cdots\unlhd G_s=G \]

such that $G_{i+1}/G_i$ is Abelian for $i=0,1,2,\cdots,s-1$.

Theorem 24. The finite group $G$ is solvable if and only if for every divisor $n$ of $|G|$ such that $\gcd(n,\dfrac{|G|}{n})=1$, $G$ has a subgroup of order $n$.

Property 25. Let $N\unlhd G$, then if $N$ and $G/N$ are solvable, then $G$ is also solvable.

Proof: Let $\bar{G}=G/N$, $1=N_0\unlhd N_1\unlhd N_2\unlhd\cdots\unlhd N_n=N$ be a chain of subgroups of $N$ such that $N_{i+1}/N_i$ is Abelian for all $0\le i\lt n$ and $\bar{1}=\bar{G_0}\unlhd\bar{G_1}\unlhd\cdots\unlhd\bar{G_m}=\bar{G}$ be a chain of subgroups of $\bar{G}$ such that $\bar{G_{i+1}}/\bar{G_i}$ is Abelian for all $0\le i<m$. By the Lattice Isomorphism Theorem, there exist subgroups $G_i$ such that $G_i/N=\bar{G_i}$ and $G_i\unlhd G_{i+1}$ for $0\le i<m$. And by the Third Isomorphism Theorem $\bar{G_{i+1}}/\bar{G_i}=(G_{i+1}/N)/(G_i/N)\cong G_{i+1}/G_i$, thus $1=N_0\unlhd N_1\unlhd\cdots\unlhd N_n=N=G_0\unlhd G_1\unlhd \cdots\unlhd G_m=G$ is a chain of subgroups of $G$ whose successive quotient groups are Abelian. Thus $G$ is solvable.

Alternating groups

Definition. A $2$-cycle is called a transposition.

Property. Every element of $S_n$ can be written as a product of transpositions.

Proof: $(a_1,a_2,\cdots,a_m)=(a_1a_m)(a_1a_{m-1})\cdots(a_1a_3)(a_1a_2)$.

Let $x_1,x_2,\cdots,x_n$ be independent variables and $\Delta$ be the polynomial $\Delta=\prod\limits_{1\le i<j\le n}(x_i-x_j)$. For $\sigma\in S_n$, let $\sigma$ act on $\Delta$ by permutating the variables the same way it permutes their indices: $\sigma(\Delta)=\prod\limits_{1\le i<j\le n}(x_{\sigma(i)}-x_{\sigma(j)})$.

For $\forall \sigma\in S_n$, define $\epsilon:S_n\to Z_2\cong\{\pm 1\}$ by:

\[\epsilon(\sigma)=\begin{cases}+1&(\sigma(\Delta)=\Delta)\\-1&(\sigma(\Delta)=-\Delta)\end{cases} \]

Define:

$\epsilon(\sigma$) is called the sign of $\sigma$.
$\sigma$ is called an even permutation if $\epsilon(\sigma)=1$, otherwise it is an odd permutation.

$\epsilon$ is a homomorphism since $\tau\sigma(\Delta)=\prod\limits_{1\le i\lt j\le n}(x_{\tau(\sigma(i))}-x_{\tau(\sigma(j))})=\epsilon(\sigma)\prod\limits_{1\le p\lt q\le n}(x_{\tau(p)}-x_{\tau(q)})=\epsilon(\sigma)\epsilon(\tau)\Delta$.

Define $A_n=\ker(\epsilon)$ called Alternating group (i.e., the set of even permutations). $|A_n|=\dfrac{n!}{2}$.

An $m$-cycle may be written as a product of $m-1$ transpositions, thus an $m$-cycle is odd permutation if and only if $m$ is even.

Similarly the permutation $\sigma$ is odd if and only if the number of cycles of even length in its cycle decomposition is odd.

3.25

Recall: A group action of $G$ on $A$ is a map from $G\times A$ to $A$ (written as $g·a$) such that:

$g_1·(g_2·a)=(g_1g_2)·a$ for $\forall g_1,g_2\in G,a\in A$。
$1·a=a$ for $\forall a\in A$。

For $\forall g\in G$ the map $\sigma_g:A\to A$ defined by $\sigma_g:a\to g·a$ is a permutation of $A$.

The homomorphism $\varphi:G\to S_A$ defined by $\varphi(g)=\sigma_g$ is called the permutation presentation associated to the given action.

Definition:

The kernel of the action: $\{g\in G|g·a=a,\forall a\in A\}$
The stabilizer of $a\in A$: $G_a=\{g\in G|g·a=a\}$.
An action is faithful if and only if its kernel is the identity.

Note that the kernel of an action is precisely the same as the kernel of the associated permutation presentation. In particular, the kernel is a normal subgroup of $G$.

Two group elements induce the same permutation on $A$ if and only if they are in the same coset of the kernel.

Property 1. For any group $G$ and non-empty set $A$, exists a bijection between actions of $G$ on $A$ and the homomorphism of $G$ onto $S_A$.

Definition. For any homomorphism $\varphi:G\to S_A$, we define $g·a=\varphi(g)(a)$ for $\forall g\in G,a\in A$.

Property 2. Let $G$ be a group acting on the non-empty set $A$. The relation on $A$ defined by $a\sim b$ if and only if $a=g·b$ for some $g\in G$, is an equivalence relation. For each $a\in A$, the number of elements in $\bar{a}$ is $|G:G_a|$.

Proof: The relation $\sim$ is an equivalence relation since

$a=1·a$, so $a\sim a$.

If $a\sim b$, then $a=g·b$ for some $g\in G$, $g^{-1}·a=b$, so $b\sim a$.

If $a\sim b,b\sim c$, then $a=g·b,b=h·c$ for some $g,h\in G$, so $a=g·(h·c)=(g·h)·c$, $a\sim c$.

Thus $\sim$ is an equivalence equation.

To prove the last statement, we construct a bijection between the left cosets of $G_a$ in $G$ and the elements of $\bar{a}=\{g·a|g\in G\}$. Suppose $b=g·a$, then $gG_a$ is a left coset of $G_a$ in $G$. The map $b\to gG_a$ is surjective since $\forall g\in G$, $g·a\in\bar{a}$. Moreover, $g·a=h·a$ if and only if $h^{-1}g\in G_a$ if and only if $gG_a=hG_a$, thus the map is also injective, hence a bijection.

Definition: Let $G$ be a group acting on $A\ne\varnothing$

The equivalence class $\{g·a|g\in G\}$ is called the orbit of $G$ containing $a$.
The action is called transitive if there is only one orbit.

Examples: Let $G$ be a group acting on $A$.

If $G$ acts trivially, then $G_a=G$ for all $a\in A$, and the orbits are the elements of $A$. This action is transitive if and only if $|A|=1$.
$S_n$ acts transitively on $A=\{1,2,3,4,\cdots,n\}$. $\forall i\in A$, the stabilizer in $G$ of $i$ has index $n$.
When $G$ acts on $A$, any subgroup of $G$ also acts on $A$. If $G$ is transitive on $A$, then a subgroup of $G$ NEED NOT be transitive on $A$.

Subgroup of symmetric groups are called permutation groups.

We consider $G$ acting on itself by left multiplication $g·a=ga$ for $\forall a,g\in G$.

Let $H\le G$ and $A$ be the set of all left cosets of $H$ in $G$, define an action of $G$ on $A$ by $g·(aH)=(ga)·H$ for all $g\in G,a\in A$.

Theorem 3. Let $G$ be a group, $H\le G $ and $G$ acts by left multiplication on the set $A$ of left cosets of $H$ in $G$. Let $\pi_H$ be the associated permutation representation afforded by this action, then:

$G$ acts transitively on $A$.
The stabilizer in $G$ of the point $1H\in A$ is the subgroup $H$.
The kernel of the action $\ker\pi_H$ is $\cap_{x\in G}xHx^{-1}$, and $\ker\pi_H$ is the largest normal subgroup of $G$ contained in $H$.

Proof:

To see that $G$ acts transitively on $A$, let $aH,bH$ be two elements of $A$, and $g=ba^{-1}$, then $gaH=bH$, and so two arbitrary elements lie in the same orbit, which proves 1.

The stabilizer of the point $1H$, by definition, $\{g\in G|g·1H=1H\}=H$.

By the definition of $\pi_H$ we have

\[\begin{aligned} \ker\pi_H&=\{g\in G|gxH=xH,\forall x\in G\}\\ &=\{g\in G|(x^{-1}gx)H=H,\forall x\in G\}\\ &=\{g\in G|g\in xHx^{-1},\forall x\in G\}\\ &=\cap_{x\in G}xHx^{-1} \end{aligned} \]
In addition, $\ker\pi_H\unlhd G$ and $\ker\pi_H\unlhd H$. If now $N$ is any normal subgroup of $G$ contained in $H$, then we have $N=xNx^{-1}\le xHx^{-1}$, so that $N\le\cap_{x\in G}xHx^{-1}=\ker\pi_H$.

Corollary 4. (Cayley's Theorem) Every group is Isomorphism to a subgroup of some symmetric group.

Proof. Let $H=1$ and apply Theorem 3 to obtain a homomorphism of $G$ onto $S_G$. Since the kernel of this homomorphism is $H=1$, $G$ is Isomorphism to its image in $S_G$.

Corollary 5. If $G$ is a finite group of order $n$ and $p$ is the smallest prime dividing $|G|$, then any subgroup of index $p$ is normal.

Proof. Suppose $H\le G$ and $|G:H|=p$. Let $\pi_H$ be the permutation representation afforded by multiplication on the set of left cosets of $H$ in $G$. Let $K=\ker\pi_H$ and $|H:K|=k$, then $|G:K|=pk$. Since $H$ has $p$ cosets, $G/K$ is isomorphic to a subgroup of $S_p$ by the first Isomorphism Theorem.

By Lagrange's Theorem, $pk=|G/K|$ divides $p!$, thus $k\mid(p-1)!$, but all prime divisors of $(p-1)!$ are less than $p$ and by the minimality of $p$, $k=1$, thus $H=K\unlhd G$.

$G$ acting on itself by conjugation: $g·a=gag^{-1}$ for $\forall a,g\in G$. Two elements $a,b\in G$ are said to be conjugate if there exists some $g\in G$ such that $b=gag^{-1}$. The orbits of $G$ acting on itself by conjugation are called the conjugate classes of $G$.

Define two subsets $S$ and $T$ are said to be conjugate in $G$ if $\exists g\in G$ such that $S=gTg^{-1}$.

Examples:

If $G$ is Abelian, then the action of $G$ on itself by conjugation is trivial, and for each $a\in G$, the conjugate class of $a$ is $\{a\}$.
If $|G|>1$ and $G$ is Abelian, $G$ does not act transitively because $\{1\}$ is always a conjugate class.

Property 6. The number of conjugates of a subset $S$ in a group $G$ is the index of normalizer of $S$, $|G:N_G(S)|$. In particular, the number of conjugates of an element $s$ of $G$ is the index of the centralizer of $s$, $|G:C_G(s)|$.

Proof: By property 2, if $S$ a subset of $G$, the number of conjugates of $S$ equals to $|G:G_S|$. For the action by conjugation $G_S=\{g\in G|gSg^{-1}=S\}=N_G(S)$, so the number of conjugates of $S$ equals to $|G:N_G(S)|$.

Theorem 7. (The class equation) Let $G$ be a finite group and let $g_1,g_2,\cdots,g_r$ be representatives of the distinct conjugacy classes of $G$ not contained in $Z(G)$, then $|G|=|Z(G)|+\sum|G:C_G(g_i)|$.

Theorem 8. If $p$ is a prime and $P$ is a group of prime power order $p^{\alpha}$ for some $\alpha\ge 1$, then $p$ has a non-trivial center: $|Z(P)|\ne 1$.

Proof: By the class equation, $|P|=|Z(P)|+\sum\limits_{i=1}^r|P:C_P(g_i)|$, where $g_1,g_2,\cdots,g_r$ are representatives of the distinct non-central conjugacy classes. By definition, $C_P(g_i)\ne P$ for $i=1,2,\cdots,r$, so $p\mid|P:C_P(g_i)|$. Since $p$ also divides $|P|$, it follows that $p$ divides $|Z(P)|$, so the center must be non-trivial.

Property 10. Let $\sigma,\tau\in S_n$ and suppose $\sigma$ has cycle decomposition $\sigma=(a_1,a_2,\cdots,a_{k_1})(b_1,b_2,\cdots,b_{k_2})\cdots$, then $\tau\sigma\tau^{-1}$ has cycle decomposition $(\tau(a_1),\tau(a_2),\cdots,\tau(a_{k_1}))(\tau(b_1),\tau(b_2),\cdots,\tau(b_{k_2}))\cdots$.

Proof: Note that if $\sigma(i)=j$, then $\tau\sigma\tau^{-1}(\tau(i))=\tau(j)$.

Definition:

If $\sigma\in S_n$ is the product of distinct cycles of length $n_1,n_2,\cdots,n_r$ with $n_1\le n_2\le n_3\le\cdots\le n_r$, then the integers $n_1,n_2,\cdots,n_r$ are called the cycle type of $\sigma$.
If $n\in\mathbb{Z}^+$, a partition is any nondecreasing sequence of position integers whose sum is $n$.

Property 11. Two elements of $S_n$ are conjugate if and only if they have the same cycle type. The number of conjugacy classes of $S_n$ equals the number of partitions of $n$.

4.1

Automorphism: $\varphi:G\to G$ which is also an isomorphism. The set of all automorphisms of $G$ is denoted by $\text{Aut}(G)$:

$\text{Aut}(G)$ is a group under composition of automorphisms.
$\text{Aut}(G)$ is a subgroup of $S_G$.

Property 13. Let $H\unlhd G$, then $G$ acts by conjugation on $H$ as automorphisms of $H$. More specifically, the action of $G$ on $H$ by conjugation is defined for each $g\in G$ by $h\to ghg^{-1}$ for each $h\in H$. For each $g\in G$, conjugation by $g$ is an automorphism of $H$. The permutation representation afforded by this action is a homomorphism of $G$ into $\text{Aut}(H)$ with kernel $C_G(H)$. In particular, $G/C_G(H)$ is isomorphic to a subgroup of $\text{Aut}(H)$.

Proof: Let $\varphi_g$ be the conjugation by $g$. $\varphi_g(H)=H$ since $H\unlhd G$. We have already seen that conjugation defines an action, it follows that $\varphi_1=1$ and $\varphi_a\circ\varphi_b=\varphi_{ab}$ for $\forall a,b\in G$. Thus each $\varphi_g$ gives a bijection from $H$ to itself since it has a 2-sided inverse $\varphi_{g^{-1}}$. In addition: $\varphi_g(hk)=g(hk)g^{-1}=ghg^{-1}ggkg^{-1}=\varphi_g(h)\varphi_g(k)$ for $\forall h,k\in H$, hence $\varphi_g$ is also a homomorphism. This proves that conjugation by any fixed element of $G$ defines an automorphism of $H$. The permutation representation $\psi:G\to S_H$ defined by $\psi(g)=\varphi(g)$ has imaged contained in the subgroup $\text{Aut}(H)$ of $S_H$. Finally, $\ker\psi=C_G(H)$. By the first automorphism theorem $G/C_G(H)\cong\psi(G)\le\text{Aut}(H)$.

Corollary 14. If $K\le G$, then $\forall g\in G$, $K\cong gKg^{-1}$.

Corollary 15. $\forall H\le G$, $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\text{Aut}(H)$. In particular, $G/Z(G)$ is isomorphic to a subgroup of $\text{Aut}(G)$.

Proof: Since $H\unlhd N_G(H)$, by Property 13 with $N_G(H)$ playing the role of $G$, the first assertion follows directly. The second assertion is obvious by letting $H=G$.

Definition. Let $G$ be a group, $g\in G$. Conjugation by $g$ is called an an inner automorphism of $G$. The subgroup of $\text{Aut}(G)$ consisting of all inner automorphisms is denoted by $\text{Inn}(G)$.

By Corollary 15, $\text{Inn}(G)\cong G/Z(G)$.

Examples:

Since $Z(Q_8)=\lang -1\rang$, we have $\text{Inn}(Q_8)\cong V_4$.
Since $Z(D_8)=\lang -1\rang$, we have $\text{Inn}(D_8)\cong V_4$.

Definition. A subgroup $H\le G$ is called characteristic in $G$, denoted by $H\text{char} G$, if every automorphism of $G$ maps $H$ to itself, i.e, $\sigma(H)=H$, $\forall\sigma\in \text{Auto}(G)$.

Characteristic subgroups are normal, since $\forall g\in G,gHg^{-1}=H$.
If $H$ is the unique subgroup of $G$ of a given order then $H\text{char} G$.
If $K\text{char}H$ and $H\unlhd G$, then $K\unlhd G$.

Property 16. The automorphism group of the cyclic group of order $n$ is isomorphic to $(Z/nZ)^{\times}$, an Abelian group of order $\varphi(n)$.

Proof: Reminder, $(Z/nZ)^{\times}=\{a\in Z/nZ|(a,n)=1\}$. Let $Z_n=\lang x\rang$, if $\psi\in\text{Aut}(Z_n)$, then $\psi(x)=x^{a}$ for some $a\in\mathbb{Z}$, each $a$ determines the $\psi$, denoted by $\psi_a$. $|x|=n$, then integer $a$ is only defined by $\bmod n$. And since $|\psi(a)|=n$, $(a,n)=1$. The map $x\to x^a$ is an automorphism of $Z_n$. Hence, we have a surjective map $\Psi:\text{Aut}(Z_n)\to(Z/nZ)^{\times},\psi_a\to a\bmod n$. The map $\Psi$ is a homomorphism because $\psi_a\cdot\psi_b(x)=\psi_a(x^b)=x^{ab}=\psi_{ab}(x)$ for $\forall\psi_a,\psi_b\in\text{Aut}(Z_n)$. So $\Psi(\psi_a\psi_b)=\Psi(\psi_{ab})=ab\bmod n=\Psi(\psi_a)\Psi(\psi_b)$, $\Psi$ is an isomorphism.

Sylow Theorem

Definition: Let $G$ be a group and $p$ be a prime:

A group of order $p^{\alpha}$ for some $\alpha\ge 0$ is called a $p$-group.
If $|G|=p^{\alpha}m$, where $p\nmid m$, then a subgroup of order $p^{\alpha}$ is called is called a Sylow $p$-subgroup.
The set of Sylow $p$-subgroups of $G$ will be denoted by $\text{Syl}_p(G)$ and the number of Sylow $p$-subgroups of $G$ will be denoted by $n_p(G)$.

Theorem 18 (Sylow's Theorem):

Let $G$ be a group of order $p^{\alpha}m$, where $p$ is prime not dividing $m$

Sylow $p$ subgroup exists.
If $P$ is a Sylow $p$-subgroup of $G$ and $Q$ is any $p$-subgroup of $G$, then $\exists g\in G$, such that $Q\le gPg^{-1}$.
$n_p(G)\equiv 1\pmod{p}$ and $n_p\mid m$.

Corollary 20. Let $P$ be a Sylow $p$-subgroup of $G$, the following are equivalent:

$P$ is the unique Sylow $p$-subgroup of $G$, i.e., $n_p(G)=1$.
$P\unlhd G$.
$P\text{char}G$.
All subgroups generated by elements of $p$-power order are $p$-groups, i.e, if $X$ is any subset of $G$, s.t., $|x$| is a power of $p$ for $\forall x\in X$, then $\lang X\rang$ is a $p$-group.

Proof:

(1)->(2): $gPg^{-1}=P$ for $\forall g\in G$, thus $P\unlhd G$.

(2)->(1): If $P\unlhd G$, then $\forall Q\in\text{Syl}_p(G)$, $\exists g\in G$, such that $Q=gPg^{-1}=P$, so $Q=P$.

(3)->(2): Trivially holds because characteristic subgroups are always normal.

(2)->(3): Since (2) implies (1), $P$ is the unique group with order $p^{\alpha}$, so $P\text{char}G$.

(1)->(4): Suppose $X$ is a subset of $G$ such that $|x|$ is a power of $p$ for all $x\in X$, by the conjugacy part of the Sylow's Theorem, for each $x\in X$ there is some $g\in G$ that $x\in gPg^{-1}=P$, thus $X\subseteq P$, so $\lang X\rang\le P$, according to the Lagrange's Theorem, $|X|\mid|P|$, so $X$ is a $p$-group.

(4)->(1): Let $X$ be the union of all Sylow $p$ subgroups of $G$. If $P$ is any $p$-subgroup, $P$ is a subgroup of the $p$-subgroup $\lang X\rang$. Since $P$ is a $p$-subgroup of the maximal order, we must have $P=\lang X\rang$.

Product of Groups

Definition. The direct product $G_1\times G_2\times G_3\times\cdots\times G_n$ of the groups $G_1,G_2,\cdots,G_n$ with operators $*_1,*_2,\cdots,*_n$ respectively, is the group of $n$-tuples $(g_1,g_2,\cdots,g_n)$ with operation defined as $(g_1,g_2,\cdots,g_n)*(h_1,h_2,\cdots,h_n)=(g_1*_1h_1,g_2*_2h_2,\cdots,g_n*_nh_n)$.

Property 1. If $G=G_1\times G_2\times G_3\times\cdots G_n$, then $|G|=|G_1||G_2||G_3|\cdots|G_n|$.

Property 2.

$G_i\cong\{(1,1,1,\cdots,g_i,1,\cdots,1)\}$, $\{(1,1,1,\cdots,g_i,1,\cdots,1)\}\unlhd G$, and $G/G_i\cong G_1\times G_2\cdots\times G_{i-1}\times G_{i+1}\times\cdots\times G_n$.
For each fixed $i$, define $\pi_i:G\to G_i$ by $\pi(g_1,g_2,\cdots,g_n)=g_i$, then $\pi_i$ is a surjective homomorphism, with $\ker\pi_i\cong\times G_{i-1}\times G_{i+1}\times\cdots\times G_n$
Under the identification in part (1), if $x\in G_i,y\in G_j$ for some $i\ne j$, then $xy=yx$.

4.8

Definition. Let $G$ be a group, $\forall x,y\in G,A,B\subseteq G$.

Commutator of $x$ and $y$, $[x,y]=x^{-1}y^{-1}xy$.
$[A,B]=\lang [a,b]|a\in A,b\in B\rang$.
$G'=\lang[x,y]|x,y\in G\rang$

$[x,y]=1$ if and only if $xy=yx$. Commutators measure the "difference" between $xy$ and $yx$.

Property 7. Let $G$ be a group, $x,y\in G$ and $H\le G$, then:

$xy=yx[x,y]$.
$H\unlhd G$ if and only if $[H,G]\le H$.
$\sigma([x,y])=[\sigma(x),\sigma(y)]$ for any automorphism of $\sigma$. $G'\text{char}G$ and $G/G'$ is Abelian.
If $H\unlhd G$ and $G/H$ is Abelian, then $G'\le H$. Conversely, if $G'\le H$, then $H\unlhd G$ and $G/H$ is Abelian.
If $\varphi:G\to A$ is any homomorphism into an Abelian group $A$, then $\varphi$ factors through $G'$, i.e., $G'\le\ker\varphi$.

Proof:

Trivial.

$H\unlhd G\Leftrightarrow\forall g\in G,h\in H,ghg^{-1}\in H\Leftrightarrow h^{-1}g^{-1}hg\in H\Leftrightarrow[H,G]\le H$

Let $\sigma\in\text{Aut}(G)$, then $\forall x,y\in G$, $\sigma([x,y])=\sigma(x^{-1}y^{-1}xy)=\sigma(x)^{-1}\sigma(y)^{-1}\sigma(x)\sigma(y)=[\sigma(x),\sigma(y)]$. Thus for every $[x,y]$ of $G'$, $\sigma[x,y]$ is again a commutator, i.e., $\sigma(G')=G'$. So $G'\text{char}G$. Let $xG',yG'\in G/G'$, then $(xG')(yG')=(xy)G'=(yx[x,y])G'=(yx)G'=(yG')(xG')$. So $G/G'$ is Abelian.

Suppose $H\unlhd G$, and $G/H$ is Abelian, then for $\forall x,y\in G$, $(xH)(yH)=(yH)(xH)$, so $1H=(xH)^{-1}(yH)^{-1}(xH)(yH)$ since $xH$ and $yH$ commute, but $(xH)^{-1}(yH)^{-1}(xH)(yH)=[x,y]H$, so $[x,y]\in H$, thus $G'\le H$. Conversely, if $G'\le H$, then since $G/G'$ is Abelian, every subgroup of $G/G'$ is normal. In particular, $H/G'\unlhd G/G'$, so $H\unlhd G$. According to the third Isomorphism Theorem, $G/H\cong (G/G')/(H/G')$ is Abelian.

This theorem is (4) phrased in language of homomorphism.

Example:

$G$ is Abelian if and only if $G'=1$.
$D_8'=\lang r^2\rang,Q_8'=\lang -1\rang$

Property 8. Let $H,K$ be subgroups of the group $G$. The number of distinct ways of writing each element of $HK$ in the form $hk$ for some $h\in H,k\in K$ is $|H\cap K|$. In particular, if $|H\cap K|=1$, then each element can be written uniquely as a product of $hk$.

Property 9. (recognition theorem) Suppose $H,K\le G$, such that $H,K\unlhd G$, and $H\cap K=1$, then $HK\cong H\times K$.

Proof: Since we know that if $K\unlhd G$, then $\forall H\le G$, $HK\le G$, thus $HK\le G$ in this case. Let $h\in H,k\in K$, since $H\unlhd G$, $k^{-1}hk\in H$, so $h^{-1}(k^{-1}hk)\in H$. Similarly $(h^{-1}k^{-1}h)k\in K$. Since $H\cap K=1$ it follows that $hk=kh$. By Property 8, each element of $HK$ can be written uniquely as a product $hk$, with $h\in H,k\in K$. Thus the map $\varphi:HK\to H\times K$ as $\varphi(hk)=(h,k)$ is well defined. In addition $\varphi(h_1k_1h_2k_2)=(h_1h_2,k_1k_2)=\varphi(h_1k_1)\varphi(h_2k_2)$, so $\varphi$ is a homomorphism. Obviously, $\varphi$ is a bijective, thus an isomorphism.

Semidirect Product

Theorem 10. Let $H,K$ be groups, and $\varphi:K\to\text{Aut}(H)$ be a homomorphism. Let $\cdot$ denote the left action of $K$ act on $H$ determined by $\varphi$, i.e. $k\cdot h=\varphi(k)(h)$ for $k\in K,h\in H$. Let $G$ be the set of ordered pairs $(h,k)$ with $h\in H,k\in K$ and define $(h_1,k_1)(h_2,k_2)=(h_1(k_1·h_2),k_1k_2)$.

This multiplication makes $G$ into a group of order $|K||H|$.
The sets $\{(h,1)|h\in H\}$ and $\{(1,k)|k\in K\}$ are subgroups of $G$, and the maps $h\to(h,1)$ and $k\to(1,k)$ are the isomorphisms of these groups with $H,K$, respectively. Identifying $H,K$ with their isomorphic copies in $G$ we have:
$H\unlhd G$.
$H\cap K=1$
$\forall h\in H$ and $k\in K$, $khk^{-1}=k·h=\varphi(k)(h)$.

Proof: It is straightforward to check that $G$ is a group under this multiplication using the fact that $\cdot$ is an action of $K$ on $H$:

Associativity:

\[\begin{aligned} ((a,x)(b,y))(c,z)&=(ax\cdot b,xy)(c,z)\\ &=(ax\cdot bx\cdot(y\cdot c),xyz)\\ &=(ax\cdot (by\cdot c),xyz)\\ &=(a,x)(by\cdot c,yz)\\ &=(a,x)((b,y),(c,z)) \end{aligned} \]

Identity: $(1,1)$.

Inverse: $(h,k)^{-1}=(k^{-1}\cdot h^{-1},k^{-1})$ since $(h,k)(k^{-1}\cdot h^{-1},k^{-1})=(hk\cdot(k^{-1}\cdot h^{-1}),kk^{-1})=(h1\cdot h^{-1},1)=(1,1)$.

Obviously $|G|=|H||K|$, which proves (1).

Denote $\bar{H}=\{(h,1)|h\in H\}$ and $\bar{K}=\{(1,k)|k\in K\}$. We have $(a,1)(b,1)=(ab,1)$ for $\forall a,b\in H$ and $(1,x)(1,y)=(1,xy)$ for $\forall x,y\in K$. Thus $\bar{H},\bar{K}\le G$ and the maps in (2) are isomorphisms.

It is clear that $\bar{H}\cap\bar{K}=1$, which is (4). Now $(1,k)(h,1)(1,k)^{-1}=(k\cdot h,k)(1,k^{-1})=(k\cdot h,1)$. So identifying $(h,1)$ with $h$ and $(1,k)$ with $k$ we have $khk^{-1}=k\cdot h$, which is (5). Finally, from above, $K\le N_G(H)$. Since $G=HK$ and certainly $H\le N_G(H)$, we have $N_G(H)=H$, i.e., $H\unlhd G$, which proves (3).

The group described above is called the semidirect product of $H$ and $K$ with respect to $\varphi$, denoted by $H\rtimes_{\varphi}K$ ($H\rtimes K$ is $\varphi$ is clear).

Property 11. Let $H,K$ be groups, and $\varphi:K\to\text{Aut}(H)$ be homomorphism. The following are equivalent:

The identity map between $H\rtimes K$ and $H\times K$ is a group homomorphism.
$\varphi$ is the trivial homomorphism from $K$ to $\text{Aut}(H)$.
$K\unlhd H\rtimes K$.

Proof:

(1)->(2): By the definition of $H\rtimes K$, $(h_1,k_1)(h_2,k_2)=(h_1k_1\cdot h_2,k_1k_2)$ for $\forall h_1,h_2\in H$ and $k_1,k_2\in K$. By (1), $(h_1,k_1)(h_2,k_2)=(h_1h_2,k_1k_2)$, so $h_1k_1\cdot h_2=h_1h_2$, $k_1\cdot h_2=h_2$. So $K$ acts trivially on $H$, which proves (2).

(2)->(3): If $\varphi$ is trivial, then the action of $K$ on $H$ is trivial, so that the elements of $H$ commute with those by $K$ by Theorem 10(5). In particular, $H$ normalizes $K$. Since $K$ normalizes itself, $G=HK$ normalizes $K$.

(3)->(1): If $K\unlhd H\rtimes K$, then for $\forall h\in H,k\in K$, $[h,k]\in H\cap K=1$. Thus $hk=kh$ and the action of $K$ on $H$ is trivial. The multiplication in the semidirect product is the same as that in the direct product, which prove (1).

posted @ 2024-02-28 22:20 tzc_wk 阅读(169) 评论(2) 编辑收藏举报

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