用户: Endeavour/抽象代数/Groups

1Sets, functions and relations

Definition 1.1. A semigroup is a nonempty set $G$ together with a binary operation on $G$ which is
(i) associative: $a(bc)=(ab)c$ for all $a,b,c\in G$ a monoid is a semigroup $G$ which contains $a$ and
(ii) two sided identity element: $e\in G$ such that $ae=ea=a$ for all $a\in G$
a group is a monid $G$ such that
(iii)inverse element: for every $a\in G$ there exists a two-sided inverse element $a^{-}1a=a{a}^{-}1=e$ for all $a\in G$
A semigroup is said to be abelian or commutative if its binary operation satisfies
(iv) commutative: $ab=ba$ for all $a,b\in G$
The order of a group $G$ is the cardinal number $|G|$

Theorem 1.2. Properties of groups:
(i)$c\in G$ and $cc=c⟹c=e$
(ii)for all$a,b,c\in Gab=ac⟹b=c$ and $ba=ca⟹b=c$(left and right cancellation)
(iii)for each $a\in G$, the inverse element $a^{-1}$ is unique
(iv)for each $a\in G,(a^{-1}{)}^{-1}=a$
(v)for $a,b\in G,(ab{)}^{-1}=b^{-1}{a}^{-1}$
(vi)for $a,b\in G$ the equations $ax=b$ and $ya=b$ have unique solutions in $G:x=a^{-1}b$ and $y=b{a}^{-1}$
(vii)Let $G$ be a semigroup, Then $G$ is a group if and only if the following hold:
(1)there exists an element $e\in G$ such that $ea=a$ for all $a\in G$
(2)for each $a\in G$ there exists an element $a^{-1}\in G$ such that $a^{-1}a=e$
(viii)Let $G$ be a semigroup. Then $G$ is a group if and only if for all $a,b\in G$ the equations $ax=b$ and $ya=b$ have solutions in $G$

Exercise 1.3. Examples of Group:
1.Group of symmetries of the square $D_4=\{R,R^2,R^3,I,T_x,T_y,T_{1,3},T_{2,4}\}$
2.Group of permutations $A(S)$, the set of all bijections $S\to S$ under the composition of functions operation. In particular when $S=\{1,2,...,n\}$, $A(S)$ is called the symmetric group on n letters and denoted $S_n$.
3.Group of rationals modulo one $\mathbb{Q}/\mathbb{Z}$ with congruence relation$$a\sim b⟺a-b\in \mathbb{Z}$$
4.Group of integers modulo m $\mathbb{Z}/m\mathbb{Z}=\{\bar{0},\bar{1},...,\bar{m}\}$
5.Prufergroup $\mathbb{Z}(p^{\infty })=\{\overline{a/b}\in \mathbb{Q}/\mathbb{Z}|a,b\in \mathbb{Z}andb=p^{i}forsomei\ge 0\}$, an infinite group under the addition of $\mathbb{Q}/\mathbb{Z}$
6.The Klein-four group ${\mathbb{Z}}_2\oplus {\mathbb{Z}}_2=\{(\bar{0},\bar{1}),(\bar{1},\bar{1}),(\bar{0},\bar{0}),(\bar{1},\bar{0})\}$

A recurrence definition of the product of elements:
The standard n product ${\prod }_{i=1}^n{a}_{i}of{a}_1,...,a_n$:$$\prod _{i=1}^1{a}_i=a_1;andforn>1,\prod _{i=1}^n{a}_i=(\prod _{i=1}^{n-1}{a}_i)a_n$$

2Section 2

Exercise 2.1. Examples of morphism:
1.The map $f:\mathbb{Z}\to {\mathbb{Z}}_m$ given by $x\mapsto \bar{x}$ and $g:\mathbb{Q}\to \mathbb{Q}/Z$ given by $r\mapsto \bar{r}$ is an epimorphism of additive groups. (canonical epimorphism of $\mathbb{Z}\to {\mathbb{Z}}_m$)
2.If A is an abelian group, then $f:A\to A,a\mapsto a^{-1}$ is an automorphism of A, $a\mapsto a^2$ is an endomorphism of A.
3.Let $1<m,k\in \mathbb{N},themapg:{\mathbb{Z}}_m\to {\mathbb{Z}}_{m,k},\bar{x}\mapsto \overline{kx}$ is a monomorphism.

Definition 2.2. Let $f:G\to H$ be a homomorphism of groups. The kernel of $f$ is$$\ker f=\{a\in G|f(a)=e_H\}$$if A is a subset of $G$, then the image of $A$ is$$f(A)=\{b\in H|b=f(a)\text{mbox}forsomea\in A\}$$$f(G)$ is called the image of $f$ and deonted $\text{mbox}imf$.

Theorem 2.3. Let $f:G\to H$ be a homomorhpism of groups, then
(i)$f$ is a monomorphism if and only if $\ker f=e$.
(ii)$f$ is an isomorphism if and only if there is a homomorphism $f^{-1}:H\to G$ such that $f{f}^{-1}=1_H$ and $f^{-1}f=1_G$.

Sketch of proof:
(i)If $\ker f=\{e_H\}$ and $f(a)=f(b)$, then$$e_H=f(a)f^{-1}(b)=f(a{b}^{-1})\in \ker f$$so that $a{b}^{-1}\in \ker f$, therefore $a{b}^{-1}=e$ e.g. $a=b$ and by definition f is a monomorphism.
(ii)Trivial.

Theorem 2.4. Properties of subgroup: Let $H$ be a nonempty subset of a group G. Then H is a subgroup of G if and only if $a{b}^{-1}\in H$ for all $a,b\in H$.

Proof:
The identity $e=a{a}^{-1}$, so for any $b$ there exists $b^{-1}=e{b}^{-1}\in H$, if $a,b\in H$, then $b^{-1}\in H$ and hence $ab=a{b}^{-1}\in H$, since $G$ is a group thus the product is associative.

Exercise 2.5. Examples of subgroup:
(i)The set of all multiples of some fixed integer $n$ is a subgroup of $\mathbb{Z}$, which is isomorphic to $\mathbb{Z}$ (Let $S=\{na|a=1,2,...\}$, let $f:S\to \mathbb{Z}$)
(i)
(ii)

Definition 2.6. Let $H$ be a subgroup of a group $G$ and $a,b\in G$. $a$ is right congruent to $b$ modulo $H$, denoted $a{\equiv }_{r}b(\mathrm{m}\mathrm{o}\mathrm{d}H)$ if $a{b}^{-1\in H}$. $a$ is left congruent to $b$ modulo $H$, denoted $a{\equiv }_{l}b(\mathrm{m}\mathrm{o}\mathrm{d}H)$ if $a^{-1}b\in H$.

Definition 2.7. Let $H$ be a subgroup of a group $G$.
(i)Right [resp. left] congruence modulo $H$ is an equivalence relation on $G$.
(ii)The equivalence class of $a\in G$ under right [resp. left] congruence modulo $H$ is the set $Ha=\{ha|h\in H\}$ [resp. $aH=\{ah|h\in H\}$].
(iii)$|Ha|=|H|=|aH|$ for all $a\in G$.

e.g. Consider the group of residule class modulo $m$, where $G=\mathbb{Z}$, $H={\mathbb{Z}}_m$, the operation is $a\mathrm{m}\mathrm{o}\mathrm{d}m$, then $Ha=\{a\mathrm{m}\mathrm{o}\mathrm{d}h|h\in {\mathbb{Z}}_m\}$. for instance, $\bar{0}=\{0\mathrm{m}\mathrm{o}\mathrm{d}0,0\mathrm{m}\mathrm{o}\mathrm{d}1,...,0\mathrm{m}\mathrm{o}\mathrm{d}(m-1)\}$, this implies $\mathbb{Z}/m\mathbb{Z}=\{\bar{0},\bar{1},...,\overline{m-1}\}$. hence $Ha$ is clearly an equivalence class of $a\in G$.

Corollary 2.8. Let $H$ be a subgroup of a group $G$.
(i) $G$ is the union of the right [resp. left] cosets of $H$ in $G$.
(ii) Two right [resp. left] cosets of $H$ in $G$ are either disjoint or equal.
(iii) For all $a,b\in G$, $Ha=Hb⟺a{b}^{-1}\in H$ and $aH=bH⟺a^{-1}b\in H$.
(iv) If $\mathcal{R}$ is the set of distinct right cosets of $H$ in $G$ and $\mathcal{L}$ is the set of distinct left cosets of $H$ in $G$, then $|\mathcal{R}|=|\mathcal{L}|$.

Proof: (i) to (iii) are trivial. (iv) Consider the map $\mathcal{R}\to \mathcal{L}$ by $Ha\mapsto a^{-1}H$.

Definition 2.9. Let $H$ be a subgroup of a group $G$. The index of $H$ in $G$, denoted $[G:H]$ is the cardinal number of the set of distinct right [resp. left] cosets of $H$ in $G$.

Lemma 2.10 (Lemma to prove the first isomorphism thoerem).
If $f:G\to H$ is a homomorphism of groups, $N⊴G$, $N\subset \ker f$, then there exists an unique $\bar{f}:G/N\to H$ s.t. $\bar{f}(aN)=f(a)$ for all $a\in G$, $\text{im}f=\text{im}\bar{f}$ and $\ker \bar{f}=(\ker f)/N$, their relation can be expressed as the following commuataive diagram:$\bar{f}$ is an isomorphism $⟺f$ surjective and $N=\ker f$.

Corollary 2.11. Let $f:G\to H$ be a homomorphism of groups, $N⊴G,M⊴H$ and $f(N)\le M$, then there exists an unique homomorphism $\bar{f}:G/N\to H/M$ given by $aN\mapsto f(a)M$, in other words the following diagram is commutative: \xymatrix{ G \ar[r]^f \ar[d] & H \ar[d] \\ G/N \ar[r]_{\bar{f}} & H/M } $\bar{f}$ is an isomorphism $⟺$ $\text{im}f\vee M=⟨\text{im}f\cup M⟩=H$ and $f^{-1}(M)\subset N$.
In particular, $f$ is an epimorphism s.t. $f(N)=M$ and $\ker f\subset N⟹\bar{f}$ is an isomorphism.

The corollary states that quotient then map is equivalent to map then quotient.

Intuition about the first isomorphism theorem
First isomorphism theorem: If $f:G\to H$ is a homomorphism of groups, then $f$ induces an isomorphism $G/\ker f\cong \text{im}f$
Intuition: The first isomorphism theorem could construct isomorphisms from homomorphisms.
If $f:G\to H$ is a homomorphism but not injective, then we will construct a monomorphism $\bar{f}:N\to \text{im}f$ where N is a group, since any homomorphism $g$ has the property $\ker g=e⟺g$ is an monomorphism, so $N=G/H,H=\ker f$. By the lemma, $\bar{f}$ is a isomorphism since $\text{im}\tilde{f}=\text{im}\bar{f}$ and $H=\ker f$, where $\tilde{f}:G\to \text{im}f$ is an epimorphism, therefore we constructed an isomorphism $\bar{f}$ such that $G/\ker f\cong \text{im}f$.