Abstract alg-2, Normal subgp.
Given $H \le G$ : gp,
ex) $\mathbb{Z}=G$, $H=3 \mathbb{Z}$. $\Rightarrow ( \mathbb{Z}/3 \mathbb{Z} , + )$.
$G/H = \{ gH \mid g \in G \}$
$\Rightarrow gH=\{gh \mid h \in H \}$=the "set" of left cosets of $H$ in $G$. (but not a "gp", in general)
Want : make $G/H$ to be a group.
i.e., $\forall gH$, $g'H \in G/H$,
$gH \cdot g'H \overset{def}{:=} gg'H$.
ex) $4 \mathbb{Z} \le \mathbb{Z}$, $\mathbb{Z} / 4 \mathbb{Z} = \{ [0] , [1] , [2] , [3] \}$.
$[0] + [1] = [1]$
$[1] + [2] = [3]$
$[1] + [3] = [4] = [0]$
Question.
given $\begin{cases} xH = x'H & x \ne x' \\ yH = y'H & y \ne y' \end{cases}$
$\Rightarrow$
$xH \cdot yH \overset{\text{must be}}{=} x'H \cdot y'H$
$xyH \overset{(?)}{=} x'y'H$
Prop.
Let $H \le G$. Then the group operation in $G/H$
is well-defined if and only if $yxy^{-1} \in H$, $\forall y \in G$, $\forall x \in H$
Proof.
($\Rightarrow$) Let $xH=H$. ($xH=\{xh \mid h \in H \} \rightarrow x \cdot e = x \in H$)
By assumption,
$\rightarrow xH \cdot yH = xyH$ (gp operation is well-defined)
$\rightarrow xH \cdot yH = H \cdot yH = yH$ ($xH=H$)
$\Rightarrow xyH = yH$
$y^{-1}xyH = H$
$\iff y^{-1}xy \in H$ (Coset 파트 Prop.)
i.e., $\forall x \in H$, $\forall y \in G$, $y^{-1} xy \in H$.
($\Leftarrow$) Let $xH=x'H$, $yH=y'H$. $\iff x(x')^{-1} \in H$, $y(y')^{-1} \in H$.
claim : $xyH = x'y'H$.
NTS : if $x(x')^{-1}$, $y(y')^{-1} \in H$, then $xy(x'y')^{-1} \in H$.
$xy(x'y')^{-1}$
$=xy(y')^{-1}(x')^{-1}$
$=xy(y')^{-1}x^{-1}x(x')^{-1}$
$\Rightarrow \underbrace{x\underbrace{y(y')^{-1}}_{\in H}x^{-1}}_{\in H \text{ by } (\Leftarrow)} \underbrace{x(x')^{-1}}_{\in H} \in H$. (증명 완료)
Define.
We say $H \le G$ is normal-subgp
if $\forall g \in G$, $\forall h \in H$, $ghg^{-1} \in H$. Write $H \trianglelefteq G$.
Prop.
$H \trianglelefteq G \iff \forall g \in G$, $gHg^{-1} = h$ ($gHg^{-1} = \{ ghg^{-1} \mid h \in H \}$)
$\iff \forall g \in G, gH =Hg$ $(gH = \{gh \mid h \in H \}$, $Hg = \{ hg \mid h \in H \})$
Left Coset = Right Coset
Proof. Exercise.
Prop.
Let $\phi : G \to G'$ be a gp homo.
$\text{Ker } \phi \trianglelefteq G$.
Proof.
NTS : $\forall g \in G$, $\forall h \in \text{Ker } \phi$, $ghg^{-1} \overset{?}{\in} \text{Ker } \phi$.
$\iff \phi (ghg^{-1})$
$= \phi(g) \phi(h) \phi(g^{-1})$ $(\phi(h)=e \ \ (\because h \in \text{Ker } \phi))$
$= \phi(g) \phi(g^{-1})$
$= \phi(g) (\phi(g))^{-1} = e$. (커널 파트 Prop.) (증명 완료)
Q. Given $f : G \to H$ gp homo, $\text{Ker } f \le G$, $\text{Im } f \le H$.
By previous prop, $\text{Ker } f \trianglelefteq G$.
Then $\text{Im } f \overset{?}{\trianglelefteq} H$
Ans. No, in general.
counter_example) Ler $S_3 = \{ f : \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \to \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \mid f \text{ 1-1, onto} \}$ (symmetry gp)
Take $\sigma \overset{denote}{=} (1 \ 2) \in S_3$.
$(1 \ 2)$ : 1은 2로 보내고 2는 1로 보내고 나머지는 자기 자신으로 보낸다는 의미
$\rightarrow <\sigma> = \{ \sigma, \text{id} \} \le S_3$.
(2번 합성하면 자기 자신으로 돌아오기 때문에 원소가 2개뿐)
claim : $<\sigma>$ is not a normal subgp of $S_3$.
i.e., $<\sigma> \ntrianglelefteq S_3$.
Take $\tau = (1 \ 2 \ 3)$
$(1 \ 2 \ 3) \sigma (1 \ 2 \ 3 )^{-1} = \tau \sigma \tau^{-1} $
(아래 그림에서 왼쪽은 $\sigma$, 오른쪽은 $\tau$)
$= (2 \ 3) \notin <\sigma>$.
If any $\text{Im }f \trianglelefteq H$, for $H \le G$,
consider
$i : H \to G$ (포함함수 : inclusion map - 자기 자신으로 대응하는 함수) (1-1, gp homo)
$x \mapsto x$
Then $\text{Im } i = H \overset{if}{\trianglelefteq} G$. i.e., any subgp $H$ of $G$ must be a normal subgp of $G$. (???)
Abstract alg-2, Normal subgp-2.
Recall that given sets $A$, $B$, $C \subset A$, $D \subset B$,
$f : A \to B$ a fn.
Then
1) $f^{-1}(f(C)) \ge C$
2) $f^{-1}(f(C)) = C$ if $f$ is 1-1
Sometimes, even though $f$ is not 1-1, $f^{-1}(f(C))=C$.
Thm1.
$f : G \to G'$ gp homo.
Assume $\text{Ker }f \subset H \le G$ then $f^{-1}(f(H)) = H$.
Proof.
($\supset$) True, in general.
($\subset$) : Let $x \in f^{-1}(f(H))$ $(f(H)= \{y \mid f(y) \in f(H) \})$
$\overset{def}{\iff} f(x) \in f(H)$
$\iff \exists h \in H$ s.t $f(x)=f(h)$.
$\iff \underset{=f(xh^{-1})}{f(x)f(h)^{-1}}=e$
$\Rightarrow xh^{-1} \in \text{Ker } f \subset H$.
$\Rightarrow \exists h' \in H$ s.t $xh^{-1} = h'$.
$\Rightarrow x=hh' \in H$.
Hence if $x \in f^{-1}(f(H))$ then $x \in H$. (증명 완료)
Thm2.
Let $\phi : G \twoheadrightarrow G'$ a gp homo and onto.
Consider
$A := \{H \subset G \mid \text{Ker } \phi \subset H \le G \}$
$B := \{H' \subset G' \mid H' \le G' \}$.
Define
$\Phi : A \to B$
$H \mapsto \phi(H)$
Then $\Phi$ is 1-1, onto.
i.e., if $H' \in B$, then $\Phi^{-1}(H') = \phi^{-1}(H')$.
Proof.
$\forall H \le G$, $\phi(H) \le G'$.
Hence $\Phi$ is well-defined.
claim : $\Phi$ is 1-1.
Let $\Phi(H_1)=\Phi(H_2) \iff \phi(H_1) = \phi(H_2)$
by Thm1, $\begin{cases} \phi^{-1}(\phi(H_1))=H_1 \\ \phi^{-1}(\phi(H_2))=H_2 \end{cases}$
i.e., $H_1=H_2$.
claim : $\Phi$ is onto.
Take any $H' \le G'$.
(We need to find $H \le G$ s.t $\text{Ker } \phi \subset H \le G$ and
$\phi(H)=H'$).
choose $H_o = \phi^{-1} (H')$
Note that $\Phi^{-1}(H') \le G$ and $\text{Ker }\phi \subset \phi^{-1}(H')$
$\because$ If $x \in \text{Ker }\phi$, $\phi(x)=e_{G'} \in H'$ i.e., $x \in \phi^{-1}(H')$
Thus $H_o \in A$.
By thm1, $\phi^{-1}(\phi(H_o))=H_o$.
By surjectivity of $\phi$, $\phi(\underbrace{\phi^{-1}(H')}_{H_o}) = H'$
Thus $\Phi(H_o) \overset{def}{=} \phi(H_o) = H'$
Thus $\Phi^{-1}$ is well-defined and $\Phi^{-1}(H')=H_o=\phi^{-1}(H')$. (증명 완료)
Rmk.
Let $\phi : G \twoheadrightarrow G'$ be a gp homo.
$\tilde{A} : \{ H \subset G \mid \text{Ker } \phi \subset H \trianglelefteq G \}$
$\tilde{B} : \{ H' \subset G' \mid H' \trianglelefteq G' \}$
Define
$\Phi : \tilde{A} \to \tilde{B}$
$H \mapsto \phi(H)$
$\text{Ker }\phi \subset H \trianglelefteq G \Rightarrow \underset{=\Phi(H)}{\phi(H)} \trianglelefteq \phi(G) \overset{\phi : onto}{=} G'$
This shows $\Phi$ is well-defined.
By the similar argument, $\Phi$ is 1-1, onto.
i.e., $\exists$ 1-1 correspondence between
$\{ H \subset G \mid \text{Ker }\phi \subset H \trianglelefteq G \} \leftrightarrow \{H' \subset G' \mid H' \trianglelefteq G' \}$
$H \mapsto \phi(H)$
provided $\phi : G \twoheadrightarrow G'$ surj. gp. homo.
Def.
We say a group $G$ is simple if there is
no normal subgp except $\{e\}$ and $G$.
Def.
$M \trianglelefteq G$ is a maximal normal subgp $M$ of $G$.
if $M \ne G$, and $\exists N \subset G$ s.t $M \subsetneq N \trianglelefteq G \Rightarrow N=G$.
Prop.
Let $M \trianglelefteq G$.
$G/M$ is simple if and only if $M$ is a maximal normal subgp.
Lemma.
Let $N \trianglelefteq G$,
$\pi : G \twoheadrightarrow G/N$ surj. gp. homo.
$g \mapsto gN$
($\text{Ker } \pi = \{ g \in G \mid gN=N \} = N$)
Then
$\Phi : \{ H \subset G \mid \underset{=N}{\text{Ker }\pi} \subset H \trianglelefteq G \}$
$\rightarrow \{H' \subset G/N \mid H' \trianglelefteq G/N \}$ is 1-1, onto.
Proof.
It follows from the above Rmk. (증명 완료)
$\{ \text{(normal) subgps of } G \text{ containing N} \}$
$\underset{correspondence}{\overset{1-1}{\longleftrightarrow}} \{ \text{(normal) subgp of }G/N \}$
Proof of Prop.
$\pi : G \to G/M$
$g \mapsto gM$
Let $G/M$ be simple.
i.e., there are only two normal subgps.
$\{ e \} = M$ and $G/M$ of $G/M$,
By the 1-1 correspondence
there are only two normal subgps of $G$
Containing $M$ $\Rightarrow$ $M$, $G$.
This means $M$ is a maximal noraml subgp of $G$. (증명 완료)
Isomorphism theorems.
Statements of isomorphism thms
1st isomorphism thm.
Let $\varphi : G \to H$ be a gp homo.
($\text{Ker }\varphi \trianglelefteq G \Rightarrow G/ \text{Ker }\varphi$ is a gp) (set of cosets)
Then
$\bar{\varphi} : G / \text{Ker }\varphi \cong \text{Im }\varphi$ (isomorphic)
$g \text{Ker }\varphi \mapsto \varphi(g)$
In particular, $\bar{\varphi}$ is well-defined and is a gp isomorphism.
2nd isomorphism thm.
Let $H \le G$ and $K \trianglelefteq G$
Then $H / H \cap K \cong HK/K$
3rd isomorphism thm.
$H, K \trianglelefteq G, K \le H ( \iff K \trianglelefteq H)$
$\Rightarrow G/K \Big/ H/K \cong G/H$
Proof of 1st iso thm.
claim : $\bar{\varphi}$ is well-defined.
Let $g \text{Ker }\varphi = g' \text{Ker }\varphi \iff g(g')^{-1} \in \text{Ker }\varphi$
$\iff \varphi(g(g')^{-1}) = e_H$
$\iff \varphi(g)\varphi(g')^{-1} = e_H$
$\iff \varphi(g)=\varphi(g')$ (well-defined)
claim : $\bar{\varphi}$ is a gp homo.
write $g \text{Ker }\varphi = \bar{g}$.
$\bar{\varphi}(\bar{g} \cdot \bar{g'}) = \bar{\varphi} (\bar{gg'}) \overset{\text{def of }\bar{\varphi}}{=}\varphi(gg') \overset{\varphi \text{ gp homo}}{=} \varphi(g) \varphi(g') = \bar{\varphi}(\bar{g})\bar{\varphi}(\bar{g'})$
claim : $\bar{\varphi}$ is 1-1.
If $\bar{\varphi}(\bar{g}) = \bar{\varphi}(\bar{g'})$, then $\varphi(g) = \varphi(g')$
$\Rightarrow \varphi (g(g')^{-1})= e_H$, $g(g')^{-1} \in \text{Ker }\varphi \iff \underbrace{g \text{Ker }\varphi}_{=\bar{g}} = \underbrace{g' \text{Ker }\varphi}_{\bar{g'}}$
claim : $\bar{\varphi}$ is onto. its $\text{Im }\varphi$
Take any $\varphi(g) \in \text{Im }\varphi$. Then $\bar{\varphi}(\bar{g}) = \varphi(g)$.
e.g) $G=(\mathbb{C} - \{ 0 \}, \times)$ gp.
Let
$\varphi : (\mathbb{R} , +) \to S^1= \{ \mathbb{Z} \in \mathbb{C} \mid |z|=1 \}$
$x \to e^{2 \pi i x}$
Then $\varphi$ is a gp homo.
$\varphi(x + y) = e^{2\pi i (x+y)} = e^{2\pi i (x)} \cdot e^{2\pi i (y)}=\varphi(x) \varphi(y)$
$\text{Ker } \varphi$
$= \{ x \in \mathbb{R} \mid \varphi(x) = 1 \}$ ($S^1$의 identity =1)
$= \{ x \in \mathbb{R} \mid e^{2\pi i x} = 1 \} = \{x \in \mathbb{R} \mid \cos 2 \pi x + i \sin 2 \pi x = 1 \}$ (오일러 공식)
$=\mathbb{Z}$.
By 1st iso thm, $\mathbb{R}/\mathbb{Z} \cong S^1$. (onto map이여서 이미지가 전체)
Prop.
$N \trianglelefteq G$, $f:G \to H$ be a gp homo.
Then $\bar{f} : G/N \to H$ is well-defined
$gN \mapsto f(g)$
if and only if $N \subset \text{Ker }f$.
Proof.
($\Rightarrow$, well-defined) Let $g \in N \Rightarrow gN=N$
($\Leftarrow$) Let $N \subset \text{Ker }f$.
Let $g_1 N = g_2 N \iff g_1 g_2^{-1} \in N \subset \text{Ker }f$
$\Rightarrow f(g_1 g_2^{-1}) = e_H$
$\iff f(g_1)=f(g_2)$
$\iff \bar{f}(g_1 N) = \bar{f} (g_2 N)$
Cor.
$f : G \to H$ gp homo. $M \trianglelefteq G$, $K \trianglelefteq H$.
Then
$\bar{f} : G/M \to H/K$
$gM \mapsto f(g)K$
is well-defined iff $f(M) \subset K$.
Pf. $\bar{f}$ is well-defined
iff $g_1 M = g_2 M \Rightarrow f(g_1) K = f(g_2)K$
$\iff g_1 g_2^{-1} \in M \Rightarrow f(g_1)f(g_2)^{-1} \in K$
$\iff g_1 g_2^{-1} \in M \Rightarrow f(g_1 g_2^{-1}) \in K$
$\iff f(M) \subset K$.
Proof of 2nd isomorphism thm.
Let $H \le G$ and $K \trianglelefteq G$
Consider
$i : H \to G$ (inclusion map)
$\pi : G \to G/K$ (projection map) gp homos.
Let $\pi \circ i : \underset{\xrightarrow{\pi \circ i}}{H \xrightarrow{i} G \xrightarrow{\pi} G/K}$ : gp homo.
Then
$\text{Ker }(\pi \circ i) = \{ h \in H \mid hK = K \}$
($\because$
$(\pi \circ i)(h) = e_{G/K}$ ($e_{G/K}$는 $G/K$의 항등원이므로 $e_{G/K} = K$)
$\pi(i(h))= \pi(h)$(inclusion map)
$\pi(h)=hK=e_{G/K}$
$\therefore$
$hK=K$)
$= H \cap K \trianglelefteq H$ ($hK=K$인 $h$를 집합으로 만든것은 결국 H와 K의 교집합을 의미)
Then by 1st iso thm,
$H / H \cap K \cong \text{Im }(\pi \circ i)= \{ hK \mid h \in H\}$
$= \{h_k K \mid h \in H , k \in K \}$
$= HK/K$
i.e., $H/H \cap K \cong HK/K$.
Proof of 3rd isomorphism Thm.
Let $H, K \trianglelefteq G$, $K \le H (\leftrightarrow K \trianglelefteq H)$
Consider $\text{id}_G : G(\trianglelefteq K) \to G(\trianglelefteq H)$
Then by Cor,
$\bar{\text{id}} : G/K \to G/H$ is well-defined
$gK \mapsto gH$ iff $\text{id}(K) \subset H \iff K \subset H$.
Note that $\bar{\text{id}}$ is surjective.
and $\text{Ker}(\bar{\text{id}}) = \{ gK \mid gH = H \}$
$= \{ gK \in G/K \mid g \in H \}$
$= H/K$.
Thus by 1st iso thm,
$G/K \Big/ \underset{=\text{Ker}(\bar{\text{id}})}{H/K} \cong \underset{=\text{Im}(\bar{\text{id}})}{G/h}$.
