Group, review

The aim of this blog is that give a brief review and lifting some concepts in group theory.

Pre

One important source of the group comes from a description of the symmetry of an object.

Let $\mathcal{C}$$\mathcal C$ be a locally small category, for example, $\mathrm{Set},\mathrm{Top},{\mathrm{Vect}}_{\mathbb{F}}...$$\mathrm{Set},\mathrm{Top},\mathrm{Vect}_{\mathbb F}...$ and their subcategory.

Let $X\in \mathrm{Ob}(\mathcal{C}),{\mathrm{Aut}}_{\mathcal{C}}(X)$$X\in\mathrm{Ob}(\mathcal C),\mathrm{Aut}_{\mathcal C}(X)$ form a group, and this group describe the symmetry of $X$$X$ in $\mathcal{C}$$\mathcal C$.

Let ${\mathcal{C}}^{\prime }\subset \mathcal{C}$$\mathcal C'\subset \mathcal C$ be a subcategory and $X\in \mathrm{Ob}(C^{\prime })$$X\in\mathrm{Ob}(C')$ as well, ${\mathrm{Aut}}_{C^{\prime }}(X)$$\mathrm{Aut}_{C'}(X)$ is a subgroup of ${\mathrm{Aut}}_C(X)$$\mathrm{Aut}_C(X)$.

Let $\mathcal{D}$$\mathcal D$ be another category. $\mathcal{F}:\mathcal{C}\to \mathcal{D}$$\mathcal F:\mathcal C\to\mathcal D$ be a functor.

Then $\mathcal{F}:{\mathrm{Aut}}_{\mathcal{C}}(X)\to {\mathrm{Aut}}_D(\mathcal{F}X)$$\mathcal F:\mathrm{Aut}_\mathcal{C}(X)\to\mathrm{Aut}_D(\mathcal FX)$ is a group homomorphism.

Hom functor

Let $\mathcal{C}$$\mathcal C$ be a category, ${\mathrm{Hom}}_{\mathcal{C}}(-,-):{\mathcal{C}}^{\mathrm{op}}\times \mathcal{C}\to \mathrm{Set}$$\mathrm{Hom}_{\mathcal C}(-,-):\mathcal C^{\mathrm{op}}\times\mathcal C\to\mathrm{Set}$, $(X,Y)⟼{\mathrm{Hom}}_{\mathcal{C}}(X,Y)$$(X,Y)\longmapsto\mathrm{Hom}_{\mathcal C}(X,Y)$ define a bifunctor.

For the morphism $(f,g):(X,Y)\to (X^{\prime },Y^{\prime })$$(f,g):(X,Y)\to(X',Y')$,

$$\begin{array}{}\text{(1)}& \begin{array}{c}{\mathrm{Hom}}_{\mathcal{C}}(X,Y)⟶{\mathrm{Hom}}_{\mathcal{C}}(X^{\prime },Y^{\prime })\\ \varphi ⟼g\varphi f.\end{array}\end{array}$$

Definition. Bi-Coset.

Let $G$$G$ be a group, $H,K\subseteq G$$H,K\subseteq G$. Then $HxK$$HxK$ is called bi-coset

Proposition. $HxK\cap HyK\ne \mathrm{\varnothing }⟺HxK=HyK$$HxK\cap HyK\ne\empty\iff HxK=HyK$.

Proof. If $HxK\cap HyK\ne \mathrm{\varnothing }$$HxK\cap HyK\ne\empty$. Let $hxk=h^{\prime }y{k}^{\prime }$$hxk=h'yk'$, $x=h^{-1}{h}^{\prime }y{k}^{\prime }{k}^{-1}\in HyK$$x=h^{-1}h'yk'k^{-1}\in HyK$. Hence $HxK\subseteq HyK.$$HxK\subseteq HyK.$ By symmetry, $HxK=HyK$$HxK=HyK$.

Proposition. For each $HxK$$Hx K$, pick one element $x$$x$ to represent that, then $G=\underset{x}{⨆}HxK$$G=\bigsqcup_x HxK$.

Proof. Obviously.

Definition. Centre

Let $G$$G$ be a group, the centre of $G$$G$ is

$$\begin{array}{}\text{(2)}& Z(G):=\{z\in G|\mathrm{\forall }x\in G,zx=xz\}\end{array}$$

Notice that $Z(-):\mathrm{G}\mathrm{r}\mathrm{p}\to \mathrm{A}\mathrm{b}$$Z(-):\rm Grp\to Ab$ is a functor.

Definition. Inner automorphism.

Let $G$$G$ be a group, for $x\in G$$x\in G$,

$$\begin{array}{}\text{(3)}& \begin{array}{c}{\mathrm{Ad}}_x:G⟶G\\ g{⟼}^{x}g:=xg{x}^{-1}\end{array}\end{array}$$

Give us an element of ${\mathrm{Aut}}_{\mathrm{Grp}}(G)$$\mathrm{Aut}_{\mathrm{Grp}}(G)$

We call that inner automorphism of $G$$G$.

Notice that

$$\begin{array}{}\text{(4)}& {\mathrm{Ad}}_1={\mathrm{id}}_G,{\mathrm{Ad}}_{xy}(g)=xyg{y}^{-1}{x}^{-1}={\mathrm{Ad}}_x\circ {\mathrm{Ad}}_y(g)\end{array}$$

Hence we induce a group homomorphism

$$\begin{array}{}\text{(5)}& \begin{array}{c}\mathrm{Ad}:G\to \mathrm{Aut}(G)\\ x⟼{\mathrm{Ad}}_x\end{array}\end{array}$$

The group of inner automorphism is

$$\begin{array}{}\text{(6)}& \mathrm{Inn}(G)\end{array}$$

The kernel of $\mathrm{A}\mathrm{d}$$\rm Ad$ is $Z(G)$$Z(G)$, since if $x\in Z(G)$$x\in Z(G)$, then $\mathrm{\forall }g\in G,{\mathrm{Ad}}_x(g)=xg{x}^{-1}=g$$\forall g\in G,\mathrm{Ad}_x(g)=xgx^{-1}=g$.

By the first isomorphism theorem,

$$\begin{array}{}\text{(7)}& \mathrm{Inn}(G)\cong \frac{G}{Z(G)}\end{array}$$

Proposition.

$$\begin{array}{}\text{(8)}& \mathrm{Inn}(G)◃{\mathrm{Aut}}_{\mathrm{Grp}}(G)\end{array}$$

Proof.

We need to prove that $\varphi \mathrm{Inn}(G){\varphi }^{-1}=\mathrm{Inn}(G)$$\phi\mathrm{Inn}(G)\phi^{-1}=\mathrm{Inn}(G)$.

$$\begin{array}{}\text{(9)}& \varphi \circ {\mathrm{Ad}}_x\circ {\varphi }^{-1}(y)=\varphi (x{\varphi }^{-1}(y)x^{-1})=\varphi (x)y\varphi (x{)}^{-1}={\mathrm{Ad}}_{\varphi (x)}\in \mathrm{Inn}(G)◻\end{array}$$

Let $\mathcal{C}\mathcal{,}\mathcal{D}$$\mathcal{C,D}$ be two locally small category and let $X\in \mathrm{Ob}(\mathcal{C}),Y\in \mathrm{Ob}(\mathcal{D})$$X\in\mathrm{Ob}(\mathcal C),Y\in\mathrm{Ob}({\mathcal D})$.

Let $F,G:\mathcal{C}\to \mathcal{D}$$F,G:\mathcal C\to\mathcal D$ be two functors, in particular, let us consider two subcategory ${\mathrm{Aut}}_{\mathcal{C}}(X),{\mathrm{Aut}}_{\mathcal{D}}(Y)$$\mathrm{Aut}_{\mathcal C}(X),\mathrm{Aut}_\mathcal D(Y)$.

Assume $F(X)=Y=G(X)$$F(X)=Y=G(X)$, then $F,G$$F,G$ induce two group homomorphism from ${\mathrm{Aut}}_{\mathcal{C}}(X)$$\mathrm{Aut}_{\mathcal C}(X)$ to ${\mathrm{Aut}}_{\mathcal{D}}(Y)$$\mathrm{Aut}_\mathcal D(Y)$​.

Easy to see that ${\mathrm{Aut}}_{\mathcal{C}}(X)$$\mathrm{Aut}_{\mathcal C}(X)$ and ${\mathrm{Aut}}_{\mathcal{D}}(Y)$$\mathrm{Aut}_{\mathcal D}(Y)$ are two categories. Now let us view $F,G:{\mathrm{Aut}}_{\mathcal{C}}(X)\to {\mathrm{Aut}}_{\mathcal{D}}(Y).$$F,G:\mathrm{Aut}_{\mathcal C}(X)\to\mathrm{Aut}_\mathcal D(Y).$

If there exists a natural transformation between $F,G$$F,G$, then it should looks like

$$\begin{array}{}\text{(10)}& \begin{array}{c}\begin{array}{ccc}F(X)& \stackrel{F(f)}{\to }& F(X)\\ g\downarrow & & \downarrow g& \\ G(X)& \stackrel{G(f)}{\to }& G(X)\end{array}\end{array}\end{array}$$

Here $g\in \mathrm{Mor}({\mathrm{Aut}}_{\mathcal{D}}(Y))$$g\in\mathrm{Mor}(\mathrm{Aut}_{\mathcal D}(Y))$, hence invertible. i.e. every natural transformation is natural isomorphism.

$$\begin{array}{}\text{(11)}& Gg=gF⟹G=gF{g}^{-1}\end{array}$$

Definition. centralizers and normalizers.

Let $E$$E$ be a subset of $G$$G$, the centralizer of $E$$E$ is

$$\begin{array}{}\text{(12)}& Z_G(E):=\{z\in G:\mathrm{\forall }x\in E,zx=xz\}\end{array}$$

The nomoralizer of $E$$E$ is

$$\begin{array}{}\text{(13)}& N_G(E):=\{n\in G:{\mathrm{Ad}}_n(E)=E\}\end{array}$$

Proposition. Both $Z_G(E)$$Z_G(E)$ and $N_G(E)$$N_G(E)$ are subgroups of $G$$G$.

Proof.

Both $Z_G(E)$$Z_G(E)$ and $N_G(E)$$N_G(E)$ is nonempty, since they both involve identity.

Let $x,y\in Z_G(E),\mathrm{\forall }a\in E,xya=x(ya)=x(ay)=(xa)y=axy$$x,y\in Z_G(E),\forall a\in E,xya=x(ya)=x(ay)=(xa)y=axy$

$$\begin{array}{}\text{(14)}& a=ea=(x^{-1}x)a=x^{-1}(xa)=x^{-1}ax=a⟹x^{-1}a=a{x}^{-1}\end{array}$$

Notice that $Z_G(E)$$Z_G(E)$ is not necessary to be an abelian group.

For example, Suppose that $G\notin \mathrm{Ab}$$G\notin\mathrm{Ab}$, let $E=Z(G)$$E=Z(G)$, Then $Z_G(E)=\{z\in G:\mathrm{\forall }x\in Z(G),zx=xz\}=G$$Z_G(E)=\{z\in G:\forall x\in Z(G),zx=xz\}=G$

For $N_G(E)$$N_G(E)$, Let $x,y\in N_G(E),{\mathrm{Ad}}_{xy}(E)={\mathrm{Ad}}_x\circ {\mathrm{Ad}}_y(E)=E$$x,y\in N_G(E),\mathrm{Ad}_{xy}(E)=\mathrm{Ad}_x\circ\mathrm{Ad}_y(E)=E$. $◻$$\square$

Easy to see that

$$\begin{array}{}\text{(15)}& Z_G(E)=\bigcap _{x\in E}{Z}_G(x)\end{array}$$

Hence

$$\begin{array}{}\text{(16)}& E_1\subseteq E_2⟹Z_G(E_1)\supseteq Z_G(E_2)\end{array}$$