Hw1：证明 “$-$−” 在$\mathbb{R},\mathbb{Q},\mathbb{Z}$R,Q,Z上不是交换或结合的二元运算。

证明：
$\left\{ \begin{aligned} & \varphi \left( a,b \right)=b-a \\ & \varphi \left( b,a \right)=a-b \\ \end{aligned} \right.${​φ(a,b)=b−aφ(b,a)=a−b​
$\Rightarrow \varphi \left( a,b \right)-\varphi \left( b,a \right)=2\left( a-b \right)\cancel{\equiv }0$⇒φ(a,b)−φ(b,a)=2(a−b)≡​0,
thus ‘$-$−’ is not commutative on $\mathbb{R},\mathbb{Q},\mathbb{Z}$R,Q,Z.
$\left\{ \begin{aligned} & \varphi \left( \varphi \left( a,b \right),c \right)=\left( a-b \right)-c=a-b-c \\ & \varphi \left( a,\varphi \left( b,c \right) \right)=a-\left( b-c \right)=a-b+c \\ \end{aligned} \right.${​φ(φ(a,b),c)=(a−b)−c=a−b−cφ(a,φ(b,c))=a−(b−c)=a−b+c​
$\Rightarrow \varphi \left( \varphi \left( a,b \right),c \right)-\varphi \left( a,\varphi \left( b,c \right) \right)=-2c\cancel{\equiv }0,$⇒φ(φ(a,b),c)−φ(a,φ(b,c))=−2c≡​0,
thus ‘$-$−’ is not associative on $\mathbb{R},\mathbb{Q},\mathbb{Z}$R,Q,Z.

Hw2：$\Re$ℜ是$A$A上的一个等价关系，令${{\left\{ {{K}_{\alpha }} \right\}}_{\alpha }}${Kα​}α​为$\Re$ℜ所给出的等价类构成的集合，证明${{\left\{ {{K}_{\alpha }} \right\}}_{\alpha }}${Kα​}α​是$A$A的一个分划。

证明：

1. To prove $\forall {{K}_{i}}\in {{\left\{ {{K}_{\alpha }} \right\}}_{\alpha \in A}}$∀Ki​∈{Kα​}α∈A​, ${{K}_{i}}\ne \varnothing$Ki​​=∅.
   $\forall {{K}_{i}}\in \left\{ {{K}_{\alpha }} \right\}$∀Ki​∈{Kα​}, we see the definition of ${{K}_{i}}$Ki​ is
   ${{K}_{i}}:=\left\{ \left. x\in A \right|x\Re i \right\}.$Ki​:={x∈A∣xℜi}.
   Because $\Re$ℜ is self-reflexive, we have
   $i\Re i\text{ }\Rightarrow \text{ }i\in {{K}_{i}}\text{ }\Rightarrow \text{ }{{K}_{i}}\ne \varnothing.$iℜi ⇒ i∈Ki​ ⇒ Ki​​=∅.
2. To prove that $\forall {{K}_{i}},{{K}_{j}}\in \left\{ {{K}_{\alpha }} \right\},\text{ }i\ne j,\text{ }{{K}_{i}}\bigcap {{K}_{j}}=\varnothing .$∀Ki​,Kj​∈{Kα​}, i​=j, Ki​⋂Kj​=∅.
   Directly we have
   $i\in {{K}_{i}}\text{ }\wedge \text{ }j\in {{K}_{j}}.$i∈Ki​ ∧ j∈Kj​.
   And because ${{K}_{i}}\ne {{K}_{j}}$Ki​​=Kj​, we have
   $i\notin {{K}_{j}}\text{ }\wedge \text{ }j\notin {{K}_{i}}.$i∈/​Kj​ ∧ j∈/​Ki​.
   Thus firstly we have
   $i,j\notin \left( {{K}_{i}}\bigcap {{K}_{j}} \right).$i,j∈/​(Ki​⋂Kj​).
   Then consider $\forall k\in A\backslash \left\{ i,j \right\}$∀k∈A\{i,j}, all the possibilities are below.
     2.1:
   $\left( k\notin {{K}_{i}}\wedge k\notin {{K}_{j}} \right)\text{ }\vee \text{ }\left( k\in {{K}_{i}}\wedge k\notin {{K}_{j}} \right)\text{ }\vee \text{ }\left( k\notin {{K}_{i}}\wedge k\in {{K}_{j}} \right)$(k∈/​Ki​∧k∈/​Kj​) ∨ (k∈Ki​∧k∈/​Kj​) ∨ (k∈/​Ki​∧k∈Kj​)
   $\Rightarrow \text{ }k\notin \left( {{K}_{i}}\bigcap {{K}_{j}} \right)$⇒ k∈/​(Ki​⋂Kj​)
     2.2:
   $k\in {{K}_{i}}\wedge k\in {{K}_{j}}\text{ }$k∈Ki​∧k∈Kj​ 
   $\begin{aligned} & \Rightarrow \text{ }kRi\text{ }\wedge \text{ }kRj\text{ } \\ & \Rightarrow \text{ }iRk\wedge kRj \\ & \Rightarrow \text{ }iRj \\ & \Rightarrow \text{ }i\in {{K}_{j}} \\ \end{aligned}$​⇒ kRi ∧ kRj ⇒ iRk∧kRj⇒ iRj⇒ i∈Kj​​
   which is contradictious with $i\notin {{K}_{j}}$i∈/​Kj​.
   Thus $\forall k\in A\backslash \left\{ i,j \right\}$∀k∈A\{i,j}, $k\notin \left( {{K}_{i}}\bigcap {{K}_{j}} \right)$k∈/​(Ki​⋂Kj​).
   $\left\{ \begin{aligned} & {{K}_{i}}\bigcap {{K}_{j}}\subset A \\ & i,j\notin \left( {{K}_{i}}\bigcap {{K}_{j}} \right) \\ & k\notin \left( {{K}_{i}}\bigcap {{K}_{j}} \right),\text{ }\forall k\in A\backslash \left\{ i,j \right\} \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }{{K}_{i}}\bigcap {{K}_{j}}=\varnothing$⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​​Ki​⋂Kj​⊂Ai,j∈/​(Ki​⋂Kj​)k∈/​(Ki​⋂Kj​), ∀k∈A\{i,j}​ ⇒ Ki​⋂Kj​=∅.
3. To prove that $A=\underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}$A=α⋃​Kα​. The proposition is equal to
   $A\subset \underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}\text{ }\wedge \text{ }\underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}\subset A.$A⊂α⋃​Kα​ ∧ α⋃​Kα​⊂A.
     3.1 To prove that $\underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}\subset A.$α⋃​Kα​⊂A.
   $\begin{aligned} & \forall t\forall i\left( t\in {{K}_{i}}\text{ }\to \text{ }t\in A \right) \\ & \Rightarrow \text{ }\forall {{K}_{i}}\in \left\{ {{K}_{\alpha }} \right\},\text{ }{{K}_{i}}\subset A \\ & \Rightarrow {\mathop{\bigcup }}\,{{K}_{\alpha }}\subset A \\ \end{aligned}$​∀t∀i(t∈Ki​ → t∈A)⇒ ∀Ki​∈{Kα​}, Ki​⊂A⇒⋃Kα​⊂A​.
     3.2 To prove that $A\subset \underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}.$A⊂α⋃​Kα​.
     Due to $\underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}\subset A$α⋃​Kα​⊂A proved in 3.1, we divide $A$A into $A-\underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}$A−α⋃​Kα​ & $\underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}$α⋃​Kα​, and we have
   $A=\left( A-\underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }} \right)\bigcup \left( \underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }} \right).$A=(A−α⋃​Kα​)⋃(α⋃​Kα​).
     Consider $\forall k\in A$∀k∈A, there are two possibilities:
       3.2.1 $\exists {{K}_{i}}\in \left\{ {{K}_{\alpha }} \right\},\text{ }k\in {{K}_{i}}$∃Ki​∈{Kα​}, k∈Ki​, then obviously we have
   $k\in \underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}\subset A.$k∈α⋃​Kα​⊂A.
       3.2.2 $\forall {{K}_{i}}\in \left\{ {{K}_{\alpha }} \right\},\text{ }k\notin {{K}_{i}}$∀Ki​∈{Kα​}, k∈/​Ki​, then we have
   $k\notin \underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}\text{ }\Rightarrow \text{ }k\in \left( A-\underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }} \right)$k∈/​α⋃​Kα​ ⇒ k∈(A−α⋃​Kα​)
       So we can let ${{K}_{k}}:=\left\{ \left. x\in \left( A\backslash \underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }} \right) \right|x\Re k \right\}$Kk​:={x∈(A\α⋃​Kα​)∣∣∣∣​xℜk}, then we have
   $k\in {{K}_{k}}\text{ }\Rightarrow \text{ }k\in \underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }},$k∈Kk​ ⇒ k∈α⋃​Kα​,
       which is contradictious with $k\notin \underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}$k∈/​α⋃​Kα​.
       Thus we prove that $A\subset \underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}$A⊂α⋃​Kα​.
     Thus we prove that $A=\underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }}$A=α⋃​Kα​.
   In conclusion, we have proved
   $\left\{ \begin{aligned} & {{K}_{i}}\ne \varnothing ,\text{ }\forall {{K}_{i}}\in {{\left\{ {{K}_{\alpha }} \right\}}_{\alpha \in A}} \\ & {{K}_{i}}\bigcap {{K}_{j}}=\varnothing ,\text{ }\forall {{K}_{i}},{{K}_{j}}\in \left\{ {{K}_{\alpha }} \right\},\text{ }i\ne j \\ & A=\underset{\alpha }{\mathop{\bigcup }}\,{{K}_{\alpha }} \\ \end{aligned} \right.$⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​​Ki​​=∅, ∀Ki​∈{Kα​}α∈A​Ki​⋂Kj​=∅, ∀Ki​,Kj​∈{Kα​}, i​=jA=α⋃​Kα​​
   thus ${{\left\{ {{K}_{\alpha }} \right\}}_{\alpha \in A}}${Kα​}α∈A​ is a partition of $A$A.

Hw3：画出$S_3$S3​的群表。

On $\left\{ 1,2,3 \right\}${1,2,3} define permutation ${{f}_{ijk}}:=\left\{ \begin{aligned} & 1\to i \\ & 2\to j \\ & 3\to k \\ \end{aligned} \right.,\text{ }i\ne j\ne k$fijk​:=⎩⎪⎨⎪⎧​​1→i2→j3→k​, i​=j​=k.
Then we have the group table of ${{S}_{3}}$S3​:
$\begin{matrix} {} & {{f}_{123}} & {{f}_{132}} & {{f}_{213}} & {{f}_{231}} & {{f}_{312}} & {{f}_{321}} \\ {{f}_{123}} & {{f}_{123}} & {{f}_{132}} & {{f}_{213}} & {{f}_{231}} & {{f}_{312}} & {{f}_{321}} \\ {{f}_{132}} & {{f}_{132}} & {{f}_{123}} & {{f}_{231}} & {{f}_{213}} & {{f}_{321}} & {{f}_{312}} \\ {{f}_{213}} & {{f}_{213}} & {{f}_{312}} & {{f}_{123}} & {{f}_{321}} & {{f}_{132}} & {{f}_{231}} \\ {{f}_{231}} & {{f}_{231}} & {{f}_{321}} & {{f}_{132}} & {{f}_{312}} & {{f}_{123}} & {{f}_{213}} \\ {{f}_{312}} & {{f}_{312}} & {{f}_{213}} & {{f}_{321}} & {{f}_{123}} & {{f}_{231}} & {{f}_{132}} \\ {{f}_{321}} & {{f}_{321}} & {{f}_{231}} & {{f}_{312}} & {{f}_{132}} & {{f}_{213}} & {{f}_{123}} \\ \end{matrix}$f123​f132​f213​f231​f312​f321​​f123​f123​f132​f213​f231​f312​f321​​f132​f132​f123​f312​f321​f213​f231​​f213​f213​f231​f123​f132​f321​f312​​f231​f231​f213​f321​f312​f123​f132​​f312​f312​f321​f132​f123​f231​f213​​f321​f321​f312​f231​f213​f132​f123​​

Hw4：$G$G是一个群，证明若${{H}_{1}},{{H}_{2}}<G$H1​,H2​<G，则${{H}_{1}}\bigcap {{H}_{2}}<G$H1​⋂H2​<G.

证明：

1. ${{H}_{1}},\text{ }{{H}_{2}}<G\text{ }\Rightarrow \text{ }{{H}_{1}},{{H}_{2}}\subseteq G\text{ }\Rightarrow \text{ }{{H}_{1}}\bigcap {{H}_{2}}\subseteq G$H1​, H2​<G ⇒ H1​,H2​⊆G ⇒ H1​⋂H2​⊆G；
2. ${{H}_{i}}<G\text{ }\Rightarrow \text{ }\forall {{a}_{i}},{{b}_{i}}\in {{H}_{i}},\text{ }{{a}_{i}}{{b}_{i}}\in {{H}_{i}},\text{ }i=1,2$Hi​<G ⇒ ∀ai​,bi​∈Hi​, ai​bi​∈Hi​, i=1,2,
   thus $\forall a,b\in \left( {{H}_{1}}\bigcap {{H}_{2}} \right)$∀a,b∈(H1​⋂H2​) we have
   $a,b\in {{H}_{1}}\text{ }\wedge \text{ }a,b\in {{H}_{2}}\text{ }\Rightarrow \text{ }ab\in {{H}_{1}}\text{ }\wedge \text{ }ab\in {{H}_{2}}\text{ }\Rightarrow \text{ }ab\in \left( {{H}_{1}}\bigcap {{H}_{2}} \right)$a,b∈H1​ ∧ a,b∈H2​ ⇒ ab∈H1​ ∧ ab∈H2​ ⇒ ab∈(H1​⋂H2​)；
3. ${{H}_{1}},{{H}_{2}}<G\text{ }\Rightarrow \text{ }e\in {{H}_{1}}\wedge \text{ }e\in {{H}_{2}}\text{ }\Rightarrow \text{ }e\in \left( {{H}_{1}}\bigcap {{H}_{2}} \right)$H1​,H2​<G ⇒ e∈H1​∧ e∈H2​ ⇒ e∈(H1​⋂H2​)；
4. ${{H}_{1}}<G\text{ }\Rightarrow \text{ }\forall {{a}_{1}},{{b}_{1}},{{c}_{1}}\in {{H}_{1}},\text{ }\left( {{a}_{1}}{{b}_{1}} \right){{c}_{1}}={{a}_{1}}\left( {{b}_{1}}{{c}_{1}} \right),$H1​<G ⇒ ∀a1​,b1​,c1​∈H1​, (a1​b1​)c1​=a1​(b1​c1​),
   thus $\forall a,b,c\in \left( {{H}_{1}}\bigcap {{H}_{2}} \right)$∀a,b,c∈(H1​⋂H2​) we have
   $a,b,c\in {{H}_{1}}\text{ }\Rightarrow \text{ }\left( ab \right)c=a\left( bc \right)$a,b,c∈H1​ ⇒ (ab)c=a(bc)；
5. $\forall a\in \left( {{H}_{1}}\bigcap {{H}_{2}} \right)$∀a∈(H1​⋂H2​), we have
   $a\in {{H}_{1}}\text{ }\wedge a\in {{H}_{2}}\text{ }\wedge a\in G.$a∈H1​ ∧a∈H2​ ∧a∈G.
   $G$G is a group and ${{H}_{i}}<G\left( i=1,2 \right)$Hi​<G(i=1,2), we have
   $\forall a\in \left( {{H}_{1}}\cap {{H}_{2}} \right),\text{ }\exists !{{\left( {{a}^{-1}} \right)}_{0}}\in G,\text{ }{{\left( {{a}^{-1}} \right)}_{1}}\in {{H}_{1}}\subseteq G,\text{ }{{\left( {{a}^{-1}} \right)}_{2}}\in {{H}_{2}}\subseteq G\text{ },s.t.$∀a∈(H1​∩H2​), ∃!(a−1)0​∈G, (a−1)1​∈H1​⊆G, (a−1)2​∈H2​⊆G ,s.t.
   $a{{\left( {{a}^{-1}} \right)}_{0}}=a{{\left( {{a}^{-1}} \right)}_{1}}=a{{\left( {{a}^{-1}} \right)}_{2}}=e,$a(a−1)0​=a(a−1)1​=a(a−1)2​=e,
   and we have
   ${{\left( {{a}^{-1}} \right)}_{0}}={{\left( {{a}^{-1}} \right)}_{1}}\text{ }\wedge \text{ }{{\left( {{a}^{-1}} \right)}_{0}}={{\left( {{a}^{-1}} \right)}_{2}}$(a−1)0​=(a−1)1​ ∧ (a−1)0​=(a−1)2​
   $\Rightarrow {{\left( {{a}^{-1}} \right)}_{0}}={{\left( {{a}^{-1}} \right)}_{1}}={{\left( {{a}^{-1}} \right)}_{2}},$⇒(a−1)0​=(a−1)1​=(a−1)2​,
   thus $\exists !{{a}^{-1}}={{\left( {{a}^{-1}} \right)}_{1}}={{\left( {{a}^{-1}} \right)}_{2}}\in \left( {{H}_{1}}\cap {{H}_{2}} \right),\text{ }s.t.$∃!a−1=(a−1)1​=(a−1)2​∈(H1​∩H2​), s.t. $a{{a}^{-1}}=e.$aa−1=e.

By proving 1) to 5), we have proved the proposition.

Hw5：证明阿贝尔群的子群都是正规子群。

证明：
Let $G$G be any abelian group, $S$S be any subgroup of $G$G.
$G$G is an abelian group
$\Rightarrow \forall {{a}_{1}},{{a}_{2}}\in G,\text{ }{{a}_{1}}{{a}_{2}}={{a}_{2}}{{a}_{1}}$⇒∀a1​,a2​∈G, a1​a2​=a2​a1​
$\Rightarrow \forall a\in G,\text{ }S<G,\text{ }aS=Sa$⇒∀a∈G, S<G, aS=Sa.
$S<G$S<G
$\Rightarrow \forall t\in S,\text{ }\exists {{t}^{-1}}\in S,\text{ }s.t.\text{ }t\centerdot {{t}^{-1}}=e$⇒∀t∈S, ∃t−1∈S, s.t. t⋅t−1=e.
For any $t,{{t}^{-1}}\in S\subseteq G$t,t−1∈S⊆G, we have
$t\centerdot S\centerdot {{t}^{-1}}=t\centerdot {{t}^{-1}}\centerdot S=S,$t⋅S⋅t−1=t⋅t−1⋅S=S,
thus $S$S is a normal subgroup of $A$A.

Hw6：证明$G$G是一个群，则$G$G一定不是两个真子群的并集。

证明：
Assume that group $G={{G}_{1}}\bigcup {{G}_{2}}$G=G1​⋃G2​, ${{G}_{1}},{{G}_{2}}$G1​,G2​ satisfy conditions:
$1){{G}_{1}},{{G}_{2}}\subset G;\ \ 2){{G}_{1}},{{G}_{2}}<G;\ \ 3){{G}_{1}}\ne {{G}_{2}}.$1)G1​,G2​⊂G;  2)G1​,G2​<G;  3)G1​​=G2​.
Let $e$e be the identity element of $G$G, then obviously we have:
$e\in {{G}_{1}}\bigcap {{G}_{2}}\bigcap {{G}_{3}}.$e∈G1​⋂G2​⋂G3​.
$G={{G}_{1}}\bigcup {{G}_{2}}\text{ }\wedge \text{ }{{G}_{1}}\ne {{G}_{2}}$G=G1​⋃G2​ ∧ G1​​=G2​
$\Rightarrow \exists {{g}_{1}},{{g}_{2}}\in G,\text{ }s.t.\text{ }\left\{ \begin{aligned} & {{g}_{1}}\in {{G}_{1}}\text{ }\wedge \text{ }{{g}_{1}}\notin {{G}_{2}} \\ & {{g}_{2}}\notin {{G}_{1}}\text{ }\wedge \text{ }{{g}_{2}}\in {{G}_{2}} \\ \end{aligned} \right..$⇒∃g1​,g2​∈G, s.t. {​g1​∈G1​ ∧ g1​∈/​G2​g2​∈/​G1​ ∧ g2​∈G2​​.
Consider all $G's$G′s cosets on ${{G}_{2}}$G2​:$G/{{G}_{2}}:=\left\{ \left. t{{G}_{2}} \right|\forall t\in G \right\}$G/G2​:={tG2​∣∀t∈G}.
$\forall {{t}_{1}}\in {{G}_{1}},\text{ }{{t}_{2}}\in {{G}_{2}}$∀t1​∈G1​, t2​∈G2​, we have:
1)$\left\{ \begin{aligned} & {{g}_{1}}\notin {{G}_{2}} \\ & {{t}_{2}}\in {{G}_{2}}\text{ }\Rightarrow \text{ }{{t}_{2}}{{G}_{2}}\subset {{G}_{2}} \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }{{g}_{1}}\notin {{t}_{2}}{{G}_{2}};${​g1​∈/​G2​t2​∈G2​ ⇒ t2​G2​⊂G2​​ ⇒ g1​∈/​t2​G2​;
2)$\left\{ \begin{aligned} & {{g}_{1}}\in {{g}_{1}}{{G}_{2}}\text{ }\wedge \text{ }{{g}_{1}}\notin {{t}_{2}}{{G}_{2}} \\ & {{t}_{1}}{{G}_{2}}={{t}_{2}}{{G}_{2}}\text{ }or\text{ }{{t}_{1}}{{G}_{2}}\bigcap {{t}_{2}}{{G}_{2}}=\varnothing \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }{{g}_{1}}{{G}_{2}}\bigcap {{t}_{2}}{{G}_{2}}=\varnothing ;$⎩⎨⎧​​g1​∈g1​G2​ ∧ g1​∈/​t2​G2​t1​G2​=t2​G2​ or t1​G2​⋂t2​G2​=∅​ ⇒ g1​G2​⋂t2​G2​=∅;
3)$\left\{ \begin{aligned} & {{t}_{2}}\in {{t}_{2}}{{G}_{2}}\text{ }\Rightarrow \text{ }{{G}_{2}}\subset \underset{{{t}_{2}}\in {{G}_{2}}}{\mathop{\bigcup }}\,{{t}_{2}}{{G}_{2}} \\ & {{t}_{2}}{{G}_{2}}\subset {{G}_{2}}\text{ }\Rightarrow \text{ }\underset{{{t}_{2}}\in {{G}_{2}}}{\mathop{\bigcup }}\,{{t}_{2}}{{G}_{2}}\subset {{G}_{2}} \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }{{G}_{2}}=\underset{{{t}_{2}}\in {{G}_{2}}}{\mathop{\bigcup }}\,{{t}_{2}}{{G}_{2}};$⎩⎪⎪⎪⎨⎪⎪⎪⎧​​t2​∈t2​G2​ ⇒ G2​⊂t2​∈G2​⋃​t2​G2​t2​G2​⊂G2​ ⇒ t2​∈G2​⋃​t2​G2​⊂G2​​ ⇒ G2​=t2​∈G2​⋃​t2​G2​;
4)$\left\{ \begin{aligned} & {{g}_{1}}{{G}_{2}}\bigcap {{t}_{2}}{{G}_{2}}=\varnothing \\ & \\ & {{G}_{2}}=\underset{{{t}_{2}}\in {{G}_{2}}}{\mathop{\bigcup }}\,{{t}_{2}}{{G}_{2}} \\ & {{g}_{1}}{{G}_{2}}\subset G \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }{{g}_{1}}{{G}_{2}}\subset G-{{G}_{2}}\subset {{G}_{1}}.$⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧​​g1​G2​⋂t2​G2​=∅G2​=t2​∈G2​⋃​t2​G2​g1​G2​⊂G​ ⇒ g1​G2​⊂G−G2​⊂G1​.
So we have for ${{g}_{2}}\in {{G}_{2}}$g2​∈G2​, $\exists {{g}_{1}}'\in {{G}_{1}},\text{ }s.t.\text{ }{{g}_{1}}{{g}_{2}}={{g}_{1}}'$∃g1​′∈G1​, s.t. g1​g2​=g1​′.
$\Rightarrow {{g}_{1}}^{-1}{{g}_{1}}{{g}_{2}}={{g}_{1}}^{-1}{{g}_{1}}'$⇒g1​−1g1​g2​=g1​−1g1​′
$\Rightarrow \left( {{g}_{1}}^{-1}{{g}_{1}} \right){{g}_{2}}={{g}_{1}}^{-1}{{g}_{1}}'$⇒(g1​−1g1​)g2​=g1​−1g1​′
$\Rightarrow {{g}_{2}}={{g}_{1}}^{-1}{{g}_{1}}'\notin {{G}_{1}}$⇒g2​=g1​−1g1​′∈/​G1​,
which is contradictious to the operation closure of ${{G}_{1}}$G1​.
Thus the proposition is proved.

Hw7：在$R/I\ne \left\{ 0 \right\}$R/I​={0}的情况下：

1）若$R/I$R/I是交换环，$R$R不一定也是交换环。

Proof：
$R/I$R/I is a commutative ring
$\begin{aligned} & \Rightarrow \left( a+I \right)\left( b+I \right)=\left( b+I \right)\left( a+I \right),\text{ }\forall a,b\in R \\ & \Rightarrow ab+I=ba+I \\ & \Rightarrow ab=ba\text{ }or\text{ }ab\ne ba \\ \end{aligned}$​⇒(a+I)(b+I)=(b+I)(a+I), ∀a,b∈R⇒ab+I=ba+I⇒ab=ba or ab​=ba​

1. e.g.1 Let $R={{M}_{n}}\left( \mathbb{C} \right)$R=Mn​(C). $\forall M={{\left( {{m}_{ij}} \right)}_{n\times n}}\in R,\text{ }{{m}_{ij}}\in \mathbb{C}$∀M=(mij​)n×n​∈R, mij​∈C, note that
   $\left| M \right|={{\left( \operatorname{Re}\left[ {{m}_{ij}} \right] \right)}_{n\times n}}.$∣M∣=(Re[mij​])n×n​.
   $\forall P={{\left( {{p}_{ij}} \right)}_{n\times n}},\text{ }Q={{\left( {{q}_{ij}} \right)}_{n\times n}}\in {{M}_{n}}\left( \mathbb{C} \right),\text{ }{{p}_{ij}},{{q}_{ij}}\in \mathbb{C},$∀P=(pij​)n×n​, Q=(qij​)n×n​∈Mn​(C), pij​,qij​∈C, define that
   $P\times Q:={{\left( {{t}_{ij}} \right)}_{n\times n}},\text{ }{{t}_{ij}}=\sum\limits_{k=1}^{n}{{{p}_{ik}}{{q}_{kj}}}$P×Q:=(tij​)n×n​, tij​=k=1∑n​pik​qkj​
   $P\otimes Q:=\left| P \right|\times \left| Q \right|={{\left( {{w}_{ij}} \right)}_{n\times n}},\text{ }{{w}_{ij}}=\sum\limits_{k=1}^{n}{\operatorname{Re}\left[ {{p}_{ik}} \right]\operatorname{Re}\left[ {{q}_{kj}} \right]},$P⊗Q:=∣P∣×∣Q∣=(wij​)n×n​, wij​=k=1∑n​Re[pik​]Re[qkj​],
   $P+Q:={{\left( {{v}_{ij}} \right)}_{n\times n}},\text{ }{{v}_{ij}}={{p}_{ij}}+{{q}_{ij}}.$P+Q:=(vij​)n×n​, vij​=pij​+qij​.
   First we prove that $\left( R,\otimes ,+ \right)$(R,⊗,+) is a ring but not commutative for $\otimes$⊗.
   Obviously $\left( R,+ \right)$(R,+) is an abelian group.
   $\forall U,V,W\in R$∀U,V,W∈R,
   $U\otimes V=\left| U \right|\times \left| V \right|\in {{M}_{n}}\left( \mathbb{R} \right)\subseteq {{M}_{n}}\left( \mathbb{C} \right),$U⊗V=∣U∣×∣V∣∈Mn​(R)⊆Mn​(C),
   thus $\otimes$⊗ is closed.
   $\begin{aligned} & \text{ }\left( U\otimes V \right)\otimes W \\ & =\left| \text{ }\left| U \right|\times \left| V \right|\text{ } \right|\times \left| W \right| \\ & =\left( \left| U \right|\times \left| V \right| \right)\times \left| W \right| \\ & =\left| U \right|\times \left( \left| V \right|\times \left| W \right| \right) \\ & =\left| U \right|\times \left| \text{ }\left| V \right|\times \left| W \right|\text{ } \right| \\ & =U\otimes \left( V\otimes W \right) \\ \end{aligned},$​ (U⊗V)⊗W=∣ ∣U∣×∣V∣ ∣×∣W∣=(∣U∣×∣V∣)×∣W∣=∣U∣×(∣V∣×∣W∣)=∣U∣×∣ ∣V∣×∣W∣ ∣=U⊗(V⊗W)​,
   thus $\otimes$⊗ is associated.
   $\begin{aligned} & \text{ }\left( U+V \right)\otimes W \\ & =\left| U+V \right|\times \left| W \right| \\ & =\left( \left| U \right|+\left| V \right| \right)\times \left| W \right| \\ & =\left| U \right|\times \left| W \right|+\left| V \right|\times \left| W \right| \\ & =U\otimes W+V\otimes W \\ \end{aligned},$​ (U+V)⊗W=∣U+V∣×∣W∣=(∣U∣+∣V∣)×∣W∣=∣U∣×∣W∣+∣V∣×∣W∣=U⊗W+V⊗W​,
   $U\otimes \left( V+W \right)=\left| U \right|\times \left| V+W \right|=\left| U \right|\times \left| V \right|+\left| U \right|\times \left| W \right|=U\otimes V+U\otimes W,$U⊗(V+W)=∣U∣×∣V+W∣=∣U∣×∣V∣+∣U∣×∣W∣=U⊗V+U⊗W,
   thus distribution law is satisfied.
   So we have proved that $\left( R,\otimes ,+ \right)$(R,⊗,+) is a ring, but $\otimes$⊗ is obviously not commutative.
   Let $I={{M}_{n}}\left( \mathbb{R} \right)$I=Mn​(R), then obviously we have $\left( I,+ \right)$(I,+) is an abelian group and
   $I\otimes R\subseteq I\text{ }\And \text{ }R\otimes I\subseteq I,$I⊗R⊆I & R⊗I⊆I,
   thus $I$I is an ideal of $R$R.
   So we have the quotient ring $R/I\ne \left\{ 0 \right\}$R/I​={0}. Then we want to prove that $R/I$R/I is commutative for $\otimes$⊗.
   $\forall P,Q\in R$∀P,Q∈R, we have
   $P\otimes Q,\text{ }Q\otimes P\in I={{M}_{n}}\left( \mathbb{R} \right)$P⊗Q, Q⊗P∈I=Mn​(R)
   Take $P~\otimes Q$P ⊗Q as an example, we point out that $P\otimes Q+I=I.$P⊗Q+I=I.
   ① $P\otimes Q\in I$P⊗Q∈I & $I$I is an ideal (also is a subring) $\Rightarrow P\otimes Q+I\subseteq I$⇒P⊗Q+I⊆I.
   ② $\left( I,+ \right)$(I,+) is an abelian group $\Rightarrow$⇒ $\exists \left( -P\otimes Q \right)\in I$∃(−P⊗Q)∈I, and we have $-P\otimes Q+I\subseteq I$−P⊗Q+I⊆I
   $\Rightarrow I\subseteq P\otimes Q+I$⇒I⊆P⊗Q+I.
   Thus we have $P\otimes Q+I=I.$P⊗Q+I=I.
   So we have
   $P\otimes Q+I=Q\otimes P+I=I,$P⊗Q+I=Q⊗P+I=I,
   that is
   $\left( P+I \right)\otimes \left( Q+I \right)=\left( Q+I \right)\otimes \left( P+I \right),$(P+I)⊗(Q+I)=(Q+I)⊗(P+I),
   thus $R/I$R/I is commutative for $\otimes$⊗.
   Thus we have the case where $R$R is not commutative but $R/I$R/I is commutative.
2. e.g. 2. $\left( \mathbb{Z},+,\times \right)$(Z,+,×) obviously is a ring and a commutative ring.
   Consider the ideal $I=2\mathbb{Z}$I=2Z.
   We have $\forall a,b\in \mathbb{Z},\text{ }ab=ba$∀a,b∈Z, ab=ba$\Rightarrow ab+I=ba+I$⇒ab+I=ba+I, that is
   $\left( a+I \right)\left( b+I \right)=\left( b+I \right)\left( a+I \right)$(a+I)(b+I)=(b+I)(a+I)
   which shows that $\mathbb{R}/I$R/I is a commutative ring.
   Thus we have the case where $R/I\text{ }\And \text{ }R$R/I & R are both commutative rings.

2）若$R/I$R/I是含幺环，$R$R不一定也是含幺环。

Proof:
$R/I$R/I is a unital ring
$\begin{aligned} & \Rightarrow \exists c\in R,\text{ }s.t.\text{ }\left( a+I \right)\left( c+I \right)=a+I,\text{ }\forall a\in R \\ & \Rightarrow ac+I=a+I \\ & \Rightarrow ac=a\text{ }or\text{ }ac\ne a \\ \end{aligned}$​⇒∃c∈R, s.t. (a+I)(c+I)=a+I, ∀a∈R⇒ac+I=a+I⇒ac=a or ac​=a​

1. e.g. 1. $\left( 3\mathbb{Z},+,\times \right)$(3Z,+,×) is obviously a ring but not a unital ring. Consider the ideal $I=6\mathbb{Z}$I=6Z, we have
   $3\mathbb{Z}/6\mathbb{Z}=\left\{ \left. a+6\mathbb{Z} \right|a\in 3\mathbb{Z} \right\}=\left\{ \left. \text{ }\left\{ 3\left( {{n}_{1}}+2{{n}_{2}} \right) \right\}\text{ } \right|{{n}_{1}},{{n}_{2}}\in \mathbb{Z}\text{ } \right\},$3Z/6Z={a+6Z∣a∈3Z}={ {3(n1​+2n2​)} ∣n1​,n2​∈Z },
   that is
   $3\mathbb{Z}/6\mathbb{Z}=\left\{ \overline{0},\overline{1} \right\},$3Z/6Z={0,1},
   in which $\forall b\in \overline{k}\left( k=0,1,2 \right)$∀b∈k(k=0,1,2), $\frac{b}{3}\equiv k\left( \bmod 2 \right)$3b​≡k(mod2). Obviously the unit of $3\mathbb{Z}/6\mathbb{Z}$3Z/6Z is $\overline{1}$1.
   Thus we have the case where $R$R is not a unital ring but $R/I$R/I is a unital ring.
2. Consider $R=\mathbb{Z}$R=Z，$I=3\mathbb{Z}$I=3Z,
   obviously $\left( R,+,\times \right)$(R,+,×) is a unital ring.
   $R/I=\left\{ \overline{0},\overline{1},\overline{2} \right\}\ne \left\{ 0 \right\}$R/I={0,1,2}​={0} is also a ring and has unit $\overline{1}$1, thus is also a unital ring.
   Thus we have the case where both $R$R and $R/I$R/I are unital rings.

Hw8：证明当$P$P是素数时$\mathbb{Z}/P$Z/P是域 ；

证明：

1. 首先有$\forall P\in {{\mathbb{Z}}_{>0}}$∀P∈Z>0​，$\left( \mathbb{Z}/P,+ \right)$(Z/P,+)是一个阿贝尔群。
   $\mathbb{Z}/P=\left\{ \overline{0},\overline{1},...,\overline{P-1} \right\}$Z/P={0,1,...,P−1​}
   运算封闭性显然满足.
   结合律和交换律继承实数运算的结合律和交换律.
   加法单位元：$\overline{0}$0.
   可逆性：$\forall \overline{a}\in \mathbb{Z}/P,\text{ }a\in \left\{ 0,1,...,P-1 \right\}$∀a∈Z/P, a∈{0,1,...,P−1}，$\exists \text{ }\overline{P-a}\in \mathbb{Z}/P$∃ P−a​∈Z/P（注：$\overline{P}=\overline{0}$P=0）,使得
   $\overline{a}+\overline{P-a}=\overline{P}=\overline{0}.$a+P−a​=P=0.
2. 然后有$\left( \mathbb{Z}/P,\times \right)$(Z/P,×)是一个阿贝尔幺半群。
   运算封闭性显然成立。
   结合律和交换律：继承实数加法的结合律和交换律。
   单位元：$\overline{1}$1。
   所以$\left( \mathbb{Z}/P,\times \right)$(Z/P,×)是一个阿贝尔幺半群。
3. 环的分配律条件继承实数运算的分配律。
4. 当$P$P是素数时，$\left( \mathbb{Z}/P\backslash \left\{ \overline{0} \right\},\times \right)$(Z/P\{0},×)是一个群。
   1. 运算封闭性：$\forall \overline{a},\overline{b}\in \left[ \left( \mathbb{Z}/P \right)\backslash \left\{ \overline{0} \right\} \right]$∀a,b∈[(Z/P)\{0}]，$a,b\in \left\{ 1,2,...,P-1 \right\}$a,b∈{1,2,...,P−1}，根据算术基本定理，$ab$ab可以被分解成若干质因子的乘积且在不考虑顺序的意义下形式唯一。设${{p}_{ab}}$pab​是$ab$ab的任意一个质因子，则有${{p}_{ab}}\le \min \left\{ a,b \right\}<P$pab​≤min{a,b}<P，所以有${{p}_{ab}}\ne P$pab​​=P，即$P$P不是$ab$ab的质因子。$\forall n\in {{\mathbb{Z}}_{>0}}$∀n∈Z>0​，素数$P$P是$nP$nP的质因子，因此$ab\ne nP$ab​=nP，即$ab\ne 0\left( \bmod P \right)$ab​=0(modP)，即$\overline{a}\overline{b}\ne \overline{0}$ab​=0，所以有$\overline{a}\overline{b}\in \left[ \left( \mathbb{Z}/P \right)\backslash \left\{ \overline{0} \right\} \right]$ab∈[(Z/P)\{0}]
   2. 结合律：继承$\left( \mathbb{Z}/P,\times \right)$(Z/P,×)的结合律。
   3. 单位元：$\overline{1}$1。
   4. 可逆性：$\forall \overline{a}\in \left[ \left( \mathbb{Z}/P \right)\backslash \left\{ \overline{0} \right\} \right],\text{ }a\in \left\{ 1,2,...,P-1 \right\}$∀a∈[(Z/P)\{0}], a∈{1,2,...,P−1}，
      当$a=1$a=1时显然存在$\overline{1}$1，使得$\overline{a}\centerdot \overline{1}=\overline{1}\centerdot \overline{a}=\overline{1}$a⋅1=1⋅a=1；
      当$a>1$a>1时，考虑集合
      $G:=\left\{ \left. nP=ba+c \right|n=1,2,...,a-1,\text{ }1\le b\le P-2,\text{ }0\le c\le a-1,\text{ }b,c\in \mathbb{Z} \right\}$G:={nP=ba+c∣n=1,2,...,a−1, 1≤b≤P−2, 0≤c≤a−1, b,c∈Z}
      （注意$b\ne P-1$b​=P−1，因为$c=\max \left\{ G \right\}-\left( P-1 \right)a=\left( a-1 \right)P-\left( P-1 \right)a=a-P<0$c=max{G}−(P−1)a=(a−1)P−(P−1)a=a−P<0）
      考虑$G$G中的元素$nP=ba+c$nP=ba+c：
      
      我们指出，对于${{n}_{1}}P={{b}_{1}}a+{{c}_{1}}$n1​P=b1​a+c1​和${{n}_{2}}P={{b}_{2}}a+{{c}_{2}}$n2​P=b2​a+c2​，若${{n}_{1}}\ne {{n}_{2}}$n1​​=n2​，则有${{c}_{1}}\ne {{c}_{2}}$c1​​=c2​。
      不妨令${{n}_{2}}>{{n}_{1}}$n2​>n1​且${{c}_{1}}={{c}_{2}}$c1​=c2​，则有
      $\begin{matrix} {{n}_{2}}P-{{n}_{1}}P=\left( {{b}_{2}}a+{{c}_{2}} \right)-\left( {{b}_{1}}a+{{c}_{1}} \right) \\ \Rightarrow \left( {{n}_{2}}-{{n}_{1}} \right)P=\left( {{b}_{2}}-{{b}_{1}} \right)a,\text{ }1\le {{b}_{2}}-{{b}_{1}}\le P-2 \\ \end{matrix}$n2​P−n1​P=(b2​a+c2​)−(b1​a+c1​)⇒(n2​−n1​)P=(b2​−b1​)a, 1≤b2​−b1​≤P−2​
      这与前面已证明的“素数$P$P不是$\left( {{b}_{2}}-{{b}_{1}} \right)a$(b2​−b1​)a的质因子”产生矛盾。
      所以在集合$G$G里我们可构建单射
      $f:\begin{matrix} \left\{ n \right\} & \mapsto & \left\{ c \right\} \\ n & \mapsto & c \\ \end{matrix}$f:{n}n​↦↦​{c}c​
      $\Rightarrow \#\left\{ n \right\}\le \#\left\{ c \right\}$⇒#{n}≤#{c}
      在$G=\left\{ 1,2,...,\left( a-1 \right)P \right\}$G={1,2,...,(a−1)P}中，显然有$\#\left\{ n \right\}=\#G=a-1$#{n}=#G=a−1.
      由$P$P不是$ab$ab的质因数以及$c=nP-ba\in \mathbb{Z}$c=nP−ba∈Z, $0\le c\le a-1$0≤c≤a−1得到
      $c\in \left\{ 1,2,...,a-1 \right\}\Rightarrow \#\left\{ c \right\}\le a-1$c∈{1,2,...,a−1}⇒#{c}≤a−1
      所以有$a-1\le \#\left\{ c \right\}\le a-1\text{ }\Rightarrow \text{ }\!\!\#\!\!\text{ }\left\{ c \right\}=a-1$a−1≤#{c}≤a−1 ⇒ # {c}=a−1，即有
      $\left\{ c \right\}=\left\{ 1,...,a-1 \right\}${c}={1,...,a−1}
      所以$\exists n'\in \left\{ 1,2,...,a-1 \right\},b'\in \left\{ 1,2,...,P-2 \right\},$∃n′∈{1,2,...,a−1},b′∈{1,2,...,P−2},使得
      $\begin{matrix} n'P=b'a+\left( a-1 \right) \\ \Rightarrow n'P+1=\left( b'+1 \right)a,\text{ }b'+1\in \left\{ 2,3,...,P-1 \right\} \\ \end{matrix}$n′P=b′a+(a−1)⇒n′P+1=(b′+1)a, b′+1∈{2,3,...,P−1}​
      所以$a\left( b'+1 \right)=\left( b'+1 \right)a\equiv 1\left( \bmod P \right)$a(b′+1)=(b′+1)a≡1(modP)，即存在$\overline{b'+1}$b′+1​，使得$\overline{a}\centerdot \overline{b'+1}=\overline{b'+1}\centerdot \overline{a}=\overline{1}$a⋅b′+1​=b′+1​⋅a=1。
      可逆性得证。

Hw9：$End\left( R \right):=\left\{ \left. f:R\to R \right|f\text{ is a ring morphism} \right\}$End(R):={f:R→R∣f is a ring morphism} 是含幺环。

证明：
Proof:
$\left( R,+,\times \right)$(R,+,×) is a ring and $End\left( R \right)=\left\{ \left. f:R\to R \right|f\text{ is a ring morphism} \right\}$End(R)={f:R→R∣f is a ring morphism}. $\forall {{f}_{1}},{{f}_{2}}\in End\left( R \right)$∀f1​,f2​∈End(R), define operations $\otimes$⊗ and $\oplus$⊕ as below:
${{f}_{1}}\otimes {{f}_{2}}:\begin{matrix} R & \to & R \\ x & \to & {{f}_{1}}\left[ {{f}_{2}}\left( x \right) \right] \\ \end{matrix},$f1​⊗f2​:Rx​→→​Rf1​[f2​(x)]​,
${{f}_{1}}\oplus {{f}_{2}}:\begin{matrix} R & \to & R \\ x & \to & {{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right) \\ \end{matrix}.$f1​⊕f2​:Rx​→→​Rf1​(x)+f2​(x)​.

1. First we prove that $\left( End\left( R \right),\oplus \right)$(End(R),⊕) is an abelian group.
   Operation closure:
   $\forall {{f}_{1}},{{f}_{2}}\in End\left( R \right),\text{ }\forall x\in R,\text{ }\left( {{f}_{1}}\oplus {{f}_{2}} \right)\left( x \right)={{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)$∀f1​,f2​∈End(R), ∀x∈R, (f1​⊕f2​)(x)=f1​(x)+f2​(x), in which ${{f}_{1}}\left( x \right)\in R$f1​(x)∈R and
   ${{f}_{2}}\left( x \right)\in R$f2​(x)∈R. According to the closure of operation $+$+ of $R$R, we have
   ${{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)\in R,$f1​(x)+f2​(x)∈R,
   which means ${{f}_{1}}\oplus {{f}_{2}}:R\to R$f1​⊕f2​:R→R.
   Associative law & commutative law:
   $\forall {{f}_{1}},{{f}_{2}},{{f}_{3}}\in End\left( R \right),\forall x\in R$∀f1​,f2​,f3​∈End(R),∀x∈R, we have
   $\left\{ \begin{aligned} & \left[ \left( {{f}_{1}}\oplus {{f}_{2}} \right)\oplus {{f}_{3}} \right]\left( x \right)=\left( {{f}_{1}}\oplus {{f}_{2}} \right)\left( x \right)+{{f}_{3}}\left( x \right)={{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)+{{f}_{3}}\left( x \right) \\ & \left[ {{f}_{1}}\oplus \left( {{f}_{2}}\oplus {{f}_{3}} \right) \right]\left( x \right)={{f}_{1}}\left( x \right)+\left( {{f}_{2}}\oplus {{f}_{3}} \right)\left( x \right)={{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)+{{f}_{3}}\left( x \right) \\ \end{aligned} \right.${​[(f1​⊕f2​)⊕f3​](x)=(f1​⊕f2​)(x)+f3​(x)=f1​(x)+f2​(x)+f3​(x)[f1​⊕(f2​⊕f3​)](x)=f1​(x)+(f2​⊕f3​)(x)=f1​(x)+f2​(x)+f3​(x)​
   $\Rightarrow \left( {{f}_{1}}\oplus {{f}_{2}} \right)\oplus {{f}_{3}}={{f}_{1}}\oplus \left( {{f}_{2}}\oplus {{f}_{3}} \right),$⇒(f1​⊕f2​)⊕f3​=f1​⊕(f2​⊕f3​),
   $\left\{ \begin{aligned} & \left( {{f}_{1}}\oplus {{f}_{2}} \right)\left( x \right)={{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right) \\ & \left( {{f}_{2}}\oplus {{f}_{1}} \right)\left( x \right)={{f}_{2}}\left( x \right)+{{f}_{1}}\left( x \right) \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }\left( {{f}_{1}}\oplus {{f}_{2}} \right)\left( x \right)=\left( {{f}_{2}}\oplus {{f}_{1}} \right)\left( x \right)${​(f1​⊕f2​)(x)=f1​(x)+f2​(x)(f2​⊕f1​)(x)=f2​(x)+f1​(x)​ ⇒ (f1​⊕f2​)(x)=(f2​⊕f1​)(x)
   Unit element:
   Consider map ${{f}_{+}}\left( x \right)=0,\text{ }\forall x\in R$f+​(x)=0, ∀x∈R. Obviously ${{f}_{+}}\in End\left( R \right)$f+​∈End(R), and we have:
   $\forall f\in End\left( R \right),\text{ }\forall x\in R$∀f∈End(R), ∀x∈R,
   $\left\{ \begin{aligned} & \left( f\oplus {{f}_{+}} \right)\left( x \right)=f\left( x \right)+{{f}_{+}}\left( x \right)=f\left( x \right)+0=f\left( x \right) \\ & \left( {{f}_{+}}\oplus f \right)\left( x \right)={{f}_{+}}\left( x \right)+f\left( x \right)=0+f\left( x \right)=f\left( x \right) \\ \end{aligned} \right.${​(f⊕f+​)(x)=f(x)+f+​(x)=f(x)+0=f(x)(f+​⊕f)(x)=f+​(x)+f(x)=0+f(x)=f(x)​
   $\Rightarrow f\oplus {{f}_{+}}={{f}_{+}}\oplus f=f,$⇒f⊕f+​=f+​⊕f=f,
   which means ${{f}_{+}}\in End\left( R \right)$f+​∈End(R) is the unit element.
   Invertibility:
   $\forall {{f}_{1}}\in End\left( R \right),$∀f1​∈End(R), consider the map ${{f}_{2}}\in End\left( R \right),\text{ }{{f}_{2}}\left( x \right)=-{{f}_{1}}\left( x \right)$f2​∈End(R), f2​(x)=−f1​(x), then obviously ${{f}_{2}}$f2​ is the inverse of ${{f}_{1}}$f1​.
   Thus $\left( End\left( R \right),\oplus \right)$(End(R),⊕) is an abelian group.
2. Then we prove that $\left( End\left( R \right),\otimes \right)$(End(R),⊗) is a unital group.
   Operation closure:
   $\forall {{f}_{1}},{{f}_{2}}\in End\left( R \right),\text{ }\forall x\in R$∀f1​,f2​∈End(R), ∀x∈R, ${{f}_{2}}\left( x \right)\in R\text{ }\Rightarrow \text{ }\left( {{f}_{1}}\otimes {{f}_{2}} \right)\left( x \right)={{f}_{1}}\left[ {{f}_{2}}\left( x \right) \right]\in R$f2​(x)∈R ⇒ (f1​⊗f2​)(x)=f1​[f2​(x)]∈R
   $\Rightarrow \text{ }{{f}_{1}}\otimes {{f}_{2}}\in End\left( R \right)$⇒ f1​⊗f2​∈End(R).
   Associative law:
   $\forall {{f}_{1}},{{f}_{2}},{{f}_{3}}\in End\left( R \right),\text{ }\forall x\in R$∀f1​,f2​,f3​∈End(R), ∀x∈R, we have
   $\left\{ \begin{aligned} & \left[ \left( {{f}_{1}}\otimes {{f}_{2}} \right)\otimes {{f}_{3}} \right]\left( x \right)=\left( {{f}_{1}}\otimes {{f}_{2}} \right)\left[ {{f}_{3}}\left( x \right) \right]={{f}_{1}}\left[ {{f}_{2}}\left[ {{f}_{3}}\left( x \right) \right] \right] \\ & \left[ {{f}_{1}}\otimes \left( {{f}_{2}}\otimes {{f}_{3}} \right) \right]\left( x \right)={{f}_{1}}\left[ \left( {{f}_{2}}\otimes {{f}_{3}} \right)\left( x \right) \right]={{f}_{1}}\left[ {{f}_{2}}\left[ {{f}_{3}}\left( x \right) \right] \right] \\ \end{aligned} \right.${​[(f1​⊗f2​)⊗f3​](x)=(f1​⊗f2​)[f3​(x)]=f1​[f2​[f3​(x)]][f1​⊗(f2​⊗f3​)](x)=f1​[(f2​⊗f3​)(x)]=f1​[f2​[f3​(x)]]​
   $\Rightarrow \left( {{f}_{1}}\otimes {{f}_{2}} \right)\otimes {{f}_{3}}={{f}_{1}}\otimes \left( {{f}_{2}}\otimes {{f}_{3}} \right).$⇒(f1​⊗f2​)⊗f3​=f1​⊗(f2​⊗f3​).
   Unit element:
   Obviously $I\left( x \right)=x$I(x)=x is the unit element.
   Thus $\left( End\left( R \right),\oplus \right)$(End(R),⊕) is an abelian group.
3. Last we prove $End\left( R \right)$End(R) has distribution law.
   $\forall {{f}_{1}},{{f}_{2}},{{f}_{3}}\in End\left( R \right),\text{ }\forall x\in R$∀f1​,f2​,f3​∈End(R), ∀x∈R, we have
   $\begin{aligned} & \text{ }\left[ \left( {{f}_{1}}\oplus {{f}_{2}} \right)\otimes {{f}_{3}} \right]\left( x \right) \\ & =\left( {{f}_{1}}\oplus {{f}_{2}} \right)\left[ {{f}_{3}}\left( x \right) \right] \\ & ={{f}_{1}}\left[ {{f}_{3}}\left( x \right) \right]+{{f}_{2}}\left[ {{f}_{3}}\left( x \right) \right] \\ & =\left[ {{f}_{1}}\otimes {{f}_{3}} \right]\left( x \right)+\left[ {{f}_{2}}\otimes {{f}_{3}} \right]\left( x \right) \\ \end{aligned},$​ [(f1​⊕f2​)⊗f3​](x)=(f1​⊕f2​)[f3​(x)]=f1​[f3​(x)]+f2​[f3​(x)]=[f1​⊗f3​](x)+[f2​⊗f3​](x)​,
   $\begin{aligned} & \text{ }\left[ {{f}_{1}}\otimes \left( {{f}_{2}}\oplus {{f}_{3}} \right) \right]\left( x \right) \\ & ={{f}_{1}}\left[ \left[ {{f}_{2}}\oplus {{f}_{3}} \right]\left( x \right) \right] \\ & ={{f}_{1}}\left[ {{f}_{2}}\left( x \right)+{{f}_{3}}\left( x \right) \right] \\ & ={{f}_{1}}\left[ {{f}_{2}}\left( x \right) \right]+{{f}_{1}}\left[ {{f}_{3}}\left( x \right) \right]\left( \text{Because }{{f}_{1}}\text{ is a ring morphism} \right) \\ & =\left[ {{f}_{1}}\otimes {{f}_{2}} \right]\left( x \right)+\left[ {{f}_{1}}\otimes {{f}_{3}} \right]\left( x \right) \\ \end{aligned}.$​ [f1​⊗(f2​⊕f3​)](x)=f1​[[f2​⊕f3​](x)]=f1​[f2​(x)+f3​(x)]=f1​[f2​(x)]+f1​[f3​(x)](Because f1​ is a ring morphism)=[f1​⊗f2​](x)+[f1​⊗f3​](x)​.

In conclusion, $End\left( R \right)$End(R) is a unital ring.

Hw10：证明${{S}_{3}}\cong Aut\left( {{S}_{3}} \right)=Inn\left( {{S}_{3}} \right)$S3​≅Aut(S3​)=Inn(S3​)

解：
关于此式的证明见博客Aut(S3)=Inn(S3)的证明和元素寻找。
On $\left\{ 1,2,3 \right\}${1,2,3} define permutation ${{f}_{ijk}}:=\left\{ \begin{aligned} & 1\to i \\ & 2\to j \\ & 3\to k \\ \end{aligned} \right.,\text{ }i\ne j\ne k$fijk​:=⎩⎪⎨⎪⎧​​1→i2→j3→k​, i​=j​=k.
$Aut\left( {{S}_{3}} \right)=Inn\left( {{S}_{3}} \right)=\left\{ {{g}_{123}},{{g}_{132}},{{g}_{213}},{{g}_{231}},{{g}_{312}},{{g}_{321}} \right\}$Aut(S3​)=Inn(S3​)={g123​,g132​,g213​,g231​,g312​,g321​}，其中
${{g}_{123}}\left( =id \right):\left\{ \begin{aligned} & {{f}_{123}}\to {{f}_{123}}{{f}_{123}}{{f}_{123}}={{f}_{123}} \\ & {{f}_{132}}\to {{f}_{123}}{{f}_{132}}{{f}_{123}}={{f}_{132}} \\ & {{f}_{213}}\to {{f}_{123}}{{f}_{213}}{{f}_{123}}={{f}_{213}} \\ & {{f}_{231}}\to {{f}_{123}}{{f}_{231}}{{f}_{123}}={{f}_{231}} \\ & {{f}_{312}}\to {{f}_{123}}{{f}_{312}}{{f}_{123}}={{f}_{312}} \\ & {{f}_{321}}\to {{f}_{123}}{{f}_{321}}{{f}_{123}}={{f}_{321}} \\ \end{aligned} \right.,\text{ }{{g}_{132}}:\left\{ \begin{aligned} & {{f}_{123}}\to {{f}_{132}}{{f}_{123}}{{f}_{132}}={{f}_{123}} \\ & {{f}_{132}}\to {{f}_{132}}{{f}_{132}}{{f}_{132}}={{f}_{132}} \\ & {{f}_{213}}\to {{f}_{132}}{{f}_{213}}{{f}_{132}}={{f}_{321}} \\ & {{f}_{231}}\to {{f}_{132}}{{f}_{231}}{{f}_{132}}={{f}_{312}} \\ & {{f}_{312}}\to {{f}_{132}}{{f}_{312}}{{f}_{132}}={{f}_{231}} \\ & {{f}_{321}}\to {{f}_{132}}{{f}_{321}}{{f}_{132}}={{f}_{213}} \\ \end{aligned} \right.$g123​(=id):⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​​f123​→f123​f123​f123​=f123​f132​→f123​f132​f123​=f132​f213​→f123​f213​f123​=f213​f231​→f123​f231​f123​=f231​f312​→f123​f312​f123​=f312​f321​→f123​f321​f123​=f321​​, g132​:⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​​f123​→f132​f123​f132​=f123​f132​→f132​f132​f132​=f132​f213​→f132​f213​f132​=f321​f231​→f132​f231​f132​=f312​f312​→f132​f312​f132​=f231​f321​→f132​f321​f132​=f213​​,${{g}_{213}}:\left\{ \begin{aligned} & {{f}_{123}}\to {{f}_{213}}{{f}_{123}}{{f}_{213}}={{f}_{123}} \\ & {{f}_{132}}\to {{f}_{213}}{{f}_{132}}{{f}_{213}}={{f}_{321}} \\ & {{f}_{213}}\to {{f}_{213}}{{f}_{213}}{{f}_{213}}={{f}_{213}} \\ & {{f}_{231}}\to {{f}_{213}}{{f}_{231}}{{f}_{213}}={{f}_{312}} \\ & {{f}_{312}}\to {{f}_{213}}{{f}_{312}}{{f}_{213}}={{f}_{231}} \\ & {{f}_{321}}\to {{f}_{213}}{{f}_{321}}{{f}_{213}}={{f}_{132}} \\ \end{aligned} \right.,\text{ }{{g}_{231}}:\left\{ \begin{aligned} & {{f}_{123}}\to {{f}_{231}}{{f}_{123}}{{f}_{312}}={{f}_{123}} \\ & {{f}_{132}}\to {{f}_{231}}{{f}_{132}}{{f}_{312}}={{f}_{213}} \\ & {{f}_{213}}\to {{f}_{231}}{{f}_{213}}{{f}_{312}}={{f}_{321}} \\ & {{f}_{231}}\to {{f}_{231}}{{f}_{231}}{{f}_{312}}={{f}_{231}} \\ & {{f}_{312}}\to {{f}_{231}}{{f}_{312}}{{f}_{312}}={{f}_{312}} \\ & {{f}_{321}}\to {{f}_{231}}{{f}_{321}}{{f}_{312}}={{f}_{132}} \\ \end{aligned} \right.,$g213​:⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​​f123​→f213​f123​f213​=f123​f132​→f213​f132​f213​=f321​f213​→f213​f213​f213​=f213​f231​→f213​f231​f213​=f312​f312​→f213​f312​f213​=f231​f321​→f213​f321​f213​=f132​​, g231​:⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​​f123​→f231​f123​f312​=f123​f132​→f231​f132​f312​=f213​f213​→f231​f213​f312​=f321​f231​→f231​f231​f312​=f231​f312​→f231​f312​f312​=f312​f321​→f231​f321​f312​=f132​​,
${{g}_{312}}:\left\{ \begin{aligned} & {{f}_{123}}\to {{f}_{312}}{{f}_{123}}{{f}_{231}}={{f}_{123}} \\ & {{f}_{132}}\to {{f}_{312}}{{f}_{132}}{{f}_{231}}={{f}_{321}} \\ & {{f}_{213}}\to {{f}_{312}}{{f}_{213}}{{f}_{231}}={{f}_{132}} \\ & {{f}_{231}}\to {{f}_{312}}{{f}_{231}}{{f}_{231}}={{f}_{231}} \\ & {{f}_{312}}\to {{f}_{312}}{{f}_{312}}{{f}_{231}}={{f}_{312}} \\ & {{f}_{321}}\to {{f}_{312}}{{f}_{321}}{{f}_{231}}={{f}_{213}} \\ \end{aligned} \right.,\text{ }{{g}_{321}}:\left\{ \begin{aligned} & {{f}_{123}}\to {{f}_{321}}{{f}_{123}}{{f}_{321}}={{f}_{123}} \\ & {{f}_{132}}\to {{f}_{321}}{{f}_{132}}{{f}_{321}}={{f}_{213}} \\ & {{f}_{213}}\to {{f}_{321}}{{f}_{213}}{{f}_{321}}={{f}_{132}} \\ & {{f}_{231}}\to {{f}_{321}}{{f}_{231}}{{f}_{321}}={{f}_{312}} \\ & {{f}_{312}}\to {{f}_{321}}{{f}_{312}}{{f}_{321}}={{f}_{231}} \\ & {{f}_{321}}\to {{f}_{321}}{{f}_{321}}{{f}_{321}}={{f}_{321}} \\ \end{aligned} \right.$g312​:⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​​f123​→f312​f123​f231​=f123​f132​→f312​f132​f231​=f321​f213​→f312​f213​f231​=f132​f231​→f312​f231​f231​=f231​f312​→f312​f312​f231​=f312​f321​→f312​f321​f231​=f213​​, g321​:⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​​f123​→f321​f123​f321​=f123​f132​→f321​f132​f321​=f213​f213​→f321​f213​f321​=f132​f231​→f321​f231​f321​=f312​f312​→f321​f312​f321​=f231​f321​→f321​f321​f321​=f321​​