Update Math4302_L13.md

content/CSE4303/CSE4303_L6.md

@@ -0,0 +1,5 @@
+# CSE4303 Introduction to Computer Security (Lecture 6)
+
+Refer to this lecture notes
+
+[CSE442T Lecture 3](https://notenextra.trance-0.com/CSE442T/CSE442T_L3/)

content/CSE4303/CSE4303_L7.md

@@ -0,0 +1,65 @@
+# CSE4303 Introduction to Computer Security (Lecture 7)
+
+## Cyptography in Symmetric Systems
+
+### Symmetric systems
+
+Symmetric (shared-key) encryption
+
+- Classical techniques
+- Computer-aided techniques
+- Formal reasoning
+- Realizations:
+  - Stream ciphers
+  - Block ciphers
+
+#### Stream ciphers
+
+1. Operate on PT one bit at a time (usually), as a bit "stream"
+2. Generate arbitrarily long keystream on demand
+
+Security abstraction:
+
+1. XOR transfers randomness of keystream to randomness of CT regardless of PT’s content
+2. Security depends on G being “practically” indistinguishable from random string and “practically” unpredictable
+3. Idea: shouldn’t be able to predict next bit of generator given all bits seen so far
+
+Keystream $G(k)$
+
+- Idea: shouldn’t be able to predict next bit of generator given all bits seen so far
+- Strategies and challenges: many!
+- Idea that doesn’t quite work: Linear Feedback Shift Register (LFSR)
+  - Choice of feedback: by algebra
+  - Pro: fast, statistically close to random
+  - Problem: susceptible to cryptanalysis (b/c linear)
+  - LFSR-based
+- Modifications to basic LFSR:
+  - Use non-linear combo of multiple LFSRs
+  - Use controlled clocking (e.g. only cycle the LFSR when another LFSR outputs a 1)
+  - Etc.
+- Others: mod arithmetic-based, other algebraic constructions
+
+#### Block ciphers
+
+1. Operate on PT one block at a time
+2. Use same key for multiple blocks (with caveats)
+3. Chaining modes intertwine successive blocks of CT (or not)
+
+View cipher as a Pseudo-Random Permutation (PRP)
+
+- PRP defined over $(K, X)$:
+
+$$
+E: K \times X \to X
+$$
+
+such that:
+
+1. There exists an “efficient” deterministic algorithm to evaluate $E(k,x)$.
+2. The function $E( k, \cdot )$ is one-to-one.
+3. There exists an “efficient” inversion algorithm $D(k,y)$.
+
+- i.e. a PRF that is an invertible 1-to-1 mapping from message space to
+message space
+
+

content/CSE4303/_meta.js

@@ -8,4 +8,6 @@ export default {
     CSE4303_L3: "Introduction to Computer Security (Lecture 3)",
     CSE4303_L4: "Introduction to Computer Security (Lecture 4)",
     CSE4303_L5: "Introduction to Computer Security (Lecture 5)",
+    CSE4303_L6: "Introduction to Computer Security (Lecture 6)",
+    CSE4303_L7: "Introduction to Computer Security (Lecture 7)",
 }

content/Math4201/Math4201_L34.md

@@ -21,7 +21,7 @@ If $\mathbb{R}_l$ is second countable, then for any real number $x$, there is an
 
 Any such open sets is of the form $[x,x+\epsilon)\cap A$ with $\epsilon>0$ and any element of $A$ being larger than $\min(U_x)=x$.
 
-In summary, for any $x\in \mathbb{R}$, there is an element $U_x\in \mathcal{B}$ with $(U_x)=x$. In particular, if $x\neq y$, then $U_x\neq U_y$. SO there is an injective map $f:\mathbb{R}\rightarrow \mathcal{B}$ sending $x$ to $U_x$. This implies that $\mathbb{B}$ is uncountable.
+In summary, for any $x\in \mathbb{R}$, there is an element $U_x\in \mathcal{B}$ with $(U_x)=x$. In particular, if $x\neq y$, then $U_x\neq U_y$. So there is an injective map $f:\mathbb{R}\rightarrow \mathcal{B}$ sending $x$ to $U_x$. This implies that $\mathcal{B}$ is uncountable.
 
 </details>
 

content/Math4201/Math4201_L4.md

@@ -27,7 +27,7 @@ $$
 Let $(X,\mathcal{T})$ be a topological space. Let $\mathcal{C}\subseteq \mathcal{T}$ be a collection of subsets of $X$ satisfying the following property:
 
 $$
-\forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } U\subseteq C
+\forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } C\subseteq U
 $$
 
 Then $\mathcal{C}$ is a basis and the topology generated by $\mathcal{C}$ is $\mathcal{T}$.

content/Math4202/Math4202_L10.md

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+# Math4202 Topology II (Lecture 10)
+
+## Algebraic Topology
+
+### Path homotopy
+
+
+#### Theorem for properties of product of paths
+
+1. If $f\simeq_p f_1, g\simeq_p g_1$, then $f*g\simeq_p f_1*g_1$. (Product is well-defined)
+2. $([f]*[g])*[h]=[f]*([g]*[h])$. (Associativity)
+3. Let $e_{x_0}$ be the constant path from $x_0$ to $x_0$, $e_{x_1}$ be the constant path from $x_1$ to $x_1$. Suppose $f$ is a path from $x_0$ to $x_1$.
+    $$
+    [e_{x_0}]*[f]=[f],\quad [f]*[e_{x_1}]=[f]
+    $$
+    (Right and left identity)
+4. Given $f$ in $X$ a path from $x_0$ to $x_1$, we define $\bar{f}$ to be the path from $x_1$ to $x_0$ where $\bar{f}(t)=f(1-t)$.
+    $$
+    f*\bar{f}=e_{x_0},\quad \bar{f}*f=e_{x_1}
+    $$
+    $$
+    [f]*[\bar{f}]=[e_{x_0}],\quad [\bar{f}]*[f]=[e_{x_1}]
+    $$
+
+<details>
+<summary>Proof</summary>
+
+(1) If $f\simeq_p f_1$, $g\simeq_p g_1$, then $f*g\simeq_p f_1*g_1$.
+
+Let $F$ be homotopy between $f$ and $f_1$, $G$ be homotopy between $g$ and $g_1$.
+
+We can define
+
+$$
+F*G:[0,1]\times [0,1]\to X,\quad F*G(s,t)=\left(F(-,t)*G(-,t)\right)(s)=\begin{cases}
+F(2s,t) & 0\leq s\leq \frac{1}{2}\\
+G(2s-1,t) & \frac{1}{2}\leq s\leq 1
+\end{cases}
+$$
+
+$F*G$ is a homotopy between $f*g$ and $f_1*g_1$.
+
+We can check this by enumerating the cases from definition of homotopy.
+
+---
+
+(2) $([f]*[g])*[h]=[f]*([g]*[h])$.
+
+For $f*(g*h)$, along the interval $[0,\frac{1}{2}]$ we map $x_1\to x_2$, then along the interval $[\frac{1}{2},\frac{3}{4}]$ we map $x_2\to x_3$, then along the interval $[\frac{3}{4},1]$ we map $x_3\to x_4$.
+
+For $(f*g)*h$, along the interval $[0,\frac{1}{4}]$ we map $x_1\to x_2$, then along the interval $[\frac{1}{4},\frac{1}{2}]$ we map $x_2\to x_3$, then along the interval $[\frac{1}{2},1]$ we map $x_3\to x_4$.
+
+We can construct the homotopy between $f*(g*h)$ and $(f*g)*h$ as follows.
+
+Let $f((4-2t)s)$ for $F(s,t)$,
+
+when $t=0$, $F(s,0)=f(4s)\in f*(g*h)$, when $t=1$, $F(s,1)=f(2s)\in (f*g)*h$.
+
+....
+
+_We make the linear maps between $f*(g*h)$ and $(f*g)*h$ continuous, then $f*(g*h)\simeq_p (f*g)*h$. With our homotopy constructed above_
+
+---
+
+(3) $e_{x_0}*f\simeq_p f\simeq_p f*e_{x_1}$.
+
+We can construct the homotopy between $e_{x_0}*f$ and $f$ as follows.
+
+$$
+H(s,t)=\begin{cases}
+x_0 & t\geq 2s\\
+f(2s-t) & t\leq 2s
+\end{cases}
+$$
+
+or you may induct from $f(\frac{s-t/2}{1-t/2})$ if you like.
+
+---
+
+(4) $f*\bar{f}=e_{x_0},\quad \bar{f}*f=e_{x_1}$.
+
+Note that we don't need to reach $x_1$ every time.
+
+$f_t=f(ts)$ $s\in[0,\frac{1}{2}]$.
+
+$\bar{f}_t=\bar{f}(1-ts)$ $s\in[\frac{1}{2},1]$.
+
+</details>
+
+> [!CAUTION]
+>
+> Homeomorphism does not implies homotopy automatically.
+
+#### Definition for the fundamental group
+
+The fundamental group of $X$ at $x$ is defined to be
+
+$$
+(\Pi_1(X,x),*)
+$$

content/Math4202/Math4202_L11.md

@@ -0,0 +1,132 @@
+# Math4201 Topology II (Lecture 11)
+
+## Algebraic topology
+
+### Fundamental group
+
+The $*$ operation has the following properties:
+
+#### Properties for the path product operation
+
+Let $[f],[g]\in \Pi_1(X)$, for $[f]\in \Pi_1(X)$, let $s:\Pi_1(X)\to X, [f]\mapsto f(0)$ and $t:\Pi_1(X)\to X, [f]\mapsto f(1)$.
+
+Note that $t([f])=s([g])$, $[f]*[g]=[f*g]\in \Pi_1(X)$.
+
+This also satisfies the associativity. $([f]*[g])*[h]=[f]*([g]*[h])$.
+
+We have left and right identity. $[f]*[e_{t(f)}]=[f], [e_{s(f)}]*[f]=[f]$.
+
+We have inverse. $[f]*[\bar{x}]=[e_{s(f)}], [\bar{x}]*[f]=[e_{t(f)}]$
+
+#### Definition for Groupoid
+
+Let $f,g$ be paths where $g,f:[0,1]\to X$, and consider the function of all pathes in $G$, denoted as $\mathcal{G}$,
+
+Set $t:\mathcal{G}\to X$ be the source map, for this case $t(f)=f(0)$, and $s:\mathcal{G}\to X$ be the target map, for this case $s(f)=f(1)$.
+
+We define
+
+$$
+\mathcal{G}^{(2)}=\{(f,g)\in \mathcal{G}\times \mathcal{G}|t(f)=s(g)\}
+$$
+
+And we define the operation $*$ on $\mathcal{G}^{(2)}$ as the path product.
+
+This satisfies the following properties:
+
+- Associativity: $(f*g)*h=f*(g*h)$
+
+Consider the function $\eta:X\to \mathcal{G}$, for this case $\eta(x)=e_{x}$.
+
+- We have left and right identity: $\eta(t(f))*f=f, f*\eta(s(f))=f$
+
+- Inverse: $\forall g\in \mathcal{G}, \exists g^{-1}\in \mathcal{G}, g*g^{-1}=\eta(s(g))$, $g^{-1}*g=\eta(t(g))$
+
+#### Definition for loop
+
+Let $x_0\in X$. A path starting and ending at $x_0$ is called a loop based at $x_0$.
+
+#### Definition for the fundamental group
+
+The fundamental group of $X$ at $x$ is defined to be
+
+$$
+(\Pi_1(X,x),*)
+$$
+
+where $*$ is the product operation, and $\Pi_1(X,x)$ is the set o homotopy classes of loops in $X$ based at $x$.
+
+<details>
+<summary>Example of fundamental group</summary>
+
+Consider $X=[0,1]$, with subspace topology from standard topology in $\mathbb{R}$.
+
+$\Pi_1(X,0)=\{e\}$, (constant function at $0$) since we can build homotopy for all loops based at $0$ as follows $H(s,t)=(1-t)f(s)+t$.
+
+And $\Pi_1(X,1)=\{e\}$, (constant function at $1$.)
+
+---
+
+Let $X=\{1,2\}$ with discrete topology.
+
+$\Pi_1(X,1)=\{e\}$, (constant function at $1$.)
+
+$\Pi_1(X,2)=\{e\}$, (constant function at $2$.)
+
+---
+
+Let $X=S^1$ be the circle.
+
+$\Pi_1(X,1)=\mathbb{Z}$ (related to winding numbers, prove next week).
+
+</details>
+
+A natural question is, will the fundamental group depends on the basepoint $x$?
+
+#### Definition for $\hat{\alpha}$
+
+Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$. $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$. Define $\hat{\alpha}:\Pi_1(X,x_0)\to \Pi_1(X,x_1)$ as follows:
+
+$$
+\hat{\alpha}(\beta)=[\bar{\alpha}]*[f]*[\alpha]
+$$
+
+#### $\hat{\alpha}$ is a group homomorphism
+
+$\hat{\alpha}$ is a group homomorphism between $(\Pi_1(X,x_0),*)$ and $(\Pi_1(X,x_1),*)$
+
+<details>
+<summary>Proof</summary>
+
+Let $f,g\in \Pi_1(X,x_0)$, then $\hat{\alpha}(f*g)=\hat{\alpha}(f)\hat{\alpha}(g)$
+
+$$
+\begin{aligned}
+\hat{\alpha}(f*g)&=[\bar{\alpha}]*[f]*[g]*[\alpha]\\
+&=[\bar{\alpha}]*[f]*[e_{x_0}]*[g]*[\alpha]\\
+&=[\bar{\alpha}]*[f]*[\alpha]*[\bar{\alpha}]*[g]*[\alpha]\\
+&=([\bar{\alpha}]*[f]*[\alpha])*([\bar{\alpha}]*[g]*[\alpha])\\
+&=(\hat{\alpha}(f))*(\hat{\alpha}(g))
+\end{aligned}
+$$
+
+---
+
+Next, we will show that $\hat{\alpha}\circ \hat{\bar{\alpha}}([f])=[f]$, and $\hat{\bar{\alpha}}\circ \hat{\alpha}([f])=[f]$.
+
+$$
+\begin{aligned}
+\hat{\alpha}\circ \hat{\bar{\alpha}}([f])&=\hat{\alpha}([\bar{\alpha}]*[f]*[\alpha])\\
+&=[\alpha]*[\bar{\alpha}]*[f]*[\alpha]*[\bar{\alpha}]\\
+&=[e_{x_0}]*[f]*[e_{x_1}]\\
+&=[f]
+\end{aligned}
+$$
+
+The other case is the same
+
+</details>
+
+#### Corollary of fundamental group
+
+If $X$ is path-connected and $x_0,x_1\in X$, then $\Pi_1(X,x_0)$ is isomorphic to $\Pi_1(X,x_1)$.

content/Math4202/Math4202_L12.md

@@ -0,0 +1,119 @@
+# Math4201 Topology II (Lecture 12)
+
+## Algebraic topology
+
+### Fundamental group
+
+Recall from last lecture, the $(\Pi_1(X,x_0),*)$ is a group, and for any two points $x_0,x_1\in X$, the group $(\Pi_1(X,x_0),*)$ is isomorphic to $(\Pi_1(X,x_1),*)$ if $x_0,x_1$ is path connected.
+
+> [!TIP]
+> 
+> How does the $\hat{\alpha}$ (isomorphism between $(\Pi_1(X,x_0),*)$ and $(\Pi_1(X,x_1),*)$) depend on the choice of $\alpha$ (path) we choose?
+
+#### Definition of simply connected
+
+A space $X$ is simply connected if 
+
+- $X$ is [path-connected](https://notenextra.trance-0.com/Math4201/Math4201_L23/#definition-of-path-connected-space) ($\forall x_0,x_1\in X$, there exists a continuous function $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$)
+- $\Pi_1(X,x_0)$ is the trivial group for some $x_0\in X$
+
+<details>
+<summary>Example of simply connected space</summary>
+
+Intervals are simply connected.
+
+---
+
+Any star-shaped is simply connected.
+
+---
+
+$S^1$ is not simply connected, but $n\geq 2$, then $S^n$ is simply connected.
+
+</details>
+
+#### Lemma for simply connected space
+
+In a simply connected space $X$, and two paths having the same initial and final points are path homotopic.
+
+<details>
+<summary>Proof</summary>
+
+Let $f,g$ be paths having the same initial and final points, then $f(0)=g(0)=x_0$ and $f(1)=g(1)=x_1$.
+
+Therefore $[f]*[\bar{g}]\simeq_p [e_{x_0}]$ (by simply connected space assumption).
+
+Then
+
+$$
+\begin{aligned}
+[f]*[\bar{g}]&\simeq_p [e_{x_0}]\\
+([f]*[\bar{g}])*[g]&\simeq_p [e_{x_0}]*[g]\\
+[f]*([\bar{g}]*[g])&\simeq_p [e_{x_0}]*[g]\\
+[f]*[e_{x_1}]&\simeq_p [e_{x_0}]*[g]\\
+[f]&\simeq_p [g]
+\end{aligned}
+$$
+
+</details>
+
+#### Definition of group homomorphism induced by continuous map
+
+Let $h:(X,x_0)\to (Y,y_0)$ be a continuous map, define $h_*:\Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ where $h(x_0)=y_0$. by $h_*([f])=[h\circ f]$.
+
+$h_*$ is called the group homomorphism induced by $h$ relative to $x_0$.
+
+<details>
+<summary>Check the homomorphism property</summary>
+
+$$
+\begin{aligned}
+h_*([f]*[g])&=h_*([f*g])\\
+&=[h_*[f*g]]\\
+&=[h_*[f]*h_*[g]]\\
+&=[h_*[f]]*[h_*[g]]\\
+&=h_*([f])*h_*([g])
+\end{aligned}
+$$
+
+</details>
+
+#### Theorem composite of group homomorphism
+
+If $h:(X,x_0)\to (Y,y_0)$ and $k:(Y,y_0)\to (Z,z_0)$ are continuous maps, then $k_* \circ h_*:\Pi_1(X,x_0)\to \Pi_1(Z,z_0)$ where $h_*:\Pi_1(X,x_0)\to \Pi_1(Y,y_0)$, $k_*:\Pi_1(Y,y_0)\to \Pi_1(Z,z_0)$,is a group homomorphism.
+
+<details>
+<summary>Proof</summary>
+
+Let $f$ be a loop based at $x_0$.
+
+$$
+\begin{aligned}
+k_*(h_*([f]))&=k_*([h\circ f])\\
+&=[k\circ h\circ f]\\
+&=[(k\circ h)\circ f]\\
+&=(k\circ h)_*([f])\\
+\end{aligned}
+$$
+
+</details>
+
+#### Corollary of composite of group homomorphism
+
+Let $\operatorname{id}:(X,x_0)\to (X,x_0)$ be the identity map. This induces $(\operatorname{id})_*:\Pi_1(X,x_0)\to \Pi_1(X,x_0)$.
+
+If $h$ is a homeomorphism with the inverse $k$, with
+
+$$
+k_*\circ h_*=(k\circ h)_*=(\operatorname{id})_*=I=(\operatorname{id})_*=(h\circ k)_*
+$$
+
+This induced $h_*: \Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ is an isomorphism.
+
+#### Corollary for homotopy and group homomorphism
+
+If $h,k:(X,x_0)\to (Y,y_0)$ are homotopic maps form $X$ to $Y$ such that the homotopy $H_t(x_0)=y_0,\forall t\in I$, then $h_*=k_*$.
+
+$$
+h_*([f])=[h\circ f]\simeq_p[k\circ h]=k_*([f])
+$$

content/Math4202/Math4202_L13.md

@@ -0,0 +1,59 @@
+# Math4202 Topology II (Lecture 13)
+
+## Algebraic Topology
+
+### Covering space
+
+#### Definition of partition into slice
+
+Let $p:E\to B$ be a continuous surjective map. The open set $U\subseteq B$ is said to be evenly covered by $p$ if it's inverse image $p^{-1}(U)$ can be written as the union of **disjoint open sets** $V_\alpha$ in $E$. Such that for each $\alpha$, the restriction of $p$ to $V_\alpha$ is a homeomorphism of $V_\alpha$ onto $U$.
+
+The collection of $\{V_\alpha\}$ is called a **partition** $p^{-1}(U)$ into slice.
+
+_Stack of pancakes ($\{V_\alpha\}$) on plate $U$, each $V_\alpha$ is a pancake homeomorphic to $U$_
+
+_Note that all the sets in the definition are open._
+
+#### Definition of covering space
+
+Let $p:E\to B$ be a continuous surjective map. If every point $b$ of $B$ has a neighborhood **evenly covered** by $p$, which means $p^{-1}(U)$ is partitioned into slice, then $p$ is called a covering map and $E$ is called a covering space.
+
+<details>
+<summary>Examples of covering space</summary>
+
+identity map is a covering map
+
+---
+
+Consider the $B\times \Gamma\to B$ with $\Gamma$ being the discrete topology with the projection map onto $B$.
+
+This is a covering map.
+
+---
+
+Let $S^1=\{z\mid |z|=1\}$, then $p=z^n$ is a covering map to $S^1$.
+
+Solving the inverse image for the $e^{i\theta}$ with $\epsilon$ interval, we can get $n$ slices for each neighborhood of $e^{i\theta}$, $-\epsilon< \theta< \epsilon$.
+
+You can continue the computation and find the exact $\epsilon$ so that the inverse image of $p^{-1}$ is small and each interval don't intersect (so that we can make homeomorphism for each interval).
+
+Usually, we don't choose the $U$ to be the whole space.
+
+---
+
+Consider the projection for the boundary of mobius strip into middle circle.
+
+This is a covering map since the boundary of mobius strip is winding the middle circle twice, and for each point on the middle circle with small enough neighborhood, there will be two disjoint interval on the boundary of mobius strip that are homeomorphic to the middle circle.
+
+</details>
+
+#### Proposition of covering map is open map
+
+If $p:E\to B$ is a covering map, then $p$ is an open map.
+
+<details>
+<summary>Proof</summary>
+
+Consider arbitrary open set $V\subseteq E$, consider $U=p(V)$, for every point $q\in U$, with neighborhood $q\in W$, the inverse image of $W$ is open, continue next lecture.
+
+</details>

content/Math4202/_meta.js

@@ -12,4 +12,8 @@ export default {
     Math4202_L7: "Topology II (Lecture 7)",
     Math4202_L8: "Topology II (Lecture 8)",
     Math4202_L9: "Topology II (Lecture 9)",
+    Math4202_L10: "Topology II (Lecture 10)",
+    Math4202_L11: "Topology II (Lecture 11)",
+    Math4202_L12: "Topology II (Lecture 12)",
+    Math4202_L13: "Topology II (Lecture 13)",
 }

content/Math4302/Math4302_L10.md

@@ -0,0 +1,131 @@
+# Math4302 Modern Algebra (Lecture 10)
+
+## Groups
+
+### Group homomorphism
+
+Recall the kernel of a group homomorphism is the set
+
+$$
+\operatorname{ker}(\phi)=\{a\in G|\phi(a)=e'\}
+$$
+
+<details>
+<summary>Example</summary>
+
+Let $\phi:(\mathbb{Z},+)\to (\mathbb{Z}_n,+)$ where $\phi(k)=k\mod n$.
+
+The kernel of $\phi$ is the set of all multiples of $n$.
+
+</details>
+
+#### Theorem for one-to-one group homomorphism
+
+$\phi:G\to G'$ is one-to-one if and only if $\operatorname{ker}(\phi)=\{e\}$
+
+If $\phi$ is one-to-one, then $\phi(G)\leq G'$, $G$ is isomorphic ot $\phi(G)$ (onto automatically).
+
+If $A$ is a set, then a permutation of $A$ is a bijection $f:A\to A$.
+
+#### Cayley's Theorem
+
+Every group $G$ is isomorphic to a subgroup of $S_A$ for some $A$ (and if $G$ is finite then $A$ can be taken to be finite.)
+
+<details>
+<summary>Example</summary>
+
+$D_n\leq S_n$, so $A=\{1,2,\cdots,n\}$
+
+---
+
+$\mathbb{Z}_n\leq S_n$, (use the set of rotations) so $A=\{1,2,\cdots,n\}$ $\phi(i)=\rho^i$ where $i\in \mathbb{Z}_n$ and $\rho\in D_n$
+
+---
+
+$GL(2,\mathbb{R})$. Set $A=\mathbb{R}^2$, for every $A\in GL(2,\mathbb{R})$, let $\phi(A)$ be the permutation of $\mathbb{R}^2$ induced by $A$, so $\phi(A)=f_A:\mathbb{R}^2\to \mathbb{R}^2$, $f_A(\begin{pmatrix}x\\y\end{pmatrix})=A\begin{pmatrix}x\\y\end{pmatrix}$
+
+We want to show that this is a group homomorphism.
+
+- $\phi(AB)=\phi(A)\phi(B)$ (it is a homomorphism)
+
+$$
+\begin{aligned}
+    f_{AB}(\begin{pmatrix}x\\y\end{pmatrix})&=AB\begin{pmatrix}x\\y\end{pmatrix}\\
+    &=f_A(B\begin{pmatrix}x\\y\end{pmatrix})\\
+    &=f_A(f_B(\begin{pmatrix}x\\y\end{pmatrix}))\\
+    &=(f_A\circ f_B)(\begin{pmatrix}x\\y\end{pmatrix})\\
+\end{aligned}
+$$
+
+- Then we need to show that $\phi$ is one-to-one.
+
+It is sufficient to show that $\operatorname{ker}(\phi)=\{e\}$.
+
+Solve $f_A(\begin{pmatrix}x\\y\end{pmatrix})=\begin{pmatrix}x\\y\end{pmatrix}$, the only choice for $A$ is the identity matrix.
+
+Therefore $\operatorname{ker}(\phi)=\{e\}$.
+
+</details>
+
+<details>
+<summary>Proof for Cayley's Theorem</summary>
+
+Let $A=G$, for every $g\in G$, define $\lambda_g:G\to G$ by $\lambda_g(x)=gx$.
+
+Then $\lambda_g$ is a **permutation** of $G$. (not homomorphism)
+
+- $\lambda_g$ is one-to-one by cancellation on the left.
+- $\lambda_g$ is onto since $\lambda_g(g^{-1}y)=y$ for every $y\in G$.
+
+We claim $\phi: G\to S_G$ define by $\phi(g)=\lambda_g$ is a group homomorphism that is one-to-one.
+
+First we show that $\phi$ is homomorphism.
+
+$\forall x\in G$
+
+$$
+\begin{aligned}
+    \phi(g_1)\phi(g_2)&=\lambda_{g_1}(\lambda_{g_2}(x))\\
+    &=\lambda_{g_1g_2}(x)\\
+    &=\phi(g_1g_2)x\\
+\end{aligned}
+$$
+
+This is one to one since if $\phi(g_1)=\phi(g_2)$, then $\lambda_{g_1}=\lambda_{g_2}\forall x$, therefore $g_1=g_2$.
+
+</details>
+
+### Odd and even permutations
+
+#### Definition of transposition
+
+A $\sigma\in S_n$ is a transposition is a two cycle $\sigma=(i j)$
+
+Fact: Every permutation in $S_n$ can be written as a product of transpositions. (may not be disjoint transpositions)
+
+<details>
+<summary>Example of a product of transpositions</summary>
+
+Consider $(1234)=(14)(13)(12)$.
+
+In general, $(i_1,i_2,\cdots,i_m)=(i_1i_m)(i_2i_{m-1})(i_3i_{m-2})\cdots(i_1i_2)$
+
+This is not the unique way.
+
+$$
+(12)(34)=(42)(34)(23)(12)
+$$
+
+</details>
+
+But the parity of the number of transpositions is unique.
+
+#### Theorem for parity of transpositions
+
+If $\sigma\in S_n$ is written as a product of transposition, then the number of transpositions is either always odd or even.
+
+#### Definition of odd and even permutations
+
+$\sigma$ is an even permutation if the number of transpositions is even.
+
+$\sigma$ is an odd permutation if the number of transpositions is odd.

content/Math4302/Math4302_L11.md

@@ -0,0 +1,163 @@
+# Math4302 Modern Algebra (Lecture 11)
+
+## Groups
+
+### Symmetric groups
+
+#### Definition of odd and even permutations
+
+$\sigma$ is an even permutation if the number of transpositions is even.
+
+$\sigma$ is an odd permutation if the number of transpositions is odd.
+
+#### Theorem for parity of transpositions
+
+The parity of the number of transpositions is unique.
+
+<details>
+<summary>Proof</summary>
+
+Prove using the determinant of a matrix, swapping the rows of the matrix multiply the determinant by $-1$.
+
+Consider the identity matrix $I_n$. Then the determinant is $1$, let $(ij)A$, where $i\neq j$ denote the matrix obtained from $A$ by swapping the rows $j$ and $i$, then the determinant of $(1j)A$ is $-1$.
+
+And,
+
+$$
+\det((a_1b_1)(a_2b_2)\cdots(a_nb_n)A)=(-1)^n\det(A)
+$$
+
+</details>
+
+$S_3$ has 6 permutations $\{e,(12),(13),(23),(12)(23),(13)(23)\}$, 3 of them are even $\{e,(12)(23),(13)(23)\}$ and 3 of them are odd $\{(13),(12),(23)\}$.
+
+#### Theorem for the number of odd and even permutations in symmetric groups
+
+In general, $S_n$ has $n!$ permutations, half of them are even and half of them are odd.
+
+<details>
+<summary>Proof</summary>
+
+Consider the set of odd permutations in $S_n$ and set of even permutations in $S_n$. Consider the function: $\alpha:S_n\to S_n$ where $\alpha(\sigma)=\sigma(12)$.
+
+$\sigma$ is a bijection,
+
+If $\sigma_1(12)=\sigma_2(12)$, then $\sigma_1=\sigma_2$.
+
+If $\phi$ is an even permutation, $\alpha(\phi(12))=\phi(12)(12)=\phi$, therefore the number of elements in the set of odd and even permutations are the same.
+</details>
+
+#### Definition for sign of permutations
+
+For $\sigma\in S_n$, the sign of $\sigma$ is defined by $\operatorname{sign}(\sigma)=1$ if sigma is even and $-1$ if sigma is odd.
+
+Then $\beta: S_n\to \{1,-1\}$ is a group under multiplication, where $\beta(\sigma)=\operatorname{sign}(\sigma)$.
+
+Then $\beta$ is a group homomorphism.
+
+#### Definition of alternating group
+
+$\ker(\beta)\leq S_n$, and $\ker(\beta)$ is the set of even permutations. Therefore the set of even permutations is a subgroup of $S_n$. We denote as $A_n$ (also called alternating group).
+
+and $|A_n|=\frac{n!}{2}$.
+
+### Direct product of groups
+
+#### Definition of direct product of groups
+
+Let $G_1,G_2$ be two groups. Then the direct product of $G_1$ and $G_2$ is defined as
+
+$$
+G_1\times G_2=\{(g_1,g_2):g_1\in G_1,g_2\in G_2\}
+$$
+
+The operations are defined by $(a_1,b_1)*(a_2,b_2)=(a_1*a_2,b_1*b_2)$.
+
+This group is well defined since:
+
+The identity is $(e_1,e_2)$, where $e_1\in G_1$ and $e_2\in G_2$. (easy to verify)
+
+The inverse is $(a_1,b_1)^{-1}=(a_1^{-1},b_1^{-1})$.
+
+Associativity automatically holds by associativity of $G_1$ and $G_2$.
+
+<details>
+<summary>Examples</summary>
+
+Consider $\mathbb{Z}_\1\times \mathbb{Z}_2$.
+
+$$
+\mathbb{Z}_\1\times \mathbb{Z}_2=\{(0,0),(0,1),(1,0),(1,1)\}
+$$
+
+$(0,0)^2=(0,0)$, $(0,1)^2=(0,0)$, $(1,0)^2=(0,0)$, $(1,1)^2=(0,0)$
+
+This is not a cyclic group, this is isomorphic to klein four group.
+
+---
+
+Consider $\mathbb{Z}_2\times \mathbb{Z}_3$.
+
+$$
+\mathbb{Z}_2\times \mathbb{Z}_3=\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\}
+$$
+
+This is cyclic ((2,3) are coprime)
+
+Consider:
+
+$$
+\langle (1,1)\rangle=\{(0,0),(1,1),(0,2),(1,0),(0,1),(1,2)\}
+$$
+
+</details>
+
+#### Lemma for direct product of cyclic groups
+
+$\mathbb{Z}_m\times \mathbb{Z}_n\simeq \mathbb{Z}_{mn}$ if and only if $m$ and $n$ have greatest common divisor $1$.
+
+<details>
+<summary>Proof</summary>
+
+First assume $\operatorname{gcd}(m,n)=d>1$
+
+Consider $(r,s)\in \mathbb{Z}_m\times \mathbb{Z}_n$.
+
+We claim that order of $(r,s)$ is at most $\frac{mn}{d}<mn$.
+
+Since $\frac{mn}{d}$ is integer, $\frac{mn}{d}=m_1dn_1$ where $m_1d$ is multiple of $m$ and $n_1d$ is multiple of $n$.
+
+Therefore $r$ combine with itself $\frac{mn}{d}$ times is $0$ in $\mathbb{Z}_m$ and $s$ combine with itself $\frac{mn}{d}$ times is $0$ in $\mathbb{Z}_n$.
+
+---
+
+Other direction:
+
+Assume $\operatorname{gcd}(m,n)=1$.
+
+Claim order of $(1,1)=mn$, so $\mathbb{Z}_m\times \mathbb{Z}_n=\langle (1,1)\rangle$.
+
+If $k$ is the order of $(1,1)$, then $k$ is a multiple of $m$ and a multiple of $n$.
+
+</details>
+
+Similarly, if $G_1,G_2,G_3,\ldots,G_k$ are groups, then
+
+$$
+G_1\times G_2\times G_3\times \cdots\times G_k=\{(g_1,g_2,\ldots,g_k):g_1\in G_1,g_2\in G_2,\ldots,g_k\in G_k\}
+$$
+
+is a group.
+
+Easy to verify by associativity. $(G_1\times G_2)\times G_3=G_1\times G_2\times G_3$.
+
+#### Some extra facts for direct product
+
+1. $G_1\times G_2\simeq G_2\times G_1$, with $\phi(a_1,a_2)=(a_2,a_1)$.
+2. If $H_1\leq G_1$ and $H_2\leq G_2$, then $H_1\times H_2\leq G_1\times G_2$.
+
+> [!WARNING]
+>
+> Not every subgroup of $G_1\times G_2$ is of the form $H_1\times H_2$.
+>
+> Consider $\mathbb{Z}_2\times \mathbb{Z}_2$ with subgroup $\{(0,0),(1,1)\}$, This forms a subgroup but not of the form $H_1\times H_2$.

content/Math4302/Math4302_L12.md

@@ -0,0 +1,135 @@
+# Math4303 Modern Algebra (Lecture 12)
+
+## Groups
+
+### Direct products
+
+$\mathbb{Z}_m\times \mathbb{Z}_n$ is cyclic if and only if $m$ and $n$ have greatest common divisor $1$.
+
+More generally, for $\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}\times \cdots \times \mathbb{Z}_{n_k}$, if $n_1,n_2,\cdots,n_k$ are pairwise coprime, then the direct product is cyclic.
+
+<details>
+<summary>Proof</summary>
+
+For the forward direction, use $\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}=\mathbb{Z}_{n_1n_2}$. if $n_1, n_2$ are coprime. 
+
+
+For the backward, suppose to the contrary that for example $\gcd(n_1,n_2)=d>1$, then $G=\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}\times H$, where any element in $H$ has order $\leq |H|$ and any element in $\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}$ has order $<\frac{n_1n_2}{d}$, therefore, all the elements in $G$ will have order strictly less than the size $n_1n_2\ldots n_k$ of the group.
+
+</details>
+
+#### Corollary for composition of cyclic groups
+
+If $n=p_1^{m_1}\ldots p_k^{m_k}$, where $p_i$ are distinct primes, then the group 
+
+$$
+G=\mathbb{Z}_n=\mathbb{Z}_{p_1^{m_1}}\times \mathbb{Z}_{p_2^{m_2}}\times \cdots \times \mathbb{Z}_{p_k^{m_k}}
+$$
+
+is cyclic.
+
+<details>
+<summary>Example for product of cyclic groups and order of element</summary>
+
+$$
+\mathbb{Z}_{8}\times\mathbb{Z}_8\times \mathbb{Z}_12
+$$
+
+the order for $(1,1,1)$ is 24.
+
+What is the maximum order of an element in this group?
+
+Guess:
+
+$8*3=24$
+
+</details>
+
+### Structure of finitely generated abelian groups
+
+#### Theorem for finitely generated abelian groups
+
+Every finitely generated abelian group $G$ is isomorphic to 
+
+$$
+Z_{p_1}^{n_1}\times Z_{p_2}^{n_2}\times \cdots \times Z_{p_k}^{n_k}\times\underbrace{\mathbb{Z}\times \ldots \times \mathbb{Z}}_{m\text{ times}}
+$$
+
+<details>
+<summary>Example</summary>
+
+If $G$ is abelian of size $8$, then $G$ is isomorphic to one of the following:
+
+- $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ (non cyclic)
+- $\mathbb{Z}_2\times \mathbb{Z}_4$ (non cyclic)
+- $\mathbb{Z}_2$ (cyclic)
+
+And any two of them are not isomorphic
+
+---
+
+Find all abelian group of order $72$.
+
+Since $72=2^3*3^2$, There are 3 possibilities for the $2^3$ part, and there are 2 possibilities for the $3^2$ part.
+
+Note that $\mathbb{Z}_8\times\mathbb{Z}_9$, where $8,9$ are coprime, $\mathbb{Z}_8\times\mathbb{Z}_9=\mathbb{Z}_{72}$, is cyclic.
+
+There are 6 possibilities in total.
+
+</details>
+
+#### Corollary for divisor size of abelian subgroup
+
+If $g$ is abelian and $|G|=n$, then for every divisor $m$ of $n$, $G$ has a subgroup of order $m$.
+
+> [!WARNING]
+>
+> This is not true if $G$ is not abelian.
+>
+> Consider $A_4$ (alternating group for $S_4$) does not have a subgroup of order 6.
+
+
+<details>
+<summary>Proof for the corollary</summary>
+
+Write $G=\mathbb{Z}_{p_1}^{n_1}\times \mathbb{Z}_{p_2}^{n_2}\times \cdots \times \mathbb{Z}_{p_k}^{n_k}$ where $p_i$ are distinct primes.
+
+Therefore $n=p_1^{m_1}\ldots p_k^{m_k}$.
+
+For any divisor $d$ of $n$, we can write $d=p_1^{m_1}\ldots p_k^{m_k}$, where $m_i\leq n_i$.
+
+Now for each $p_i$, we choose the subgroup $H_i$ of size $p_i^{m_i}$ in $\mathbb{Z}_{p_i}^{n_i}$. (recall that every cyclic group of size $r$ and any divisor $s$ of $r$, there is a subgroup of order $s$. If the group is generated by $a$, then use $a^{\frac{r}{s}}$ to generate the subgroup.)
+
+We can construct the subgroup $H=H_1\times H_2\times \cdots \times H_k$ is the subgroup of $G$ of order $d$.
+</details>
+
+### Cosets
+
+#### Definition of Cosets
+
+Let $G$ be a group and $H$ its subgroup.
+
+Define a relation on $G$ and $a\sim b$ if $a^{-1}b\in H$.
+
+This is an equivalence relation.
+
+- Reflexive: $a\sim a$: $a^{-1}a=e\in H$
+- Symmetric: $a\sim b\Rightarrow b\sim a$: $a^{-1}b\in H$, $(a^{-1}b)^{-1}=b^{-1}a\in H$
+- Transitive: $a\sim b$ and $b\sim c\Rightarrow a\sim c$ : $a^{-1}b\in H, b^{-1}c\in H$, therefore their product is also in $H$, $(a^{-1}b)(b^{-1}c)=a^{-1}c\in H$
+
+So we get a partition of $G$ to equivalence classes.
+
+Let $a\in G$, the equivalence class containing $a$
+
+$$
+aH=\{x\in G| a\sim x\}=\{x\in G| a^{-1}x\in H\}=\{x|x=ah\text{ for some }h\in H\}
+$$
+
+This is called the coset of $a$ in $H$.
+
+<details>
+<summary>Example</summary>
+
+Consider $G=S_3$
+
+</details>

content/Math4302/Math4302_L13.md

@@ -0,0 +1,143 @@
+# Math4302 Modern Algebra (Lecture 13)
+
+## Groups
+
+### Cosets
+
+Last time we see that (left coset) $a\sim b$ (to differentiate from right coset, we may denote it as $a\sim_L b$) by $a^{-1}b\in H$ defines an equivalence relation.
+
+#### Definition of Equivalence Class
+
+Let $a\in H$, and the equivalence class containing $a$ is defined as:
+
+$$
+aH=\{x|a\simeq x\}=\{x|a^{-1}x\in H\}=\{x|x=ah\text{ for some }h\in H\}
+$$.
+
+#### Properties of Equivalence Class
+
+$aH=bH$ if and only if $a\sim b$.
+
+<details>
+<summary>Proof</summary>
+
+If $aH=bH$, then since $a\in aH, a\in bH$, then for some $h$, $a=bh$, since $b^{-1}a\in H$, so $a^{-1}b\in H$, therefore $a\simeq b$.
+
+If $a\sim b$, then $aH\subseteq bH$, since anything in $aH$ is related to $a$, therefore it is related to $b$ so $a\in bH$.
+
+$bH\subseteq aH$, apply the reflexive property for equivalence relation, therefore $b\in aH$.
+
+So $aH=bH$.
+
+</details>
+
+If $aH\cap bH\neq \emptyset$, then $aH=bH$.
+
+<details>
+<summary>Proof</summary>
+If $x\in aH\cap bH$, then $x\sim a$ and $x\sim b$, so $a\sim b$, so $aH=bH$.
+
+</details>
+
+$aH=H$ if and only if $a\in H$.
+
+<details>
+<summary>Proof</summary>
+$aH=eH$ if and only if $a\sim e$, if and only if $a\in H$.
+
+</details>
+
+$aH$ is called **left coset** of $a$ in $H$.
+
+<details>
+<summary>Examples</summary>
+
+Consider $G=S_3=\{e,\rho,\rho^2,\tau_1,\tau_2,\tau_3\}$.
+
+where $\rho=(123),\rho^2=(132),\tau_1=(12),\tau_2=(23),\tau_3=(13)$.
+
+$H=\{e,\rho,\rho^2\}$.
+
+All the left coset for $H$ is $H=eH=\rho H=\rho^2H$.
+
+$$
+\tau_1\rho=(23)=\tau_2\\
+\tau_1\rho^2=(13)=\tau_3\\
+\tau_2\rho=(31)=\tau_3\\
+\tau_2\rho^2=(12)=\tau_1
+\tau_3\rho=(12)=\tau_1\\
+\tau_3\rho^2=(23)=\tau_2
+$$
+
+$$
+\tau_1H=\{\tau_1,\tau_2,\tau_3\}=\tau_2H=\tau_3H\\
+$$
+
+---
+
+Consider $G=\mathbb{Z}$ with $H=5\mathbb{Z}$.
+
+We have 5 cosets, $H,1+H,2+H,3+H,4+H$.
+
+</details>
+
+#### Lemma for size of cosets
+
+Any coset of $H$ has the same cardinality as $H$.
+
+Define $\phi:H\to aH$ by $\phi(h)=ah$.
+
+$\phi$ is an bijection, if $ah=ah'\implies h=h'$, it is onto by definition of $aH$.
+
+#### Corollary: Lagrange's Theorem
+
+If $G$ is a finite group, and $H\leq G$, then $|H|\big\vert |G|$. (size of $H$ divides size of $G$)
+
+<details>
+<summary>Proof</summary>
+
+Suppose $H$ has $r$ distinct cosets, then $|G|=r|H|$, so $|H|$ divides $|G|$.
+
+</details>
+
+#### Corollary for Lagrange's Theorem
+
+If $|G|=p$, where $p$ is a prime number, then $G$ is cyclic.
+
+<details>
+<summary>Proof</summary>
+
+Prick $e\neq a\in G$, let $H=\langle a\rangle \leq G$, then $|H|$ divides $|G|$, since $p$ is prime, then $|H|=|G|$, so $G=\langle a \rangle$.
+
+</details>
+
+If $G$ is finite and $a\in G$, then $\operatorname{ord}(a)\big\vert|G|$.
+
+<details>
+<summary>Proof</summary>
+
+Since $\operatorname{ord}(a)=|\langle a\rangle|$, and $\langle a\rangle $ is a subgroup, so $\operatorname{ord}(a)\big\vert|G|$.
+
+</details>
+
+#### Definition of index
+
+Suppose $H\leq G$, the number of distinct left cosets of $H$ is called the index of $H$ in $G$. Notation is $(G:H)$.
+
+#### Definition of right coset
+
+Suppose $H\leq G$, define the equivalence relation by $a\sim 'b$ (or $a\sim_R b$ in some textbook) if $a b^{-1}\in H$. (note the in left coset, we use $a^{-1}b \in H$, or equivalently $b^{-1}a \in H$, these are different equivalence relations)
+
+The equivalent class is defined
+
+$$
+Ha=\{x\in G|x\sim'a\}=\{x\in G|xa^{-1}\in H\}=\{x|x=ha\text{ for some }h\in H\}
+$$
+
+Some properties are the same as the left coset
+
+- $Ha=H\iff a\in H$
+- $Ha=Hb$ if and only if $a\sim'b\iff a b^{-1}\in H$.
+- $Ha\cap Hb\neq \emptyset\iff Ha=Hb$.
+
+Some exercises: Find all the left and right cosets of $G=S_3$, there should be 2 left cosets and 2 right cosets (giving different partition of $G$).

content/Math4302/_meta.js

@@ -11,4 +11,9 @@ export default {
     Math4302_L6: "Modern Algebra (Lecture 6)",
     Math4302_L7: "Modern Algebra (Lecture 7)",
     Math4302_L8: "Modern Algebra (Lecture 8)",
+    Math4302_L9: "Modern Algebra (Lecture 9)",
+    Math4302_L10: "Modern Algebra (Lecture 10)",
+    Math4302_L11: "Modern Algebra (Lecture 11)",
+    Math4302_L12: "Modern Algebra (Lecture 12)",
+    Math4302_L13: "Modern Algebra (Lecture 13)",
 }

content/index.md

@@ -1,7 +1,7 @@
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@@ -13,7 +13,7 @@ The primary audience of this project is for those challenge takers who are takin
 
 So here it is. A lite server for you to read my notes.
 
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