diff --git a/content/Math4202/Math4202_L22.md b/content/Math4202/Math4202_L22.md
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+# Math4202 Topology II (Lecture 22)
+
+## Final reading, report, presentation
+
+- Mar 30: Reading topic send email or discuss in OH.
+- Apr 3: Finalize the plan.
+- Apr 22,24: Last two lectures: 10 minutes to present.
+- Final: type a short report, 2-5 pages.
+
+## Algebraic topology
+
+### Fundamental theorem of Algebra
+
+For arbitrary polynomial $f(z)=\sum_{i=0}^n a_i x^i$. Are there roots in $\mathbb{C}$?
+
+Consider $f(z)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0$ is a continuous map from $\mathbb{C}\to\mathbb{C}$.
+
+If $f(z_0)=0$, then $z_0$ is a root.
+
+By contradiction, Then $f:\mathbb{C}\to\mathbb{C}-\{0\}\cong \mathbb{R}^2-\{(0,0)\}$.
+
+#### Theorem for existence of n roots
+
+A polynomial equation $x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0=0$ of degree $>0$ with complex coefficients has at least one complex root.
+
+There are $n$ roots by induction.
+
+#### Lemma
+
+If $g:S^1\to \mathbb{R}^2-\{(0,0)\}$ is the map $g(z)=z^n$, then $g$ is not nulhomotopic. $n\neq 0$, $n\in \mathbb{Z}$.
+
+> Recall that we proved that $g(z)=z$ is not nulhomotopic.
+
+Consider $k:S^1\to S^1$ by $k(z)=z^n$. $k$ is continuous, $k_*:\pi_1(S^1,1)\to \pi_1(S^1,1)$.
+
+Where $\pi_1(S^1,1)\cong \mathbb{Z}$.
+
+$k_*(n)=nk_*(1)$.
+
+Recall that the path in the loop $p:I\to S^1$ where $p:t\mapsto e^{2\pi it}$.
+
+$k_*(p)=[k(p(t))]$, where $n=\tilde{k\circ p}(1)$.
+
+$k_*$ is injective.
+
diff --git a/content/Math4202/_meta.js b/content/Math4202/_meta.js
index 27cf5ba..c32309c 100644
--- a/content/Math4202/_meta.js
+++ b/content/Math4202/_meta.js
@@ -25,4 +25,5 @@ export default {
     Math4202_L19: "Topology II (Lecture 19)",
     Math4202_L20: "Topology II (Lecture 20)",
     Math4202_L21: "Topology II (Lecture 21)",
+    Math4202_L22: "Topology II (Lecture 22)",
 }
diff --git a/content/Math4302/Math4302_L23.md b/content/Math4302/Math4302_L23.md
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+# Math4302 Modern Algebra (Lecture 23)
+
+## Group
+
+### Group acting on a set
+
+#### Theorem for the orbit of a set with prime power group
+
+Suppose $X$ is a $G$-set, and $|G|=p^n$ where $p$ is prime, then $|X_G|\equiv |X|\mod p$.
+
+Where $X_G=\{x\in X|g\cdot x=x\text{ for all }g\in G\}=\{x\in X|\text{orbit of }x\text{ is trivial}\}$
+
+#### Corollary: Cauchy's theorem
+
+If $p$, where $p$ is a prime, divides $|G|$, then $G$ has a subgroup of order $p$. (equivalently, $g$ has an element of order $p$)
+
+> This does not hold when $p$ is not prime.
+>
+> Consider $A_4$ with order $12$, and $A_4$ has no subgroup of order $6$.
+
+#### Corollary: Center of prime power group is non-trivial
+
+If $|G|=p^m$, then $Z(G)$ is non-trivial. ($Z(G)\neq \{e\}$)
+
+<details>
+<summary>Proof</summary>
+
+Let $G$ act on $G$ via conjugation, then $g\cdot h=ghg^{-1}$. This makes $G$ to a $G$-set.
+
+Apply the theorem, the set of elements with trivial orbit is; Let $X=G$, then $X_G=\{h\in G|g\cdot h=h\text{ for all }g\in G\}=\{h\in G|ghg^{-1}=h\text{ for all }g\in G\}=Z(G)$.
+
+Therefore $|Z(G)|\equiv |G|\mod p$.
+
+So $p$ divides $|Z(G)|$, so $|Z(G)|\neq 1$, therefore $Z(G)$ is non-trivial.
+</details>
+
+#### Proposition: Prime square group is abelian
+
+If $|G|=p^2$, where $p$ is a prime, then $G$ is abelian.
+
+<details>
+<summary>Proof</summary>
+
+Since $Z(G)$ is a subgroup of $G$, $|Z(G)|$ divides $p^2$ so $|Z(G)|=1, p$ or $p^2$.
+
+By corollary center of prime power group is non-trivial, $Z(G)\neq 1$.
+
+If $|Z(G)|=p$. If $|Z(G)|=p$, then consider the group $G/Z(G)$ (Note that $Z(G)\trianglelefteq G$). We have $|G/Z(G)|=p$ so $G/Z(G)$ is cyclic (by problem 13.39), therefore $G$ is abelian.
+
+If $|Z(G)|=p^2$, then $G$ is abelian.
+
+</details>
+
+### Classification of small order
+
+Let $G$ be a group
+
+- $|G|=1$
+  - $G=\{e\}$
+- $|G|=2$
+  - $G\simeq\mathbb{Z}_2$ (prime order)
+- $|G|=3$
+  - $G\simeq\mathbb{Z}_3$ (prime order)
+- $|G|=4$
+  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_2$
+  - $G\simeq\mathbb{Z}_4$
+- $|G|=5$
+  - $G\simeq\mathbb{Z}_5$ (prime order)
+- $|G|=6$
+  - $G\simeq S_3$
+  - $G\simeq\mathbb{Z}_3\times \mathbb{Z}_2\simeq \mathbb{Z}_6$
+<details>
+<summary>Proof</summary>
+
+$|G|$ has an element of order $2$, namely $b$, and an element of order $3$, namely $a$.
+
+So $e,a,a^2,b,ba,ba^2$ are distinct.
+
+Therefore, there are only two possibilities for value of $ab$. ($a,a^2$ are inverse of each other, $b$ is inverse of itself.)
+
+If $ab=ba$, then $G$ is abelian, then $G\simeq \mathbb{Z}_2\times \mathbb{Z}_3$.
+
+If $ab=ba^2$, then $G\simeq S_3$.
+</details>
+
+- $|G|=7$
+  - $G\simeq\mathbb{Z}_7$ (prime order)
+- $|G|=8$
+  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$
+  - $G\simeq\mathbb{Z}_4\times \mathbb{Z}_2$
+  - $G\simeq\mathbb{Z}_8$
+  - $G\simeq D_4$
+  - $G\simeq$ quaternion group $\{e,i,j,k,-1,-i,-j,-k\}$ where $i^2=j^2=k^2=-1$, $(-1)^2=1$. $ij=l$, $jk=i$, $ki=j$, $ji=-k$, $kj=-i$, $ik=-j$.
+- $|G|=9$
+  - $G\simeq\mathbb{Z}_3\times \mathbb{Z}_3$
+  - $G\simeq\mathbb{Z}_9$ (apply the corollary, $9=3^2$, these are all the possible cases)
+- $|G|=10$
+  - $G\simeq\mathbb{Z}_5\times \mathbb{Z}_2\simeq \mathbb{Z}_{10}$
+  - $G\simeq D_5$
+- $|G|=11$
+  - $G\simeq\mathbb{Z}_11$ (prime order)
+- $|G|=12$
+  - $G\simeq\mathbb{Z}_3\times \mathbb{Z}_4$
+  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3$
+  - $A_4$
+  - $D_6\simeq S_3\times \mathbb{Z}_2$
+  - ??? One more
+- $|G|=13$
+  - $G\simeq\mathbb{Z}_{13}$ (prime order)
+- $|G|=14$
+  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_7$
+  - $G\simeq D_7$
+
+#### Lemma for group of order $2p$ where $p$ is prime
+
+If $p$ is prime, $p\neq 2$, and $|G|=2p$, then $G$ is either abelian $\simeq \mathbb{Z}_2\times \mathbb{Z}_p$ or $G\simeq D_p$
+
+<details>
+<summary>Proof</summary>
+
+We know $G$ has an element of order 2, namely $b$, and an element of order $p$, namely $a$.
+
+So $e,a,a^2,\dots ,a^{p-1},ba,ba^2,\dots,ba^{p-1}$ are distinct elements of $G$.
+
+Consider $ab$, if $ab=ba$, then $G$ is abelian, then $G\simeq \mathbb{Z}_2\times \mathbb{Z}_p$.
+
+If $ab=ba^{p-1}$, then $G\simeq D_p$.
+
+$ab$ cannot be inverse of other elements, if $ab=ba^t$, where $2\leq t\leq p-2$, then $bab=a^t$, then $(bab)^t=a^{t^2}$, then $ba^tb=a^{t^2}$, therefore $a=a^{t^2}$, then $a^{t^2-1}=e$, so $p|(t^2-1)$, therefore $p|t-1$ or $p|t+1$.
+
+This is not possible since $2\leq t\leq p-2$.
+
+</details>
\ No newline at end of file
diff --git a/content/Math4302/_meta.js b/content/Math4302/_meta.js
index e4259d6..740b2f9 100644
--- a/content/Math4302/_meta.js
+++ b/content/Math4302/_meta.js
@@ -24,4 +24,6 @@ export default {
     Math4302_L19: "Modern Algebra (Lecture 19)",
     Math4302_L20: "Modern Algebra (Lecture 20)",
     Math4302_L21: "Modern Algebra (Lecture 21)",
+    Math4302_L22: "Modern Algebra (Lecture 22)",
+    Math4302_L23: "Modern Algebra (Lecture 23)",
 }