Changed PDF notes to markdown notes

content/notes/_index.md

@@ -15,6 +15,7 @@ purposes.
 ## Courses
 - [Probability and Statistical Inference I](stat381)
 - [Real Analysis Quick Facts](realanalysis)
+- [Groups in Abstract Algebra](groups-abstract-algebra)
 - [Some Abstract 2 Definitions](abstract2def)
 - [Handy Facts about Quadratic Congruences](quadraticcongruences)
 - [Bayesian Statistics](bayesianstatistics)

content/notes/groups-abstract-algebra.md

@@ -0,0 +1,46 @@
+---
+title: "Groups in Abstract Algebra"
+draft: false
+tags: []
+math: true
+---
+
+Let us have a set $G$ together with some binary operation $*$.
+We will use multipicative notation where $ab = a * b$.
+Let $x, y, z \in G$. If $\langle G, *\rangle$ has the
+following properties:
+1. $(xy)z = x(yz)$
+2. $ex = x$
+3. $x^{-1}x = e$
+
+for some fixed $e \in G$, then we say that $\langle G, *\rangle$ is a group.
+In my class, we were also told to show that $xe = x$ and $xx^{-1} = e$.
+However, these can be derived by the prior three properties.
+
+## Prove $xx^{-1} = e$
+
+$$
+\begin{align*}
+e &= (xx^{-1})^{-1}(xx^{-1}) \\\\
+  &= (xx^{-1})^{-1}(x(ex^{-1})) \\\\
+  &= (xx^{-1})^{-1}(x((x^{-1}x)x^{-1})) \\\\
+  &= (xx^{-1})^{-1}(x(x^{-1}x)x^{-1}) \\\\
+  &= (xx^{-1})^{-1}((xx^{-1})(xx^{-1})) \\\\
+  &= ((xx^{-1})^{-1}(xx^{-1}))(xx^{-1}) \\\\
+  &= e(xx^{-1}) \\\\
+  &= xx^{-1} \\\\
+\end{align*}
+$$
+
+## Prove $xe = x$
+
+We can use the last proof to solve this faster.
+
+$$
+\begin{align*}
+x &= ex \\\\
+  &= (xx^{-1})x \\\\
+  &= x(x^{-1}x) \\\\
+  &= xe
+\end{align*}
+$$