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Brandon Rozek 2022-06-11 20:09:26 -04:00
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## 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)

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---
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.
$$
\begin{align*}
x &= ex \\\\
&= (xx^{-1})x \\\\
&= x(x^{-1}x) \\\\
&= xe
\end{align*}
$$

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