website/content/blog/groupssimplified.md

title	date	draft	tags	math	medium_enabled
Simplifying the Definition of Algebraic Groups	2019-12-10T21:40:00-05:00	false	Math	true	true

This post is inspired by the book "Term Rewriting & All That" by Franz Baader and Tobias Nipkow.

Let us have a set G together with a binary operation *. We will use multiplicative notation throughout meaning ab = a * b. Let x, y, z \in G. If \langle G , * \rangle has the following properties:

1. (x y)z = x (y z)
2. ex = x
3. x^{-1} x = e

for some fixed e \in G, then we say that \langle G, * \rangle is a group.

When I was taking Abstract Algebra, we needed to also show that xe = x and xx^{-1} = e for an algebraic structure to be a group.

However, these can be derived by the prior properties.

Prove xx^{-1} = e

\begin{align*} e &= (xx^{-1})^{-1}(x x^{-1}) \\ &= (xx^{-1})^{-1} (x (ex^{-1})) \\ &= (xx^{-1})^{-1} (x ((x^{-1} x) x^{-1})) \text{ ----- (A)} \\ &= (x x^{-1})^{-1} (x (x^{-1} x)x^{-1}) \\ &= (x x^{-1})^{-1}((x x^{-1})xx^{-1}) \\ &= (x x^{-1})^{-1} ((xx^{-1}) (x x^{-1})) \\ &= ((x x^{-1})^{-1}(x x^{-1})) (x x^{-1}) \\ &= e(xx^{-1}) \\ &= xx^{-1} \end{align*}

Prove xe = x

Once we showed xx^{-1} = e, the proof of xe = e is simple. \begin{align*} x &= ex \\ &= (xx^{-1})x \\ &= x(x^{-1}x) \\ &= xe \end{align*}