Groups Simplified

Brandon Rozek

December 10, 2019

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. In class, we needed to show that $xe = x$ and $xx^{-1} = e$. 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*}