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## Brandon Rozek PhD Student @ RPI studying Automated Reasoning in AI and Linux Enthusiast.

# Simplifying the Definition of Algebraic Groups

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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*}