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45 lines
1.3 KiB
Markdown
45 lines
1.3 KiB
Markdown
---
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title: "Simplifying the Definition of Algebraic Groups"
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date: 2019-12-10T21:40:00-05:00
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draft: false
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tags: [ "Math" ]
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math: true
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medium_enabled: true
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---
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This post is inspired by the book "Term Rewriting & All That" by Franz Baader and Tobias Nipkow.
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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:
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1. $(x y)z = x (y z)$
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2. $ex = x$
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3. $x^{-1} x = e$
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for some fixed $e \in G$, then we say that $\langle G, * \rangle$ is a group.
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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.
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However, these can be derived by the prior properties.
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### Prove $xx^{-1} = e$
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\begin{align*}
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e &= (xx^{-1})^{-1}(x x^{-1}) \\\\
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&= (xx^{-1})^{-1} (x (ex^{-1})) \\\\
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&= (xx^{-1})^{-1} (x ((x^{-1} x) x^{-1})) \text{ ----- (A)} \\\\
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&= (x x^{-1})^{-1} (x (x^{-1} x)x^{-1}) \\\\
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&= (x x^{-1})^{-1}((x x^{-1})xx^{-1}) \\\\
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&= (x x^{-1})^{-1} ((xx^{-1}) (x x^{-1})) \\\\
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&= ((x x^{-1})^{-1}(x x^{-1})) (x x^{-1}) \\\\
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&= e(xx^{-1}) \\\\
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&= xx^{-1}
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\end{align*}
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### Prove $xe = x$
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Once we showed $xx^{-1} = e$, the proof of $xe = e$ is simple.
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\begin{align*}
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x &= ex \\\\
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&= (xx^{-1})x \\\\
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&= x(x^{-1}x) \\\\
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&= xe
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\end{align*}
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