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## Courses
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- [Probability and Statistical Inference I](stat381)
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- [Real Analysis Quick Facts](realanalysis)
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- [Groups in Abstract Algebra](groups-abstract-algebra)
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- [Some Abstract 2 Definitions](abstract2def)
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- [Handy Facts about Quadratic Congruences](quadraticcongruences)
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- [Bayesian Statistics](bayesianstatistics)
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content/notes/groups-abstract-algebra.md
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46
content/notes/groups-abstract-algebra.md
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---
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title: "Groups in Abstract Algebra"
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draft: false
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tags: []
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math: true
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---
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Let us have a set $G$ together with some binary operation $*$.
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We will use multipicative notation where $ab = a * b$.
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Let $x, y, z \in G$. If $\langle G, *\rangle$ has the
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following properties:
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1. $(xy)z = x(yz)$
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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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In my class, we were also told to show that $xe = x$ and $xx^{-1} = e$.
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However, these can be derived by the prior three properties.
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## Prove $xx^{-1} = e$
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$$
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\begin{align*}
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e &= (xx^{-1})^{-1}(xx^{-1}) \\\\
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&= (xx^{-1})^{-1}(x(ex^{-1})) \\\\
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&= (xx^{-1})^{-1}(x((x^{-1}x)x^{-1})) \\\\
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&= (xx^{-1})^{-1}(x(x^{-1}x)x^{-1}) \\\\
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&= (xx^{-1})^{-1}((xx^{-1})(xx^{-1})) \\\\
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&= ((xx^{-1})^{-1}(xx^{-1}))(xx^{-1}) \\\\
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&= e(xx^{-1}) \\\\
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&= xx^{-1} \\\\
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\end{align*}
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$$
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## Prove $xe = x$
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We can use the last proof to solve this faster.
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$$
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\begin{align*}
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x &= ex \\\\
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&= (xx^{-1})x \\\\
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&= x(x^{-1}x) \\\\
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&= xe
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\end{align*}
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$$
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