The Group of Invertible Elements in an Algebra
 Table of Contents

# The Group of Invertible Elements in an Algebra

 Proposition 1: Let $X$ be an algebra with unit. Then $(\mathrm{Inv}(X), \cdot )$ is a group.
• Proof: We will prove that all of the group axioms hold.
• 1. Closure under Multiplication: Let $x, y \in \mathrm{Inv}(X)$. Let $z = y^{-1}x^{-1}$. Then:
(1)
\begin{align} \quad xyz = xy(y^{-1}x^{-1}) = 1 \\ \quad zxy = (y^{-1}x^{-1})xy = 1 \end{align}
• Thus $xy \in \mathrm{Inv}(X)$ and $(xy)^{-1} = y^{-1}x^{-1}$.
• 2. Associativity of Multiplication: This follows immediately from the fact that $X$ is an algebra (and so $\cdot$ is associative).
• 3. Existence of an Identity Element: This hollows immediately from the fact that $X$ is an algebra with unit.
• 4. Existence of Inverse Elements: Again, this follows by the definition of $\mathrm{Inv}(X)$.
• Thus $(\mathrm{Inv}(X), \cdot)$ is a group. $\blacksquare$
 Definition: Let $X$ be an algebra with unit. Then $(\mathrm{Inv}, \cdot)$ is called the Group of Invertible Elements for the algebra $X$.
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