The Group of Invertible Elements in an Algebra

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 $X$.
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License