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:

\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 $X$. |