The Group of Invertible Elements in an Algebra

Proposition 1: Let $\mathfrak{A}$ be an algebra with unit. Then $(\mathrm{Inv}(\mathfrak{A}), \cdot )$ is a group.

• Proof: We will prove that all of the group axioms hold.

• 1. Closure under Multiplication: Let $a, b \in \mathrm{Inv}(\mathfrak{A})$. Let $z = b^{-1}a^{-1}$. Then:

(1)

\begin{align} \quad (ab)z = ab(b^{-1}a^{-1}) = 1 \\ \quad z(ab) = (b^{-1}a^{-1})ab = 1 \end{align}

• Thus $ab \in \mathrm{Inv}(\mathfrak{A})$ and $(ab)^{-1} = b^{-1}a^{-1}$.

• 2. Associativity of Multiplication: This follows immediately from the fact that $\mathfrak{A}$ is an algebra (and so $\cdot$ is associative).

• 3. Existence of an Identity Element: This also follows immediately from the fact that $\mathfrak{A}$ is an algebra with unit.

• 4. Existence of Inverse Elements: Again, this follows by the definition of $\mathrm{Inv}(\mathfrak{A})$.

• Thus $(\mathrm{Inv}(\mathfrak{A}), \cdot)$ is a group. $\blacksquare$

Definition: Let $\mathfrak{A}$ be an algebra with unit. Then $(\mathrm{Inv}(\mathfrak{A}), \cdot)$ is called the Group of Invertible Elements for the algebra $\mathfrak{A}$.