The Group of Invertible Elements in an Algebra

# 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:

\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 $\mathfrak{A}$. |