Inverse Operation Distributes in Groups

Theorem

For any Group $G$ then:

(a b)^{- 1} = b^{- 1} a^{- 1}

*Proof

Check the inverse property:

\begin{aligned} (a b) (b^{- 1} a^{- 1}) & = a (b b^{- 1}) a^{- 1} \\ = a (1) a^{- 1} \\ = (a 1) a^{- 1} \\ = a a^{- 1} \\ = 1 \end{aligned}

☐

As a corollary:

The product of invertible elements is invertible

$(a_{1} a_{2} \dots a_{n})^{- 1} = a_{n}^{- 1} \dots a_{1}^{- 1}$ by induction on our previous theorem.