> [!NOTE] Theorem > Let $(G,\circ)$ be a [[Groups|group]]. Let $g\in G$ with [[Inverse under a binary operation|inverse]] $g^{-1}.$ Then $(g^{-1})^{-1}=g.$ **Proof**: By definition, $a^{-1}a=a^{-1}a=1$which shows that $a^{-1}$ is the inverse of $a.$