AKA **Socks-Shoes Property**. If one thinks of $a$ as putting on socks, $b$ as putting on shoes, $a^{-1}$ as taking off socks, and $b^{-1}$ as taking off shoes, the theorem demonstrates the order in which one must perform these actions. $ab$ would represent putting on socks followed by shoes. In order to take them off, they must be removed in reverse order, that is, $b^{-1}a^{-1}.$ > [!NOTE] Theorem (Inverse of product) > For [[Groups|group]] elements $a$ and $b,$ $(ab)^{-1}=b^{-1}a^{-1}.$ *Proof*. By associativity, $(b^{-1}a^{-1})(ab)=b^{-1}(a^{-1}a)b=b^{-1}1b=1.$ Similarly $(ab)(b^{-1}a^{-1})=1.$