Let $(G,\ast)$ be a \reftext{def:group-2026a}{group}. We write $ab$ for $a\ast b$, and we use the identity element $e_G$ and the inverse $a^{-1}$ of an element $a$, both of which are well defined by \ref{thm:group-identity-inverse-uniqueness-2026a}. Then for all $a,b,c\in G$ the following hold.

1. (\textit{Cancellation}) If $ab=ac$ then $b=c$, and if $ba=ca$ then $b=c$.
2. $(a^{-1})^{-1}=a$.
3. $(ab)^{-1}=b^{-1}a^{-1}$.
4. $e_G^{-1}=e_G$.
5. (\textit{Unique solvability}) There is exactly one $x\in G$ with $ax=b$, namely $x=a^{-1}b$, and exactly one $y\in G$ with $ya=b$, namely $y=ba^{-1}$.