Uniqueness of inverses

Statement

Let $G$ be a group with identity element $e$, and let $a, b \in G$. If $ab = e$, then $b = a^{-1}$.

Remarks

In particular, each element of a group has exactly one inverse. The same argument works in a monoid: if $ba = e$ and $ac = e$, then $b = b(ac) = (ba)c = c$. Two consequences are $(a^{-1})^{-1} = a$ and $(ab)^{-1} = b^{-1} a^{-1}$.

Explanations

L1 human

In a group, you only need to check one equation to know that $b$ is the inverse of $a$. Multiply $ab = e$ on the left by $a^{-1}$, and you get $b = a^{-1}$. Thus the notation $a^{-1}$ makes sense: there is only one inverse.