Uniqueness of inverses

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 $(ab)^{-1} = b^{-1} a^{-1}$.

Going deeper

For a student

The definition of a group asks for an inverse $b$ with $ab = e$ and $ba = e$. This result shows that one of the two equations is enough, and that the inverse is unique.

The proof uses only associativity and the identity:

1. Suppose that $ab = e$.
2. Multiply on the left by $a^{-1}$: $a^{-1}(ab) = a^{-1}e$.
3. By associativity, the left side is $(a^{-1}a)b = eb = b$.
4. The right side is $a^{-1}$. Thus $b = a^{-1}$.

A worked example in $\mathbb{Z}/7\mathbb{Z}$ under multiplication of the units:

1. We look for the inverse of $3$.
2. We have $3 \cdot 5 = 15 = 2 \cdot 7 + 1$, so $3 \cdot 5 = 1$.
3. By the result, $5$ is the only inverse of $3$. There is no need to check $5 \cdot 3$.

Two standard consequences follow at once. First, $(a^{-1})^{-1} = a$, because $a^{-1}a = e$. Second, $(ab)^{-1} = b^{-1}a^{-1}$, because $(ab)(b^{-1}a^{-1}) = e$.

A common error is to write $(ab)^{-1} = a^{-1}b^{-1}$. This is false when $a$ and $b$ do not commute. In $S_3$, take $a = (1\,2)$ and $b = (1\,3)$: then $a^{-1}b^{-1} = (1\,2)(1\,3)$, but $(ab)^{-1} = (1\,3)(1\,2)$, and these differ.

The step that needs care is step 2. It uses the inverse $a^{-1}$, so the argument is for groups. In a monoid an element can have a right inverse and no left inverse.