Operator inverse

Suppose we have a binary operation G on a set S, $G:S\times S \to S$ , and suppose this operation has an identity e, so that for every $g\in S$ we have $G(e, g) = G(g, e) = g$ . An inverse to g under this operation is an element $h \in S$ such that $G(h, g) = G(g, h) = e$ .

If our operation is not commutative, we can talk separately about left inverses and right inverses. A left inverse of g would be some h such that $G(h, g) = e$ while a right inverse would be some h such that $G(g, h) = e$ .

Uniqueness (under appropriate conditions)

If the operation G is associative and an element has both a right and left inverse, these two inverses are equal.

Proof

Let g be the element with left inverse h and right inverse h', so $G(h, g) = G(g, h') = e$ . Then $G(G(h, g), h') = G(e, h') = h'$ , by the properties of e. But by associativity, $\displaystyle G(G(h, g), h') = G(h, G(g, h')) = G(h, e) = h$ , so we do indeed have $h = h'$ .

Corollary

If the operation G is associative, inverses are unique.

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Operator inverse

Uniqueness (under appropriate conditions)

Proof

Corollary