Difference between revisions of "Operator inverse"

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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 <math>G(h, g) = e</math> while a right inverse would be some h such that <math>G(g, h) = e</math>.
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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 <math>G(h, g) = e</math>, while a right inverse would be some h such that <math>G(g, h) = e</math>.
  
  

Revision as of 13:09, 18 July 2006

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 $G(h, g) = G(g, h) = e$.


Thus, informally, operating by g is the "opposite" of operating by g-inverse.


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.