View source for Operator inverse

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


Thus, informally, operating by <math>g</math> is the "opposite" of operating by <math>g</math>-inverse.


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


==Uniqueness (under appropriate conditions)==
If the operation <math>G</math> is [[associative]] and an element has both a right and left inverse, these two inverses are equal.
===Proof===
Let <math>g</math> be the element with left inverse <math>h</math> and right inverse <math>h'</math>, so <math>G(h, g) = G(g, h') = e</math>.  Then <math>G(G(h, g), h') = G(e, h') = h'</math>, by the properties of <math>e</math>.  But by associativity, <math>G(G(h, g), h') = G(h, G(g, h')) = G(h, e) = h</math>, so we do indeed have <math>h = h'</math>.

===Corollary===
If the operation <math>G</math> is associative, inverses are unique.

[[Category:Abstract algebra]]
[[Category:Definition]]