Difference between revisions of "Operator inverse"
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| − | Suppose we have a [[binary operation]] G on a set S, <math>G:S\times S \to S</math>, and suppose this operation has an [[identity]] e, so that for every <math>g\in S</math> we have <math>G(e, g) = G(g, e) = g</math>. An '''inverse to ''g | + | 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 | + | 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 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>. | + | 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)== | ==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. | + | If the operation <math>G</math> is [[associative]] and an element has both a right and left inverse, these two inverses are equal. |
===Proof=== | ===Proof=== | ||
| − | Let g be the element with left inverse h and right inverse h', 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 e. But by associativity, <math>\displaystyle G(G(h, g), h') = G(h, G(g, h')) = G(h, e) = h</math>, so we do indeed have <math>h = h'</math>. | + | 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>\displaystyle 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=== | ===Corollary=== | ||
| − | If the operation G is associative, inverses are unique. | + | If the operation <math>G</math> is associative, inverses are unique. |
Revision as of 13:13, 14 August 2006
Suppose we have a binary operation 
on a set 
, 
, and suppose this operation has an identity 
, so that for every 
we have 
. An inverse to 
under this operation is an element 
such that 
.
Thus, informally, operating by 
is the "opposite" of operating by -inverse.
If our operation is not commutative, we can talk separately about left inverses and right inverses. A left inverse of would be some 
such that 
, while a right inverse would be some such that 
.
Uniqueness (under appropriate conditions)
If the operation is associative and an element has both a right and left inverse, these two inverses are equal.
Proof
Let be the element with left inverse and right inverse 
, so 
. Then 
, by the properties of . But by associativity, 
, so we do indeed have 
.
Corollary
If the operation is associative, inverses are unique.
