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''' under this operation is an element <math>h \in S</math> such that <math>G(h, g) = G(g, h) = e</math>. | + | 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'' ''' under this operation is an element <math>h \in S</math> such that <math>G(h, g) = G(g, h) = e</math>. |
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| + | Thus, informally, operating by ''g'' is the "opposite" of operating by ''g''-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>. | 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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==Uniqueness (under appropriate conditions)== | ==Uniqueness (under appropriate conditions)== | ||
Revision as of 08:13, 18 July 2006
Suppose we have a binary operation G on a set S, 
, and suppose this operation has an identity e, so that for every 
we have 
. An inverse to g under this operation is an element 
such that 
.
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 
while a right inverse would be some h such that 
.
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 
. Then 
, by the properties of e. But by associativity, 
, so we do indeed have 
.
Corollary
If the operation G is associative, inverses are unique.
