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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>.
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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>.
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Thus, informally, operating by <math>g</math> is the "opposite" of operating by <math>g</math>-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 <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>.
  
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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>.
 
  
 
==Uniqueness (under appropriate conditions)==
 
==Uniqueness (under appropriate conditions)==
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If the operation G is [[associative]] and an element has both a right and left inverse, these two inverses are equal.
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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===
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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>.
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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===
 
===Corollary===
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If the operation G is associative, inverses are unique.
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If the operation <math>G</math> is associative, inverses are unique.
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[[Category:Abstract algebra]]
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[[Category:Definition]]

Latest revision as of 11:40, 23 November 2007

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 $\mathbf g$ 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, $G(G(h, g), h') = G(h, G(g, h')) = G(h, e) = h$, so we do indeed have $h = h'$.

Corollary

If the operation $G$ is associative, inverses are unique.

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