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Difference between revisions of "Operator inverse"

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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.
 
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 <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>.
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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===
 
If the operation <math>G</math> is associative, inverses are unique.
 
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 10:40, 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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