PlanetMath: inverse of a product

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inverse of a product

(Theorem)

Theorem 1 If

and

are arbitrary elements of the group $(G,\,*)$ , then the inverse of

$\displaystyle (a*b)^{-1} = b^{-1}*a^{-1}.$

(1)

Proof. Let the neutral element of the group, which may be proved unique, be . Using only the group postulates we obtain

$\displaystyle (a*b)*(b^{-1}*a^{-1}) = a*(b*(b^{-1}*a^{-1})) = a*((b*b^{-1})*a^{-1}) = a*(e*a^{-1}) = a*a^{-1} = e,$

$\displaystyle (b^{-1}*a^{-1})*(a*b) = b^{-1}*(a^{-1}*(a*b)) = b^{-1}*((a^{-1}*a)*b) = b^{-1}*(e*b) = b^{-1}*b = e,$

Q.E.D.

Note. The formula (1) may be by induction extended to the form

$\displaystyle (a_1*\cdots*a_n)^{-1} = a_n^{-1}*\cdots*a_1^{-1}.$

"inverse of a product" is owned by pahio.

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Other names:

inverse of a product in group, inverse of product

Keywords:

inverse, group operation

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Cross-references: induction, postulates, neutral element, proof, inverse, group

This is version 13 of inverse of a product, born on 2004-12-13, modified 2008-05-10.
Object id is 6575, canonical name is InverseFormingInProportionToGroupOperation.
Accessed 2773 times total.

Classification:

AMS MSC:	20-00 (Group theory and generalizations :: General reference works )
	20A05 (Group theory and generalizations :: Foundations :: Axiomatics and elementary properties)

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