PlanetMath: uniqueness of inverse (for groups)

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uniqueness of inverse (for groups)

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Lemma Suppose $(G,\ast)$ is a group. Then every element in has a unique inverse.

Proof. Suppose $g\in G$ . By the group axioms we know that there is an $h\in G$ such that

$\displaystyle g\ast h = h\ast g = e,$

where

. If there is also a $h'\in G$ satisfying

$\displaystyle g\ast h' = h'\ast g = e,$

then

$\displaystyle h = h\ast e = h \ast(g \ast h') =(h\ast g)\ast h' = e\ast h' = h',$

, and

has a unique inverse. $\Box$

"uniqueness of inverse (for groups)" is owned by waj. [ owner history (1) ]

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Cross-references: identity element, axioms, inverse, group
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This is version 2 of uniqueness of inverse (for groups), born on 2004-03-11, modified 2004-03-11.
Object id is 5687, canonical name is UniquenessOfInverseForGroups.
Accessed 2096 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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