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uniqueness of inverse (for groups)
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Lemma Suppose $(G,\ast)$ is a group. Then every element in $G$ has a unique inverse.
Proof. Suppose $g\in G$ . By the group axioms we know that there is an $h\in G$ such that $$ g\ast h = h\ast g = e,$$ where $e$ is the identity element in $G$ . If there is also a $h'\in G$ satisfying $$ g\ast h' = h'\ast g = e,$$ then $$ h = h\ast e = h \ast(g \ast h') =(h\ast g)\ast h' = e\ast h' = h',$$ so $h=h'$ , and $g$ has a unique inverse. 
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"uniqueness of inverse (for groups)" is owned by waj. [ owner history (1) ]
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Cross-references: identity element, axioms, proof, 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 3119 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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Pending Errata and Addenda
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