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[parent] uniqueness of inverse (for groups) (Result)

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

$\displaystyle g\ast h = h\ast g = e,$
where $ e$ is the identity element in $ G$. If there is also a $ h'\in G$ satisfying
$\displaystyle g\ast h' = h'\ast g = e,$
then
$\displaystyle h = h\ast e = h \ast(g \ast h') =(h\ast g)\ast h' = e\ast h' = h',$
so $ h=h'$, and $ g$ has a unique inverse. $ \Box$



"uniqueness of inverse (for groups)" is owned by waj. [ owner history (1) ]
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See Also: uniqueness of additive inverse in a ring, identity element is unique


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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 MSC20-00 (Group theory and generalizations :: General reference works )
 20A05 (Group theory and generalizations :: Foundations :: Axiomatics and elementary properties)

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