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Prove that \emptyset \neq H\subseteq G\, and \forall a,b\in H,ab^{{-1}}\in H\,, if and only if H\, is a subgroup of G\,.

If H\, is a subgroup of G\,, by the properties of identity, closure and inverse, it is true that \emptyset \neq H\, and \forall a,b\in H,ab^{{-1}}\in H\,.

If \forall a,b\in H\neq \emptyset ,ab^{{-1}}\in H\,, show that this implies H\, is a subgroup of G\,.

Identity: First let b=a\, so that aa^{{-1}}=e\in H\,.

Inverse: Let a=e\, so that eb^{{-1}}=b^{{-1}}\in H\,.

Closure: Let b=b^{{-1}}\, so that a(b^{{-1}})^{{-1}}=ab\in H\,.

Associativity: If \exists a,b,c\in H\mid (ab)c\neq a(bc)\,, then \exists a,b,c\in G\mid (ab)c\neq a(bc)\, which is not true by the assumption that G\, is a group.

Abstract Algebra

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