Corollary 14.5

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Let $H$ be a normal subgroup of $G$. Then the cosets of $H$ form a group $G / H$ under the binary operation $(aH)(bH) = (ab)H$.


Computing, $(aH)[(bH)(cH)]=(aH)[(bc)H]=[a(bc)]H$, and similarly, we have $[(aH)(bH)](cH)=[(ab)cH]$, so associativity in $G/H$ follows from associativity in G. Because $(aH)(eH)=(ae)H=aH=(ea)H=(eH)(aH)$, we see that $eH=H$ is the identity element in $G/H$. Finally, $(a^{-1}H)(aH)=(a^{-1}a)H=eH=(aa^{-1})H=(aH)(a^{-1}H)$ shows that $a^{-1}H=(aH)^{-1}$.

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