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Proof

(related to Theorem: Order of Subgroup Divides Order of Finite Group)

• By hypothesis, \((G,\ast)\) is a group with a finite order \(|G| < \infty\) and \(H\subseteq G\) is its subgroup.
• All left cosets of $G$ are disjoint and $G$ is the set union of them.
• Since the subgroups and their cosets are equipotent, their order is $|H|$ and there are exactly $|G/H|$ of them.
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References

Bibliography

1. Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013