Theorem 10.10

Return to Theorems, Glossary, Homework Problems.


Let $H$ be a subgroup of a finite group $G$. Then the order of $H$ is a divisor of the order of $G$.


Let $n$ be the order of $G$, and let $H$ have order $m$. Since we know every coset (left or right) of a subgroup $H$ of a group $G$ has the same number of elements as $H$, it follows that every coset of $H$ also has $m$ elements. Let $r$ be a number of cells in the partition of $G$ into left cosets of $H$. Then $n=rm$, so $m$ is indeed a divisor of $n$.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License