Lagrange's theorem
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Revision 308
7/7/2026, 3:00:04 PM · Ancient Tree
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Let $G$ be a finite [[Group|group]] and $H \subseteq G$ a [[Subgroup|subgroup]]. Then, the [[Order of a finite group|order]] of $H$ [[Divisibility|divides]] the order of $G$:2
$$|H| \;\mid\; |G|$$2
$$|H| \;\mid\; |G|.$$3
More precisely, we have : $|G| = [G:H]\cdot |H|$ where $[G,H]$ is the [[Index of a subgroup|index]] of $H$ in $G$.Revision 307
7/7/2026, 2:59:50 PM · Ancient Tree
Concept edited
Compare with revision 306No text changes
1
Let $G$ be a finite [[Group|group]] and $H \subseteq G$ a [[Subgroup|subgroup]]. Then, the [[Order of a finite group|order]] of $H$ [[Divisibility|divides]] the order of $G$:2
$$|H| \;\mid\; |G|$$3
More precisely, we have : $|G| = [G:H]\cdot |H|$ where $[G,H]$ is the [[Index of a subgroup|index]] of $H$ in $G$.Revision 306
7/7/2026, 2:57:56 PM · Ancient Tree
Concept edited
Compare with revision 874 changed lines
1
Finite groups...1
Let $G$ be a finite [[Group|group]] and $H \subseteq G$ a [[Subgroup|subgroup]]. Then, the [[Order of a finite group|order]] of $H$ [[Divisibility|divides]] the order of $G$:2
$$|H| \;\mid\; |G|$$3
More precisely, we have : $|G| = [G:H]\cdot |H|$ where $[G,H]$ is the [[Index of a subgroup|index]] of $H$ in $G$.Revision 87
6/26/2026, 7:43:10 AM · Ancient Tree
Concept created
Finite groups...