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Difference between revisions of "Lagrange's Theorem"

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By letting <math>K</math> be the [[trivial group | trivial subgroup]], we have
 
By letting <math>K</math> be the [[trivial group | trivial subgroup]], we have
 
<cmath> |G| = (G:H) |H|. </cmath>
 
<cmath> |G| = (G:H) |H|. </cmath>
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In particular, if <math>G</math> is a [[finite]] group of [[order (group theory) | order]] <math>G</math> and <math>H</math> is a subgroup of <math>G</math> of order <math>h</math>,
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In particular, if <math>G</math> is a [[finite]] group of [[order (group theory) | order]] <math>g</math> and <math>H</math> is a subgroup of <math>G</math> of order <math>h</math>,
 
<cmath> g = (G:H) h, </cmath>
 
<cmath> g = (G:H) h, </cmath>
 
so the index and order of <math>H</math> are [[divisor]]s of <math>g</math>.
 
so the index and order of <math>H</math> are [[divisor]]s of <math>g</math>.
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* [[Quotient group]]
  
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[[Category:Group theory]]
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[[Category:Group theory]] [[Category: Theorems]]
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Latest revision as of 11:02, 24 November 2024

Lagrange's theorem is a result on the indices of cosets of a group.

Theorem. Let $G$ be a group, $H$ a subgroup of $G$, and $K$ a subgroup of $H$. Then \[(G:K) = (G:H)(H:K) .\]

Proof. For any $a\in G$, note that $aK \subseteq aH$; thus each left coset mod $K$ is a subset of a left coset mod $H$; since each element of $G$ is in some left coset mod $K$, it follows that the left cosets mod $H$ are unions of left cosets mod $K$. Furthermore, the mapping $x\mapsto ba^{-1}x$ induces a bijection from the left cosets mod $K$ contained in an arbitrary $H$-coset $aH$ to those contained in an arbitrary $H$-coset $bH$. Thus each $H$-coset is a union of $K$-cosets, and the cardinality of the set of $K$-cosets contained in an $H$-coset is independent of the choice of the $H$-coset. The theorem then follows. $\blacksquare$

By letting $K$ be the trivial subgroup, we have \[|G| = (G:H) |H|.\] In particular, if $G$ is a finite group of order $g$ and $H$ is a subgroup of $G$ of order $h$, \[g = (G:H) h,\] so the index and order of $H$ are divisors of $g$.

See also

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