Difference between revisions of "Lagrange's Theorem"

Revision as of 00:30, 5 December 2017

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 $g$ .

@@ Line 8: / Line 8: @@
 By letting <math>K</math> be the [[trivial group | trivial subgroup]], we have
 <cmath> |G| = (G:H) |H|. </cmath>
-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>,
+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>
 so the index and order of <math>H</math> are [[divisor]]s of <math>g</math>.

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Difference between revisions of "Lagrange's Theorem"

Revision as of 00:30, 5 December 2017

See also