Difference between revisions of "Lagrange's Theorem"
(statement, proof, and a few remarks) |
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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> | ||
− | In particular, if <math>G</math> is a [[finite]] group of [[order (group theory) | order]] <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> | <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>. |
Revision as of 00:30, 5 December 2017
Lagrange's theorem is a result on the indices of cosets of a group.
Theorem. Let be a group,
a subgroup of
, and
a subgroup of
. Then
Proof. For any , note that
; thus each left coset mod
is a subset of a left coset mod
; since each element of
is in some left coset mod
, it follows that the left cosets mod
are unions of left cosets mod
. Furthermore, the mapping
induces a bijection from the left cosets mod
contained in an arbitrary
-coset
to those contained in an arbitrary
-coset
. Thus each
-coset is a union of
-cosets, and the cardinality of the set of
-cosets contained in an
-coset is independent of the choice of the
-coset. The theorem then follows.
By letting be the trivial subgroup, we have
In particular, if
is a finite group of order
and
is a subgroup of
of order
,
so the index and order of
are divisors of
.