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8. Lagrange's Theorem

Lagrange's theorem is the first great result of group theory. It states

THEOREM: The order of a subgroup Hof group G divides the order ofG.

First we need to define the orderof a group or subgroup

Definition: IfG is a finite group (or subgroup) then the order ofG is the number of elements ofG.

Lagrange's Theorem simply states that the number of elements in any subgroup of a finite group must divide evenly into the number of elementsin the group. Note that the {A, B} subgroup of the Atayun-HOOT! group has 2 elements while the Atayun-HOOT! group has 4 members.Also we can recall that the subgroups ofS3, the permutation group on 3 objects, that we found cosets of in the previous chapter had either 2 or 3elements -- 2 and 3 divide evenly into 6.

A consequence of Lagrange's Theorem would be, for example, that a group with 45 elements couldn't have a subgroup of 8 elements since 8does not divide 45. It could have subgroups with 3, 5, 9, or 15 elements since these numbers are all divisors of 45.

Now that we know what Lagrange's Theorem says let's prove it. We'll prove it by extablishing that the cosets of a subgroup are

z disjoint -- different cosets have no member in common, andz each have the same number of members as the subgroup.

This leads to the conclusion that a subset with n elements has kcosets each with n elements. These cosets do not overlap and together theycontain every element in the group. This means that the group must have kn elements -- a multiple ofn. We'll accomplish our proof with twolemmas.

Lemma: IfH is a finite subgroup of a group G and Hcontains n elements then any right coset ofHcontains n elements.

Proof: For any element x ofG, Hx = {h x | h is in H} defines a right coset ofH. By the cancellation law each h in Hwill give a differentproduct when multiplied on the left onto x. Thus each element ofHwill create a corresponding unique element ofHx. Thus Hx will have thesame number of elements as H.

Lemma: Two right cosets of a subgroup Hof a group G are either identical or disjoint.

Proof: Suppose Hx and Hy have an element in common. Then for some elements h1 and h2 ofH

h1 x = h2 y

This implies that x = h1

-1

h2 y. SinceH

is closed this means there is some element h3 (which equals h1

-1

h2) ofH

such that x = h3 y. Thismeans that every element ofHx can be written as an element ofHy by the correspondence

h x = (h h3) y

for every h in H. We have shown that ifHx and Hy have a single element in common then every element ofHx is in Hy. By a symmetricalargument it follows that every element ofHy is in Hx and therefore the "two" cosets must be the same coset.

Since every element g ofG is in some coset (namely it's in Hg since e, the identity element is in H) the elements ofG can be distributed amongHand its right cosets without duplication. Ifk is the number of right cosets and n is the number of elements in each coset then |G| = kn.

Alternate Proof: In the last chapter we showed that a b-1 being an element ofHwas equivalent to a and b being in the same right coset ofH.We can use this Idea establish Lagrange's Theorem.

Define a relation on G with a ~ b if and only ifa b-1 is in H. Lemma: The relation a ~ b is an equivalence relation.

Proof: We need to establish the three properties of an equivalence relation -- reflexive, symmetrical and transitive.

(1) Reflexive: Since a a-1 = e and e is in H it follows that for any a in G

a ~ a

(2) Symmetrical: Ifa ~ b then a b-1 is in H. Then the inverse ofa b-1 is in H. But the inverse ofa b-1 is b a-1 so

b ~ a

(3) Transitive: Ifa ~ b and b ~ c then both a b-1 and b c-1 are in H. Therefore their product (a b-1) (b c-1) is in H. But the product is

simply a c-1. Thus

a ~ c

And we have shown that the relation is an equivalence relation.

It remains to show that the (disjoint) equivalence classes each have as many elements as H.

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Lemma: The number of elements in each equivalence class is the same as the number of elements in H.

Proof: For any a in G the elements of the equivalence class containing a are exactly the solutions of the equation

a x-1 = h

Where h is any element ofH. By the cancellation law each member h ofHwill give a different solution. Thus the equivalence classes have thesame number of elements as H.

One of the imediate results of Lagrange's Theorem is that a group with a prime number of members has no nontrivial subgroups. (why?)

Definition: ifH is a subgroup ofG then the number of left cosets ofH is called the index ofH in G and is symbolized by (G:H). From ourdevelopment of Lagrange's theorem we know that

|G| = |H| (G:H)

Converse of Lagrange's Theorem One of the most interesting questions in group theory deals with considering the converse of Lagrange'stheorem. That is if a number n divides the order of group G does that mean that G must have a subgroup of order n? The answer is no in generalbut the special cases where it does work out are many and interesting. They are dealt with in detail in the Sylow Theorems which we will treatlater. As a tidbit we look at the following

Theorem: If the order of a group G is divisible by 2 then G has a subgroup of two elements.

Proof: The proof is left as an exercise for the student. [Hint: If an element other than the identity of a group is its own inverse then that elementtogether with the identity forms a subgroup of two elements (Prove!). The identity is its own inverse. If we remove the identity from a group of

even order must at least one of the remaining elements be its own inverse?]

Next:

9. Cyclic groups and subgroups:Back to the index:

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1998 1999 by Arfur Dogfrey

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