# Lagrange’s Theorem Lagrange’s theorem is the first great result of group theory. It states

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

First we need to define the order of a group or subgroup

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

Lagrange’s Theorem simply states that the number of elements in any subgroup of a finite group must divide evenly into the number of elements in 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 of S3, the permutation group on 3 objects, that we found cosets of in the previous chapter had either 2 or 3 elements — 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 8 does 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

• disjoint — different cosets have no member in common, and
• each have the same number of members as the subgroup.

This leads to the conclusion that a subset with n elements has k cosets each with n elements. These cosets do not overlap and together they contain every element in the group. This means that the group must have kn elements — a multiple of n. We’ll accomplish our proof with two lemmas.

Lemma: If H is a finite subgroup of a group G and H contains n elements then any right coset of H contains n elements.

Proof: For any element x of GHx = {h • x | h is in H} defines a right coset of H. By the cancellation law each h in H will give a different product when multiplied on the left onto x. Thus each element of Hwill create a corresponding unique element of Hx. Thus Hx will have the same number of elements as H.

Lemma: Two right cosets of a subgroup H of 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 of H

h1 • x = h2 • y
This implies that x = h1-1 • h2 • y. Since H is closed this means there is some element h3 (which equals h1-1 • h2) of H such that x = h3 • y. This means that every element of Hx can be written as an element of Hy by the correspondence

h • x = (h • h3) • y
for every h in H. We have shown that if Hx and Hy have a single element in common then every element of Hx is in Hy. By a symmetrical argument it follows that every element of Hy is in Hx and therefore the “two” cosets must be the same coset.

Since every element g of G is in some coset (namely it’s in Hg since e, the identity element is in H) the elements of G can be distributed among H and its right cosets without duplication. If k 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 of H was equivalent to a and b being in the same right coset of H. We can use this Idea establish Lagrange’s Theorem.

Define a relation on G with a ~ b if and only if a • b-1 is in HLemma: 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: If a ~ b then a • b-1 is in H. Then the inverse of a • b-1 is in H. But the inverse of a • b-1 is b • a-1 so

b ~ a
(3) Transitive: If a ~ 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.

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 of H. By the cancellation law each member h of H will give a different solution. Thus the equivalence classes have the same 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: if H is a subgroup of G then the number of left cosets of H is called the index of H in G and is symbolized by (G:H). From our development 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’s theorem. 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 general but the special cases where it does work out are many and interesting. They are dealt with in detail in the Sylow Theoremswhich we will treat later. 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 element together 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?] 