In this post, we will build up towards proving Lagrange's Theorem, which can be viewed as an extension of Euler's Theorem.
Let be a group with operation and consider an arbitrary subset . In general, is not a group under . In fact, may not even be a well-defined function on , since its image may lie outside . Hence, we make the following definition.
A subgroup of is a subset such that:
Note that any such becomes a group in its own right, with the operation "inherited" from
For any group itself and are subgroups of Most groups have other, non-trivial subgroups. The following are some examples of subgroups, that correspond to the example list of groups from the post Group Theory.
1) , the set of integers
2) , the set of positive real numbers.
3) , the set of functions which also satisfy .
4) , the subset of multiples of . Note that this is strictly smaller than if
5) , the set of squares modulo .
A general example of a subgroup is obtained by picking an element and taking all powers of . This is known as the subgroup generated by , and is denoted by:
Proposition. The order of the subgroup is the smallest positive for which . If such an does not exist, then the order is infinite.
As such, we define the order of element to be the smallest positive for which , and write .
Let be a subgroup of the group , and let . The set is called a (left) coset of in . In other words, the coset is the image (range) of the function where . Because this function is one-to-one (by the Cancellation Property!) Note also that
The most important property of the cosets is that and either coincide or do not intersect (see Worked Example 2 below). This implies the following classical theorem.
Lagrange's Theorem. If is a finite group and is a subgroup of , then divides .
As an immediate corollary, we get that if is a finite group and , then divides . In particular, for all .
1. Prove the proposition.
Solution: Suppose is finite. We claim that and that all these elements are distinct. For the first statement, note that for any , we can write where . Hence . To prove the second statement, suppose for , then , which contradicts the minimality of .
2. Let and be a subgroup.
(a) If , then ;
(b) If , then .
Solution: For the first statement, any element of can be written as , for some . But since . This tells us that .
Conversely, any element of can be written as and . But lies in since is a subgroup of . Hence the result follows and .
For the second statement, suppose . Then it can be written as for some . Thus which contradicts the fact that .
3. Prove Lagrange's Theorem
Solution: The above shows that the cosets partition the entire group into mutually disjoint subsets, all of which have elements. Hence, Lagrange's Theorem follows.