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
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.
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 .
Prove the proposition.
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 .
Let and be a subgroup.
(a) If , then .
(b) If , then .
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 .
Prove Lagrange's theorem.
The above shows that the cosets partition the entire group into mutually disjoint subsets, all of which have elements. Hence, Lagrange's theorem follows.
Show that if is prime, then is cyclic.
Taking any element , by Lagrange's theorem, the order of must be either 1 or . Since the only element of with order 1 is , has order . Thus, is generated by , so is cyclic.
Show that if the orders of groups and are relatively prime, then their intersection contains only the identity.
The first thing to note is that is a subgroup of both and . Then, by Lagrange's theorem, divides both and . Since , we must have . Only one subgroup of order 1 exists, which is the group consisting of only the identity, so .