The order of a subgroup of a finite group divisor of the order of the group.
Proof: Let be any subgroup of order of a finite group of order . Let us consider the coset decomposition of relative to .
We will first show that each coset consists of different elements.
Let , then are the members of , all distinct. For, we have be cancellation law of .
Since is a finite group, the number of distinct left cosets will also be finite, say . Hence the total number of elements of all cosets is which is equal to the total number of elements of . Hence
This shows that , the order of , is advisor of , the order of the group . We also see that the index is also a divisor of the order of the group.
Corollary 1: If is of finite order , then order of any divides the order of and in particular .
Proof: Let be of order is the least positive integer such that.
Then it is easy to verify that the elements of are all distinct and form a subgroup.
Since this subgroup is of order it follows, that , the order of , is a divisor of the order of the group.
We may write , where is a positive integer. Then .
Corollary 2: A finite group of prime order has no proper subgroups.
Proof: Let the order of the group be a prime number . Since is a prime, its only divisor are and . Therefore the only subgroup of are and , i.e. the group has no proper subgroup.
Corollary 3: Every group of prime order is cyclic.
Proof: Let be a group of prime order ofand let . Since the order of is s divisor of , it is either or . But , since .
Therefore, , and the cyclic subgroup of generated by is also of order . It follows that is identical with the cyclic subgroup generated by , i.e. is cyclic.
Corollary 4: Every finite group of Composite order possesses proper subgroups.
Corollary 5: If is a prime number which does not divide the integer then .