Lagrange's Theorem

Lagrange’s Theorem
The order of a subgroup of a finite group divisor of the order of the group.

Proof: Let H be any subgroup of order m of a finite group G of order n. Let us consider the coset decomposition of G relative to H.

We will first show that each coset aH consists of m different elements.

Let H = \left\{ {{h_1},{h_2}, \ldots ,{h_m}} \right\}, then a{h_1},a{h_2}, \ldots ,a{h_m} are the m members of aH, all distinct. So let a{h_i} = a{h_j} \Rightarrow {h_i} = {h_j} be the cancellation law of G.

Since G is a finite group, the number of distinct left cosets will also be finite, say k. Hence the total number of elements of all cosets is km which is equal to the total number of elements of G. Hence
n = mk

This shows that m, the order of H, is advisor of n, the order of the group G. We also see that the index k is also a divisor of the order of the group.

Corollary 1: If G is of finite order n, then the order of any a \in G divides the order of G and in particular {a^n} = e.

Proof: Let a be of order m is the least positive integer such that {a^m} = e.

Then it is easy to verify that the elements a,{a^2},{a^3}, \ldots ,{a^{m - 1}},{a^m} = e of G are all distinct and form a subgroup.
Since this subgroup is of order m, it follows that m, the order of a, is a divisor of the order of the group.

We may write n = mk, where k is a positive integer. Then {a^n} = {a^{mk}} = {\left( {{a^m}} \right)^k} = e.


Corollary 2: A finite group of prime order has no proper subgroups.

Proof: Let the order of the group G be a prime number p. Since p is a prime, its only divisors are 1 and p. Therefore the only subgroup of G are \left\{ e \right\} and G, i.e. the group G has no proper subgroup.

Corollary 3: Every group of prime order is cyclic.

Proof: Let G be a group of prime order of p and let a \ne e \in G. Since the order of a is a divisor of p, it is either 1 or p. But o\left( a \right) \ne 1, since a \ne e.
Therefore, o\left( a \right) = p, and the cyclic subgroup of G generated by a is also of order p. It follows that G is identical with the cyclic subgroup generated by a, i.e. G is cyclic.

Corollary 4: Every finite group of composite order possesses proper subgroups.

Corollary 5: If p is a prime number which does not divide the integer a then {a^{p - 1}} \equiv 1\left( {\bmod p} \right).