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 $${a^{p – 1}} \equiv 1\left( {\bmod p} \right)$$.