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