The finite and infinite groups, the Lagrange theorem, its proof and a corollary - Lesson IV

The finite and infinite groups, the Lagrange theorem, its proof and a corollary.

Today i will be discussing about the finite and infinite groups, and just like the set theory where we have finite and infinite sets, we will be transferring this properties of set theory to groups.

What is a finite and infinite group? 

A finite group can be defined as a set $G$ equipped with a binary operation $"*"$ i.e $(G,*)$ that consists of finite number of elements. If the group $(G,*)$ has infinite number of elements, then it is called an infinite group.

A very good example of the finite group is the cyclic group, permutation group, finite abelian group and the lie group. 

Lagrange's theorem 

If $G$ is a finite group and $H$ is a subgroup of $G$, then the order of $H$ divides the order of $G$. 

Proof : Let $H$ be any subgroup of the order $n$ of a finite group $G$ of order $m$. 

Let $aH$ be a left coset that comprises $n$ different elements.

Let $H=\{h_1,h_2,h_3,...,h_n\}$ then $ah_1,ah_2,ah_3,...,ah_n$ are the $n$ distinct members of $aH$. 

Suppose, $ah_i=ah_j\Rightarrow{hi}=hj$ be the cancellation law of $G$.

Since $G$ is a finite group, the number of discrete left cosets will also be finite, say $p$. So, the total number of elements of all cosets is $np$ which is equal to the total number of elements of $G$. Hence, $m=np$ and  

$p= m/n$. This shows that $n$, the order of $H$, is a divisor of $m$, the order of the finite group $G$. We also see that the index $p$ is also a divisor of the order of the group.

Hence,  $|G|=|H|$ $\square$. 

Let's take a look at a Lagrange corollary. 

Corollary 

If $G$ is a group of finite order $m$, then the order of any $a\in{G}$ divides the order of $G$ and $a^m=e$. 

Proof: Let $a$ be a subgroup of $G$ and let $p$ be the order of $a$. Suppose $p$ is the least positive integer then $a^p=e$ and $a^1,a^2,...,a^{p-1},a^p=e$ this shows that every elements in group $G$ are all distinct and forms a subgroup. 

Since the subgroup is of order $p$, then $p$ divides $m$ which is an order of group $G$. 

$\therefore{m=np}$ where $n$ is a positive integer. So, $a^m=a^{np}=(a^p)^n=e$ $\square$.