Lagrange Theorem Statment, Proof

Lagrange Theorem Statement: Both the order and Index of a subgroup of a finite group divide the order of a group.

Lagrange Theorem Proof:

Let $H$ be a subgroup of order $m$ of a finite group $G$ of order $n$ and let $k$ be the index of $H$ in $G$ .

since $\left|G\right|$ is finite , the set

{ $a_1H,a_2H,,a_3H,\dots,a_kH$ }

all of distinct left cosets of $H$ in $G$ . then we have theorem, ( Let $H$ be a subgroup of group $G$ . Then the set of all left cosets of $H$ in $G$ defines a partition of $G$

i.e, $G=\;\overset k{\underset{i=1}U}\;a_iH\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;;\;a_iH\;\bigcap\;a_{j_{}}H=\varnothing\;\;\;,i\neq j,\;j=1,2,3,…,k$

Now, the mapping $\phi:H\rightarrow a_iH$ defined by

$\phi(h)=a_iH\;\;,h\in H$

is obviously onto.

Also,

$\phi(h_1)=\phi(h_2)\;\;\;\;,h_1,h_2\in H$

Implies,

$a_ih_1=a_ih_2$

$\Rightarrowh_1=h_2$

So $\phi$ is one-one.

Hence the number of element in $H$ and $a_1H$ is the same for $i=1,2,3,...,k$ . As $H$ has $m$ , each $a_iH$ also has $m$ element for $i=1,2,3,...,k$

Now, form

$G=\;\overset k{\underset{i=1}U}\;a_iH$

$G=a_1H\;\cup\;a_2H\;\cup\;a_3H\cup…\cup\;a_kH$

$\left|G\right|=\left|H\right|+\left|{}_{}H\right|\;+\left|H\right|\;+…+\left|\;H\right|$

$n=m+m+m+...+m(k\;times)$

$n=km$

Thus both $m$ and $k$ ,the order and index of $H$ in $G$ , Divides $n$ (the order of $G$ ) i.e Lagrange Theorem