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,…,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

⇒h_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|a_1H\right|+\left|a_2H\right|\;+\left|a_3H\right|\;+…+\left|\;a_kH\right|

\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**