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