# Lagrange Theorem Statment, Proof Joseph Lagrange (1736-1813), a French mathematician. Lagrange contributed to number and gravitational theory and found solutions and proofs for various unsolved problems. His major work, Analytical Mechanics, was started when he was 19 but was not finished until 33 years later when he was 52. In it, he applied a powerful combination of two forms of calculus to general mechanical problems. Lagrange also worked on weights and measures and the development of the metric system.

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 