Theorem Let G be a finite group and let H be a subgroup. Then the order of H divides the order of G.

The proof of this theorem comes down to the fact that the cosets of H in G partition G. In fact if we write [G:H] for the number of such cosets (also called the index of H in G) we have

The Counting Formula


Corollary The order of an element of a finite group divides the order of the group.

The proof follows from the fact that if a is an element of G then the order of a is the same as the order of the cyclic subgroup it generates <a>.

In spite of the name, several theorems are called "Lagrange's". But usually it's one of these:

Group theory:
What Noether noded above. See the proof of Lagrange's theorem for groups.
Let f be a real function which is differentiable on [a,b]. Then for some a<c<b,
f'(c) = ---------.
If this seems to you like a trivial modification of Rolle's theorem, you're right.
An interesting and useful corollary of Lagrange's Theorem:
If a group G is order p (a prime) it contains no subgroups other than {e} (the identity) and G itself.
This follows directly from Noether's post above. The only divisors of a prime number p are 1 and p. So any subgroup of G must be of order 1, ie {e}, or order p, ie {G}.

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