The derivation of the matrix for a general rotation in three dimensional Euclidean space is often encountered in strange forms. This is sometimes a bit puzzling until you have found your own way around it. In this post, I will not derive the expression, but show how the useful matrix exponential of the SO(3) generators is in fact producing a rotation about an arbitrary axis through some angle. [Note: SO(3) is the group of 3x3 orthogonal matrices with unit determinant].

Let the rotation matrix about an arbitrary rotation axis through an angle be denoted . The matrix satisfies and we can therefore pick and let be arbitrary. This way any double covering is avoided.

The infinitesimal generators of the SO(3) group is can be written as:

; ;

Which for example can be obtained by considering the generators of SO(2) (which is far more easier obtained if you are starting from scratch) and generalize them to three dimension. Note that you can form Hermitian matrices by defining . The generators satisfy the Lie algebra , where is the Levi-Civita symbol.

Consider the matrix , where the repeated index i is to be summed over. Now, since the generators are antisymmetric we can write: for some matrix . This is a general result from linear algebra, that any antisymmetric matrix can be decomposed like this (c.f. http://en.wikipedia.org/wiki/Antisymmetric_matrix). Hence,

The neat thing is that the exponential matrix containing is easy to evaluate. In fact, we probably already know the answer from the derivation of the generators in the first place.

This matrix is the well known matrix which produces rotations about the z-axis. The matrix can be expressed in terms of the generator by realizing that , where is the identity matrix:

Putting this result into the relation:

Which is the expression for a general rotation in three dimensional Euclidean space! Oh, that was easy to show, if you knew all the right algebra! Cheers.

In group theory Lagrange’s theorem is one of the most important in the theory of finite groups.

Theorem 2.6 (Lagrange’s theorem) The order of a subgroup H of a finite group G is a divisor of the order of G, i.e. |H| divides |G|.

One evident, but funny, implication of the theorem is in the answer of the following question: List all of the subgroups of any group whose order is a prime number.

Solution: According to Lagrange’s theorem, the order of a subgroup H of a group G must be a divisor of |G|. Since the divisors of a prime number are only the number itself and unity, the subgroups of a group of prime order must be either the unit element alone, H = {e}, or the group G itself, H = G, both of which are improper subgroups. Therefore, a group of prime order has no proper subgroups.