I will begin with group theory, starting with group actions on sets.  Given a group G and a set S, an action of G on S is a rule that assigns to each pair (g,s) in G x S an element t of S, usually denoted gs, such that 1s = s for all s in S, where i is the identity element of G and h(gs) = (hg)s for g,h in G, s in S.  (More precisely, this defines a *left* action of G on S; had we written sg for the action of the element g on the element s, the natural axiom would have been (sg)h = s(gh); this defines a *right* action).  It is not difficult to verify that a left action of G on S is equivalent to a homomorphism \pi from G to ther permutation group Perm(S) of all bijections from S to S, where we decree that \pi(g)(s) = gs, where gs is the action of g on s.  If G acts on S and s is an element of S, then the set of g in G such that gs = s is an important subgroup of G, called the *stabilizer* of s and denoted G^s.  Likewise, we have an important subset Gs of S, defined to be the set of all gs as g runs through G and called the orbit of s.  If H = G^s, then it is easy to check that any two elements of the same left coset gH of H in G map s to the same element, and in fact two elements of G map s to the same element if and only if they lie in the same left coset of H in G.  If G is finite, it follows by Lagrange's Theorem (which I assume you have seen) that the order |G| of G equals the product of the orders |G^s||Gs| of the orders of the orbit and stabilizer of any s in S:  this important formula, called the Orbit Formula, will be used constantly in this course (and beyond).  If the set S has only one G-orbit, then we call the G-action on it {\sl transitive}; in this case we can of course replace the orbit Gs in the Orbit Formula by the entire set S.  In general, no two orbits of G in S can overlap, so another useful formula, if S is finite, is that its order equals the sum of the orders of the orbits in it.

An important but all too rarely seen example occurs when G is a finite subgroup of SO_3; that is, a finite group of 3 x 3 real matrices M of determinant 1 such that the transpose M^t of M equals its inverse M^{-1}, or equivalently such that M preserves dot products in R^3:  Mv\cdot Mw = v\cdot w for all vectors v,w in R^3.  Any such matrix M is such that det (M- I) = det (M^t - I^t) = det (M^{-1} - I) =
det (M^{-1} - I)(det M) = det (I - M) = -det (M - I) = 0, whence M has an eigenvalue 1 and must fix some nonzero vector v in R^3.  But then M preserves the plane in R^3 perpendicular to v, whence it acts by a rotation in this plane.  It follows that M is either the identity or fixes exactly two unit vectors in R^3 (each the negative of the other), acting by a rotation about the line through these vectors, which also goes through the origin.  Thus any g in G with g\ne1 has exactly two poles.  If N is the order of G, there are 2(N-1) poles in R^3 of nonidentity elements of G, counting each as often as it appears as the pole of a nonidentity element.  On the other hand, if p is a pole of some element of g and h is any other element of G, then hp is a pole of hgh^{-1}, so G acts on the set P of poles of its nonidentity elements.  Each pole p_i in P will have a stabilizer G^p consisting of finitely many rotations in G; if there are r_i such rotations, then the number n_i of elements in the orbit of p satisfies r_i n_i = N.  Counting the number of poles again, this time for each pole counting the number of nonidentity elements of G fixing it and observing that this number is the same for any pole in the orbit of p, we get \sum n_i(r_i - 1), where there is one index i for every orbit of poles in G.  Hence

2 (N - 1) = \sum_i n_i(r_i - 1)

where the number of terms in the sum equals the number of orbits.  Dividing both sides by N, we get

2(1 - 1/N) = \sum_i (1- (1/r_i))

using again that n_i r_i = N, by the Orbit Formula.  But now the left side is less than 2, while every term in the right side is at least 1/2, so there are at most 3 orbits.  More precisely, the left side is at least 1 and each term on the right side is less than 1, so there cannot be just one orbit.  If there are two orbits, then we must have r_1 = r_2 = N (since r_1 and r_2 are less than N).  In this case there are exactly two poles, each lying in an orbit by itself.  The group G must be a cyclic group, consisting of rotations by multiples of 2\pi/N about a fixed axis.  If there are three orbits, then 
(1/r_1)+(1/r_2)+()1/r_3) > 1.  We will work out the consequences of this elementary but very famous inequality and (sketch) a classification of the groups G arising in this way next time.  A reference for this material is Artin's book Algebra, on reserve (and in the stacks) of the Math Library.