- (10 points) Let G be a group acting on a set S. Show that the orbits form
a partition of S; i.e., S is the disjoint union of the orbits.

**Answer**: Let x be in orb_{G}(s) and in orb_{G}(s'). Then x = gs and x = g's' for some g and g' in G. But this means that s = g^{-1}g's'. Now let y be any element of orb_{G}(s). Then y = hs for some h in G. Hence y = hs = hg^{-1}g's' is in orb_{G}(s'). Thus orb_{G}(s') contains orb_{G}(s). Similarly, orb_{G}(s) contains orb_{G}(s'). Thus any two orbits with a single element in common are actually the same subset of S. Since s is in orb_{G}(s), every element of S is in some orbit. This means S is the union of the orbits, and the orbits are disjoint.

- Let G = { g
_{1}, ..., g_{m}} be a finite group acting on a finite set S = { s_{1}, ..., s_{n}}. Let t be the number of distinct orbits. Let D = { (g, s) in GxS | gs = s }. Let p_{1}: D -> G be defined by p_{1}((g, s)) = g, let p_{2}: D -> S be defined by p_{2}((g, s)) = s, and let f : G ->**Z**be defined by taking f(g) = | { s in S | gs = s } |.- (10 points) Show that
|stab
_{G}(s_{1})| + ... + |stab_{G}(s_{n})| = t|G|.

**Answer**: Let O_{1}, ..., O_{t}be the distinct orbits, and let O_{i}= { s_{i1}, ..., s_{iri}} be the elements of the i-th orbit. Note that |O_{i}| = r_{i}. Then |stab_{G}(s_{i1})| + ... + |stab_{G}(s_{iri})| = r_{i}|stab_{G}(s_{i1})| = |O_{i}||stab_{G}(s_{i1})| = |G|, since as we saw in class, the stabilizers for elements in the same orbit have the same order, and the order of an orbit times the order of the stabilizer is the order of the group. Now |stab_{G}(s_{1})| + ... + |stab_{G}(s_{n})| equals the sum of |stab_{G}(s_{i1})| + ... + |stab_{G}(s_{iri})| (which is |G|) over all i. Since the index i runs from 1 to t, the sum over i equals t|G|.

- (10 points) Show that
f(g
_{1}) + ... + f(g_{m}) = t|G|. [Hint: Show that f(g_{1}) + ... + f(g_{m}) = |D| = |stab_{G}(s_{1})| + ... + |stab_{G}(s_{n})|. Do this by showing that stab_{G}(s) = p_{1}(p_{2}^{-1}(s)), and f(g) = |p_{1}^{-1}(g)|.]

**Answer**: Now, p_{2}^{-1}(s) = { (g, s) : gs = s} = { (g, s) : g is in stab_{G}(s) }, so p_{1}(p_{2}^{-1}(s)) = { g : g is in stab_{G}(s) } = stab_{G}(s). For each s, we have |stab_{G}(s)| = |p_{1}(p_{2}^{-1}(s))| = |p_{2}^{-1}(s)|. Thus the sum of |stab_{G}(s)| over all s gives t|G| by the previous problem, but it also gives the sum of |p_{2}^{-1}(s)| over all s, which is just |D|. On the other hand, |D| is also the sum of |p_{1}^{-1}(g)| over all g in G. But p_{1}^{-1}(g) = { (g, s) : gs = s } and by definition |{ (g, s) : gs = s }| = f(g), hence the sum of f(g) over all g is also |D|. Thus f(g_{1}) + ... + f(g_{m}) = |D| = |stab_{G}(s_{1})| + ... + |stab_{G}(s_{n})| = t|G|.

- (10 points) Show that
|stab
- (10 points) Let H :
**R**x**C**->**C**be defined as H((t, x + iy)) = (cos(t) + i sin(t))(x + iy). Show that this defines an action of the reals**R**(regarded as a group under addition) on the complex numbers**C**.

**Answer**: We must show that h(t) defined by (h(t))(x + iy) = H((t, x + iy)) is a bijection on**C**, and that h is a homomorphism h from**R**to Bij(**C**). But h(t) is multiplication by cos(t) + i sin(t), and this is bijective since sin(t) + i cos(t) has the multiplicative inverse (cos(t) + i sin(t))^{-1}= cos(-t) + i sin(-t). And h(tt') = cos(tt') + i sin(tt') = (cos(t) + i sin(t))(cos(t') + i sin(t')) = h(t)h(t'), so h is a homomorphism.

- (10 points) For each complex number c, determine the orbit and stabilizer
of c with respect to the action defined in the previous problem.

**Answer**: If c = 0, then the orbit of c is just {0} (since 0 times any complex number is just 0), and the stabilizer of 0 is all of**R**. If c = x + iy is nonzero, write c as r(cos(s) + i sin(s)), where r = (x^{2}+ y^{2})^{1/2}and s is some angle. Then the orbit of c is the circle of radius r (since (h(t))(x + iy) = r(cos(s+t) + i sin(s+t))), and the stabilizer of c is the subgroup of**R**comprising the integer multiples of 2Pi (since sine and cosine are periodic of period 2Pi).