M417 Homework 9 Solutions Spring 2004
1. (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 orbG(s) and in orbG(s'). Then x = gs and x = g's' for some g and g' in G. But this means that s = g-1g's'. Now let y be any element of orbG(s). Then y = hs for some h in G. Hence y = hs = hg-1g's' is in orbG(s'). Thus orbG(s') contains orbG(s). Similarly, orbG(s) contains orbG(s'). Thus any two orbits with a single element in common are actually the same subset of S. Since s is in orbG(s), every element of S is in some orbit. This means S is the union of the orbits, and the orbits are disjoint.

2. Let G = { g1, ..., gm } be a finite group acting on a finite set S = { s1, ..., sn }. Let t be the number of distinct orbits. Let D = { (g, s) in GxS | gs = s }. Let p1 : D -> G be defined by p1((g, s)) = g, let p2 : D -> S be defined by p2((g, s)) = s, and let f : G -> Z be defined by taking f(g) = | { s in S | gs = s } |.
• (10 points) Show that |stabG(s1)| + ... + |stabG(sn)| = t|G|.

Answer: Let O1, ..., Ot be the distinct orbits, and let Oi = { si1, ..., siri } be the elements of the i-th orbit. Note that |Oi| = ri. Then |stabG(si1)| + ... + |stabG(siri)| = ri|stabG(si1)| = |Oi||stabG(si1)| = |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 |stabG(s1)| + ... + |stabG(sn)| equals the sum of |stabG(si1)| + ... + |stabG(siri)| (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(g1) + ... + f(gm) = t|G|. [Hint: Show that f(g1) + ... + f(gm) = |D| = |stabG(s1)| + ... + |stabG(sn)|. Do this by showing that stabG(s) = p1(p2-1(s)), and f(g) = |p1-1(g)|.]

Answer: Now, p2-1(s) = { (g, s) : gs = s} = { (g, s) : g is in stabG(s) }, so p1(p2-1(s)) = { g : g is in stabG(s) } = stabG(s). For each s, we have |stabG(s)| = |p1(p2-1(s))| = |p2-1(s)|. Thus the sum of |stabG(s)| over all s gives t|G| by the previous problem, but it also gives the sum of |p2-1(s)| over all s, which is just |D|. On the other hand, |D| is also the sum of |p1-1(g)| over all g in G. But p1-1(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(g1) + ... + f(gm) = |D| = |stabG(s1)| + ... + |stabG(sn)| = t|G|.

3. (10 points) Let H : RxC -> 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.

4. (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 = (x2 + y2)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).