Homework #5, due March 5th (oops, should have been 4th, but whatever):

Determine the conjugacy classes in the group D_{2n}. Careful: you will probably want to separate into the cases n even and n odd.

The elements come in two types: rotations {r^k}, and reflections {f r^k}, where k=0..n-1. So we have four conjugations to compute:

r^j r^k (r^j)^{-1} = r^{j + k - j} = r^k boring, they commute

(f r^j) r^k (f r^j)^{-1} = f r^j r^k r^{-j} f = f r^k f = r^{-k}

So when we conjugate r^k, the only things we can get are r^k, r^{-k}. As for the other:

r^j f r^k (r^j)^{-1} = r^j f r^{k - j} = f r^{-j} r^{k-j} = f r^{k-2j}

(f r^j) f r^k (f r^j)^{-1} = f r^j f r^k r^{-j} f = f r^j r^{-k+j} f f = f r^{2j - k}

So we can negate the exponent on r, and add an even number to it.

Now we split into cases. If n is odd, then r^k is never r^{-k} unless r^k=1, and we can get from any f r^k to any f r^m by adding an even number to k to get m. (For example, if n=5, then from 0 we can get to 2,4,6=1,8=3.) So the conjugacy classes are {1}, {r^k, r^-k} for each k=1..(n-1)/2, and one conjugacy class of all rotations.

Whereas if n is even, we do get one new case of r^k = r^{-k}, when k = n/2. "Rotating by 180 degrees is a central element." Also, we end up with two conjugacy classes of flips, namely {f r^odd} and {f r^even}.

How do you compute conjugacy classes? Compute all the conjugations!

2.2 #3. If A is contained in B contained in G, show that C_G(B) is inside C_G(A).

Proof. If g in G commutes with every element of B, it certainly commutes with every element of A. So anybody worthy of being in C_G(B) is already worthy of being in C_G(A).

4.1 #1, part 1. If h is in G_a, then (g h g^{-1}).b = (g h).(g^{-1}.a) = (g h).a = g.(h.a) = g.a = b. So g h g^{-1} is in G_b. This shows that everything in g G_a g^{-1} is in G_b, which is one direction of containment.

Now comes the mentally tricky part: apply that same statement but rename a as b, b as a, and g as g^{-1}. The statement then becomes that g^{-1} G_b g is part of G_a. So G_b is part of g G_a g^{-1}. That's the other containment.

Part 2. The kernel is the stabilizer of everything, i.e. Intersection_{b in A} G_b. If the action's transitive, then every b arises as g.a for some g. So we get the same conditions, if we rewrite as Intersection_{g in G} G_{g.a}. But that's Intersection_{g in G} g G_a g^{-1}.

#2. The only difference from the previous question is that we don't just know that G acts on A -- i.e. that there's a homomorphism from G -> S_A -- but that G is a subgroup of S_A, so that homomorphism is just an inclusion. So the kernel is just {1}.

#3. "Abelian" means that everything commutes with everything else, so g G_a g^{-1} = G_a g g^{-1} = G_a. Hence the kernel, Intersection_{g in G} g G_a g^{-1} can be rewritten as Intersection_{g in G} G_a. But now it's just an intersection of the same thing, G_a, over and over again. So the kernel is G_a. Apply the previous question, and we find out that G_a = {1}, which was the question.

Finally, |G| = |A| |G_a| = |A| 1 = |A|.

#4. There are 9 ordered pairs, which we'll call {11,12,13,21,22,23,31,32,33}. The cycle decomposition of the permutations is

Id: Id

(12): (11 22) (13 23) (21 12) (31 32)

(23): (12 13) (21 31) (22 33) (23 32)

(13): (11 33) (12 32) (13 31) (21 23)

(123): (11 22 33) (12 23 31) (13 21 32)

(132): (11 33 22) (12 31 23) (13 32 21)

From this, one can see the orbits, which are {11,22,33} and {everything else}. The stabilizer of 11 is {Id, (2 3)}. The stabilizer of 12 is {Id}.