1. Evaluate , , and .
2. Evaluate .
3. Can you explain why gives a better approximation to the limit than does?

The distance between Singapore and Tokyo is 3300 miles. On a certain wall map, this distance is represented by 11 inches. The actual distance between Mexico City and Cairo is 7700 miles. How far apart are they on the same map? [#53, p. 117]

Now, this would be an excellent problem… IF THE WORLD WERE FLAT!

A homomorphism f : G -> Q is called a split epimorphism if there exists a homomorphism s : Q -> G so that the composition fs is the identity on Q. We proved that there exists a split epimorphism f : G -> Q if and only if G is a semidirect product of K and Q1, where Q1 is isomorphic to Q. If this occurs, then we may take K = ker(f) and Q1 = sQ.

We also learned that if G is the semidirect product of K and Q (where K is normal) then Q acts on K by conjugation. In other words, there is a homomorphism from Q to Aut(K) defined by

q -> (k -> q^-1 k q)

It turns out that the semidirect product is uniquely determined up to isomorphism by the groups K and Q and the action of Q on K.

Let K and Q be subgroups of G. We say that G is the internal semidirect product of K and Q if the following conditions hold:

1. K is a normal subgroup of G,
2. G = KQ, and
3. the intersection of K and Q is the identity.

Notation: G = K Q.

We recall that, for any two subgroups U and V of G, UV is defined as {uv : u in U, v in V}. In general UV is not a subgroup, but it is a subgroup if at least one of the factors is normal. Also, UV is a subgroup of G if and only if UV = VU.

If K and Q are both normal, then G is the internal direct product of K and Q. In that case it is easily shown that the elements of K commute with the elements of Q.

Examples of semidirect products:

1. S_n = A_n C₂
2. D_2n = C_n C₂
3. S₄ = V S₃

S_n denotes the symmetric group on n letters. A_n is the alternating group, consisting of the even permutations of S_n. D_2n is the dihedral group of order 2n, i.e. the symmetry group of a regular n-sided polygon. C_n is the cyclic group of order n, and V is the Klein 4-group.

If g is an element of G, then let Fix(g) denote the number of elements of X so that g*x = x. Likewise, if x is an element of X, then let Fix(x) denote the number of elements of G so that g*x = x. Our first observation is that

sum (g in G) Fix(g) = sum (x in X) Fix(x).

This holds because both sums enumerate the pairs (g,x) so that g*x = x.

The next ingredient is the idea of a multiorbit. The multiorbit of x, denoted G(x), is the multiset [gx : g in G]. The size of G(x) is equal to |G|, and the multiplicity of x in G(x) is equal to Fix(x).

Now, the union of the distinct multiorbits has size n * |G|, where n is the number of orbits. The multiplicity of x in this union is equal to Fix(x). Since the size of a multiset is equal to the sum of the multiplicities of its distinct elements, we conclude that

sum (g in G) Fix(g) = sum (x in X) Fix(x) = n * |G|.

It follows that the number of orbits (n) is

1/|G| * sum (g in G) Fix(g)

Reference: Bogart, Kenneth P. An obvious proof of Burnside’s lemma. Amer. Math. Monthly 98 (1991), no. 10, 927–928.