Group Chains, Commutator Subgroup and Abelianization

Commutator Subgroup and Abelianization

Let G be a group, and let G′ denote the commutator subgroup of G, which is defined as follows.

For any x and y in G, xy/x/y is a commutator, and the group generated by these elements is the commutator subgroup. Notice that if x and y commute, xy/x/y is simply e. If G is abelian, G′ is trivial.

Let q be any element of G, and conjugate the commutator xy/x/y by q. The result is the commutator q(xy/x/y)/q/(xy/x/y) * (xy/x/y), which is in G′. Thus conjugation by q, for any q in G, maps G′ onto itself, and G′ is normal in G.

Actually, any automorphism moves commutators to other commutators, hence G′ is preserved by every automorphism. We say G′ is "fully invariant" in G.

Let q and r represent cosets of G′ in G. We know that qr/q/r is a commutator, a member of G′. Write qr/q/r = w, for w in G′. Hence qr = wrq. Thus qr and rq represent the same coset of G′. The quotient group G/G′ is abelian. This is called the abelianization of G.

Conversely, assume G/H is abelian, and reverse the above argument. Write Hqr = Hrq, or Hqr/q/r = H, or eqr/q/r ∈ H. Every commutator belongs to H, and H contains G′.

Conjugacy Classes

Let x and y belong to the same conjugacy class of G. Thus y = wx/w for some w. With wx/w/x in G′, y/x is in G′. This means y and x are in the same coset of G′, and represent the same element in the quotient group. Each conjugacy class maps, in its entirety, to a unique element in the abelianizzation of G. However, this map is not 1-1. Separate conjugacy classes can map to the same thing in G/G′. In other words, the conjugacy classes do not define the quotient group. Consider S₃, the symmetric group on 3 letters. Each commutator is an even permutation, and it's easy to verify G′ = A₃. Thus the quotient has order 2. However, there are three conjugacy classes: e, the two circular shifts, and the three transpositions. Three conjugacy classes cannot correspond to a quotient group of order 2. In fact, the first two conjugacy classes map to A₃, while the third class maps to the coset of A₃ in S₃.

This generalizes to any nonabelian group G. The commutator subgroup G′ is nontrivial, and e is its own conjugacy class, hence other conjugacy classes belong to G′. Conjugacy classes correspond to G/G′ iff G is abelian.

Universal Object

This is an excursion into category theory. You may skip it if you wish. We will show that the abelianization is universal. That is, any homomorphism from G into an abelian group factors uniquely through G/G'.

Let f map G onto an abelian group H, having kernel K, and let J be G/G′. We already showed that K contains G′.

Let f₁ map G onto J. This is G mod G′, as described above. Then apply f to J, to build f₂, a map from J onto H. Now f = f₂(f₁(G)), and there is only one function f₂ that satisfies f = f₁ followed by f₂.

Suppose U is another universal abelianization of G. Think of U as G/K, where K is the kernel. Since U is abelian, K contains G′. Suppose K is larger than G′, and map G, through U, onto J. This is suppose to be the same as the map from G directly to J. Let w be a coset of G′, other than G′, in K. Now w maps to e in U, and to e in J. However, the direct map from G to J carries w to a nontrivial element of J. This is a contradiction, hence K = G′, and our abelianization U is actually isomorphic to J. There is one unique abelianization of G, up to isomorphism, namely G/G′.