Commutators – Summary and Classroom Activity

First recall:

1. A commutator is an element of a group that can be written as g·h·g^-1·h^-1

2. Since it’s not obvious (and in fact false) that the product of two commutators is a commutator, we can not say the commutators form a subgroup. However since inverses of commutators are commutators we can consider the set of finite products of commutators and that is a subgroup. If G is a group we call the subgroup described in the last sentence “the commutator subgroup” or “the derived subgroup” and we use the notation G’.

3. It is quite easy to prove that G’ is a normal subgroup of G.

4. Since G’ is normal we can create a “factor group” : G/G’. Then a slightly confusing but straight-forward argument shows that G/G’ is always abelian. (We show that by showing that the commutator group of G/G’ is the identity.)

And here’s something new:

5. Note that, as the example we will now compute illustrates, that we can also compute (G’)’. (This is written as G’’.) Though one might guess (incorrectly) that G’’=G’, in fact this is not always so. Hence we can get a sequence of subgroups: G, G’, G’’, ... where each properly includes the next. This will terminate, possibly but not necessarily in the identity subgroup.

Now lets compute the commutator subgroup of S₄, the set of permutations of {1,2,3,4}.

S₄ = { (1), (1,2),(1,3),(1,4),(2,3),(2,4),(3,4),

(1,2,3),(1,3,2),(1,2,4), (1,4,2),(1,3,4),(1,4,3),(2,3,4),(2,4,3),

(1,2,3,4),(1,2,4,3),(1,3,2,4),(1,3,4,2),(1,4,2,3),(1,4,3,2),

(1,2)(3,4), (1,3)(2,4), (1,4)(2,3) }

There are 24·23 possible commutators though we can compute all those involving the identity instantly. (WHY?) Let’s each do a few of these now.

Thus we find that the commutator subgroup of S₄ is _________________ . This gives us G’.

Now let’s compute G’’.