n) we relied on the fact that addition commutes in an abelian group. Without that fact, we could not have swapped c and 4m. Can we identify a condition on a subgroup that would guarantee that the procedure results in an operation? If cosets are to act as a group, does the group have to be abelian? The key in Example 3.50 was not really that Z is abelian. Instead, the key was that we could swap 4m and c in the expression ((d +4m) + c) +4m. In a general group setting where A < G, for every c ∈ G and every an ∈ A, we would need to find a 0 ∈ A to replace ca come with a 0 c. The abelian property makes it easy to do that, but we don’t need G to be abelian; we need A to satisfy this property. Let’s emphasize that.
Since W Y and X Z set, showing that they are equal requires us to show that each is a subset of the other. First, we show that W Y ⊆ X Z. To do this, let t ∈ W Y = (wy)A. By definition of a coset, t = (wy) a for some an ∈ A. What we will do now is rewrite t by • using the fact that A is normal to move some element of a left, then right, through the representation of t; and • using the fact that W = X and Y = Z to rewrite products of the form wαˇ as xαb and yα˙ as zα¨, where αˇ,αb,α˙,α¨ ∈ A. How, precisely? By the associative property, t = w (you). By definition of a coset, ∈ you. By hypothesis, A is normal, so you = Ay; thus, ya ∈ Ay. By definition of a coset, there exists aˇ ∈ A such that you = ayˇ . By substitution, t = w (ayˇ ). By the associative property, t = (waˇ) y. By definition of a coset, waˇ ∈ WA. By hypothesis, A is normal, so we = Aw. Thus waˇ ∈ Aw. By hypothesis, W = X ; that is, we = xA. Thus waˇ ∈ xA, and by definition of a coset and this tool for conversion.