A fact regarding the interaction of powers and commutators in a group

Let G be a group and let x,y \in G such that x and y commute with [x,y]. Prove that (xy)^n = x^ny^n[y,x]^{n(n-1)/2}.


We begin with some lemmas.

Lemma 1: Let a,b \in G be group elements and n a positive integer. Then (ab)^n = a(ba)^{n-1}b. Proof: We proceed by induction on n. If n = 1, the conclusion is trivial. Suppose the conclusion holds for some n. Then (ab)^{n+1} = (ab)^n ab = a(ba)^{n-1}bab = a(ba)^nb, and the conclusion holds for all n. \square

Lemma 2: Let x and y be as in the problem statement and let n be a positive integer. For all 1 \leq k \leq n, we have (xy)^n = x^k [y,x]^{k(k-1)/2} (yx[y,x]^{k-1})^{n-k} y^k. Proof: We proceed by induction on k. For the base case k = 1, the conclusion is trivial (using Lemma 1). For the inductive step, suppose the conclusion holds for some 1 \leq k < n. Then, noting that yx = xy[y,x], we have the following.

(xy)^n = x^k [y,x]^{k(k-1)/2} (yx[y,x]^{k-1})^{n-k} y^k
= x^k [y,x]^{k(k-1)/2} (xy[y,x]^k)^{n-k} y^k
= x^k [y,x]^{k(k-1)/2} (x[y,x]^ky)^{n-k} y^k
= x^k [y,x]^{k(k-1)/2} x[y,x]^k (yx[y,x]^k)^{n-k-1} yy^k
= x^{k+1} [y,x]^{(k+1)k/2} (yx[y,x]^k)^{n-(k+1)} y^{k+1}

Thus the conclusion holds for all 1 \leq k \leq n. \square

Now the main result follows from Lemma 2 with k=n.

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