## 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 $k=n$.