TITLE: Commutators in a group
QUESTION [5 upvotes]: Let $G$ be a group and for $x,y\in G$, define $[x,y]=x^{-1}y^{-1}xy$ to be the commutator of $x$ and $y$. If $y_1,\cdots,y_n\in G$, is it true that $[x,y_1\cdots y_n]$ can be written of a product of conjugates of $[x,y_i]$ with $i=1,\cdots,n$? Namely, is it true that there exist elements $g_1,\cdots,g_n\in G$ such that $$[x,y_1\cdots y_n]=[x,y_{j_1}]^{g_1}\cdots[x,y_{j_n}]^{g_n},$$ where $\{j_1,\cdots,j_n\}=\{1,\cdots,n\}$ and $[x,y_{j_i}]^{g_i}=g_i^{-1}[x,y_{j_i}]g_i$?

REPLY [2 votes]: Yes, it is true. Consider the assertion that $[x,y_1 \cdots y_n]$ can be expressed as a product of conjugates of $[x,y_n],\ldots,[x,y_1]$, in that order.  For $n=2$, it can be verified that $[x,y_1 y_2] = [x,y_2][x,y_1]^{y_2}$.  Fix $n$ and assume the assertion holds for smaller values of $n$. Then $[x,y_1 \cdots y_n] = [x,y_n] [x,y_1 \cdots y_{n-1}]^{y_n}$.  By the induction hypothesis, this expression can be written in the form $[x,y_n]([x,y_{n-1}]^{g_{n-1}} \cdots [x,y_1]^{g_1})^{y_n} = [x,y_n][x,y_{n-1}]^{g_{n-1} y_n} \cdots [x,y_1]^{g_1 y_n}$. This proves the assertion.