I'm working on some palindromic words right now. Those are the elements of $F\_2$, the free group on two generators, which are the same if reveresed. For example $xyx, y^2, xyxxyx$ and so on.


Can you write each element of $F'$, the derived group, as a product of palindromes of EVEN length? Does this property hold if you are not in $F$, but in a group $G$ in which you know that every element is a product of at most $n$ palindromes? If $G$ has this property then $G$ has finite palindromic width.


I've puzzled around quite a bit but can't find a way of even writing $x^{-1}y^{-1}xy$ as a product of palindromes of even length.


Thanks,
Elisabeth



$\left[x,y\right]$ as a product of palindromes of even length?
The set of products of palindroms of even length $P$ is a normal subgroup, because the inverse of a palindrome is a palindrome and because $xpx^{-1} = xpx\cdot x^{-1} x^{-1}$ for all generators $x$ and all $p\in P$.


In the quotient $F(S)/P$ every generator $s\in S$ squares to 1. Therefore $F(S)/P$ is a quotient of $\langle S \mid \forall s\in S: s^2=1\rangle$. Since every palindrome of even length is of the form $x\_1 x\_2 \ldots x\_{n-1} x\_n x\_n x\_{n-1}\ldots x\_2 x\_1$, there are no further relations, so $F(S)/P \cong \langle S \mid \forall s\in S: s^2=1\rangle = \underbrace{C\_2 \ast \ldots \ast C\_2}\_{|S|\,\text{times}}$. In particular the quotient is not abelian, if $|S|>1$. Hence $F(S)'$ is not contained in $P$.


