$\left[x,y\right]$ as a product of palindromes of even length? 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
 A: Let $g=(1,2)$, $h=(2,3)$ two non-commuting involutions in (say) the symmetric group $S_3$. Consider the homomorphism $\varphi:F_2\to S_3$ with $\varphi(x)=g$, $\varphi(y)=h$.
If $w$ is any palindromic word of even length, then $w\in\ker(\varphi)$. However, $x^{-1}y^{-1}xy$ is not in $\ker(\varphi)$, and so can't be a product of palindromic words of even length.
A: 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 $P$.
