Let $F\_n$ be the free group with $n$ generators, where $n$ is an interger greater than $1$. Let $RF\_n$ be the reduced free group, which is defined to be the quotient group of $F\_n$ obtained from $F\_n$ by adding the relations that each generator of $F\_n$ commutes with all its conjugates.


Can anyone help to prove or disprove that $RF\_n$ is centerless, that is, the center of $RF\_n$ is trivial?


Your help will be much appreciated.



Centerlessness of reduced free group
The group has a center. For example, if $F\_2=\langle a,b\rangle$, $[a,b]$ is in the center of the factor-group. Indeed, $[a,b]=a\cdot (a^{-1})^b$ and so it is a product of conjugates of $a$, and commutes with $a$ in the factor-group. On the other hand, $[a,b]=b^ab^{-1}$ is a product of conjugates of $b$, so it commutes with $b$ in the factor-group. Thus $[a,b]$ commutes with $a$ and $b$, and hence is in the center of the factor-group. This example can be generalized to any rank $n$. If $F\_n=\langle a\_1,...,a\_n\rangle$, and we impose the relations that $a\_i$ commutes with all its conjugates, then in the factor-group $a\_i$ commutes with all products of its conjugates and their inverses, hence $a$ commutes with every element of the normal subgroup $N\_i=a\_i^{F\_n}$ generated by $a\_i$. Now take any element $w$ in the intersection of all $N\_i$ (that intersection is non-trivial obviously). It would belong to the center of the factor. One can easily find such $w$ which is not 1 in the factor. 


