Question

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.

Answer 1

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.