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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.

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1 Answer 1

up vote 11 down vote accepted

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.

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Thank you Mark for your answer!! Then can we conclude that the center of RF_n is the intersection of all N_i, as is defined in your answer? –  Zuriel Aug 15 '11 at 9:21
@Zuriel: I am not sure that the center is the intersection of all $N_i$ (for all $n$). It looks so, but I do not know a proof at the moment. For $n=2$ it is true because the factor-group is just free nilpotent of rank 2 and class 2. –  Mark Sapir Aug 15 '11 at 9:40
@Mark Sapir: as you mentioned in your previous answer, [a, b] is in the center of RF_2 and thus the group has a center. But do we also need to prove [a, b] is not equal to the identity element of the group? I know that it should be true; but how to prove it? Thank you again for your answer. –  Zuriel Aug 15 '11 at 10:15
If $[a,b]=1$ in the factor-group, then the factor-group is commutative. But the Heizenberg group $\langle a,b \mid [[a,b],a]=[[a,b],b]=1\rangle$ satisfies your relations and is not commutative. –  Mark Sapir Aug 15 '11 at 11:30

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