Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.