Sign up ×
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|cite|improve this question

1 Answer 1

up vote 12 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|cite|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
The center of $RF_n$ is given by Theorem 1.7.1 of Combinatorial Group Theory In Homotopy Theory I by F. Cohen, available at – Zuriel Nov 29 '14 at 13:21

Your Answer


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.