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I believe, although I can't say that I've given a rigorous proof, that for a free group $F_r$, and an element of it $a$, $C_{F_r}(\langle a \rangle)=$ the group generated by the elements $b \in F_r$ such that $a=b^n$ for some integer $n$ (I will say: the group of powers and roots of $a$).

One may similarly ask (and this is my interest in this), given $ a\in \hat{F_r}$ (the free profinite group on $r$ generators), what is $C_{\hat{F} _r}(\langle a \rangle)$? In particular, is it the profinite completion of the group of powers and roots of $a$?

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I do not know the complete answer on your question. However it is clear that the situation it the profinite case is different.

Consider a semidirect product $G=\mathbf Z_3\rtimes \mathbf Z_2$ where a generator $a$ of $\mathbf Z_2$ acts by sending elements of $\mathbf Z_3$ to their inverses. Then $G$ is a proyective profinite group (because all its Sylow subgroups are free pro-$p$) and so it is a subgroup of a free profinite group. Now the centralizer of $a^2$ contains $G$ and so it is not abelian.

Added: If $a$ is a non-trivial pro-2 element of $\hat {\mathbf Z}$ (profinite completion of $\mathbf Z$), then all the roots and powers of $a$ are also pro-2 elements.

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Hi, Andrei. It is worth mentioning that in a free pro-$p$ group every subgroup is free. Therefore, as centralizers have non-trivial center they must be cyclic. –  Yiftach Barnea Apr 24 '11 at 11:11
    
I'm not sure I understand: First, isn't this $G$ just $S_3$? And if so, then all of $G$ being in the centralizer of $a^2$ would in particular say that $S_3$ has nontrivial center, which it doesn't. Second, it seems that you conclude your contradiction by saying that the group that I described must be abelian. Why would that be true? It seems to me that different roots of the same element don't have to commute with each other. Am I wrong in this impression? –  Makhalan Duff Apr 24 '11 at 17:03
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$\mathbf Z_p$ is the pro-$p$ completion of $\mathbf Z$ (not a cyclic group of order $p$). Any root of $a^2$ is a pro-2 element, but you are right $\mathbf Z_3$ is contained in the group generated by roots of $a^2$. –  Andrei Jaikin Apr 24 '11 at 20:40
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Another partial answer:

It is a theorem of Herfort and Ribes that if $A$ and $B$ are profinite groups, and $G = A \sqcup B$ is their free profinite product, then centralizers of nontrivial elements of $A$ lie in $A$. (See Wolfgang Herfort and Luis Ribes, Torsion elements and centralizers in free products of profinite groups. J. Reine Angew. Math. 358 (1985), 155–161. )

So, at least centralizers are nice when $\langle a \rangle$ lies in a cyclic free factor.

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