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Let $G$ be a group. We can define

$$F(G)=\{g\in G\,|\,o(\operatorname{Cl}(g))<\infty\},$$

where $o(\operatorname{Cl}(g))$ is the order (cardinality) of the conjugacy class of $g$ in $G$. This set is a subgroup of $G$ because

1) $o(\operatorname{Cl}(1))=1<\infty;$

2) we have $g(xy)g^{-1}=(gxg^{-1})(gyg^{-1}),$ so $\operatorname{Cl}(xy)\subseteq\operatorname{Cl}(x)\operatorname{Cl}(y).$ Therefore, if $$o(\operatorname{Cl}(x))<\infty\text{ and }o(\operatorname{Cl}(y))<\infty,$$ then $o(\operatorname{Cl}(xy))<\infty.$

$F(G)$ is a characteristic subgroup of $G.$ Indeed, let $\alpha$ be an automorphism of $G$. Then


for all $g,x\in G.$ We have $o(\alpha(\operatorname{Cl}(x)))=o(\operatorname{Cl}(x))$ because $\alpha$ is a bijection and so, since $\{\alpha(g)\,|\,g\in G\}=G,$ we obtain


and therefore

$$\alpha(g)\in F(G).$$

I would like to ask if there is a name for this characteristic subgroup. Also, since $F(G)$ is a piece of ad hoc notation, I'd be grateful if you could tell me what the common notation is.

Is the group $G/F(G)$ important? What is it called?

And finally, where can I read about it?

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I know that the set of all elements of finite order in a group $G$ is called the torsion subgroup of $G$, but I can't tell if it relates or not to this. – Patrick Da Silva Feb 12 '12 at 1:58
Can you explain the motivation behind your question, i.e. where this subgroup arose? – Alex Becker Feb 12 '12 at 2:38
@AlexBecker Of course. I'm reading some lecture notes on group rings. This definition was there -- apparently this subgroup is useful in the theory of group rings, although I haven't gotten to the point where it manifests itself yet, so I cannot give any details. This particular lecture was about FC-groups. I've found some additional material on those, but $F(G)$ isn't there. – user23211 Feb 12 '12 at 2:49
I have never come across a name for this subgroup, but I do not recommend $F(G)$, because that has a standard meaning already - the Fitting subgroup of $G$, which is the largest normal nilpotent subgroup. – Derek Holt Feb 12 '12 at 4:54
The usual notation is $\triangle(G)$. It comes up in the group ring because finitely generated subgroups of $\triangle(G)$ have finite-index centers, and this makes their group ring very interesting. In particular, linear identities in the group ring $K[G]$ can usually be solved by first passing to the group ring $K[\triangle(G)]$. All of this (and much more!) is covered in chapter 4 of The Algebraic Structure of Group Rings by Passman. I highly recommend it. – user641 Feb 12 '12 at 7:19

1 Answer 1

up vote 1 down vote accepted

Yes, this is called the FC-centre of the group. I've seen $FC(G)$ used for this. I don't recall whether the quotient $G/FC(G)$ (or $G/F(G)$ in your notation) has a special name. There is a (nice) book on the subject by M. J. Tomkinson.

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Thanks a lot, James! – user23211 Feb 14 '12 at 21:30
@ymar You're welcome. If you can't put your hands on a copy of Tomkinson, there is a little bit on this in D.J.S. Robinson's text. – James Feb 14 '12 at 21:40

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