Are there some theorems about the splitting of infinite conjugacy classes into several conjugacy classes in a subgroup? I am mainly interested in subgroups of finite index. Thanks.

  • $\begingroup$ Take any conjugacy class in any group that splits into more than one class in a subgroup (of finite index if you like). And now just take the product of the entire situation with a group with an infinite conjugacy class? What am I missing? $\endgroup$ – Kevin Buzzard Nov 20 '09 at 19:01
  • $\begingroup$ But in general, is there a conjugacy class splits into more than one in some subgroup? In abelian group, there is no such conjugacy class. Actually, I want to know whether there some theorems related this. $\endgroup$ – yeshengkui Nov 21 '09 at 1:41
  • $\begingroup$ "in general is there a conj class that splits into more than one in some subgroup?". In general, yes. $\endgroup$ – Kevin Buzzard Nov 21 '09 at 7:14

I'm not sure what you mean by an infinite length conjugacy class. Most likely you mean that the cardinality is infinite.

Consider the Heisenberg group generated by 3 elements $x,y$ and $z$ with relations so that $z$ is central and $xyx^{-1}y^{-1}=z$. Then the conjugacy class containing $y$ consists of all elements of the form $yz^n$ for integers $n$.

If we pass to the finite index subgroup generated by $x^k,y$ and $z$ for some natural number $k$ this splits into $k$ distinct classes represented by $yz^i$ for $i=0,\ldots k-1$.

  • $\begingroup$ Perhaps I should add that if you pass to a subgroup of index n each class can split into at most n new classes and n can occur as illustrated in my answer. I suppose an orbit-stabiliser type argument will show the number of classes must divide n. $\endgroup$ – Simon Wadsley Nov 20 '09 at 22:49
  • $\begingroup$ This is a good example! $\endgroup$ – yeshengkui Nov 21 '09 at 1:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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