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.
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$\begingroup$ 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 BuzzardNov 20, 2009 at 19:01
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$\begingroup$ 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. $\endgroup$– yeshengkuiNov 21, 2009 at 1:41
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$\begingroup$ "in general is there a conj class that splits into more than one in some subgroup?". In general, yes. $\endgroup$– Kevin BuzzardNov 21, 2009 at 7:14
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1 Answer
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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 $i=0,\ldots k-1$.
answered Nov 20, 2009 at 13:48
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$\begingroup$ 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. $\endgroup$ Nov 20, 2009 at 22:49
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