A question about normal sets in a finite subgroup

Question

Let $G$ a finite non-abelian group and suppose that $N$ is a normal subset (not a subgroup) of $G$, that is, $aN=Na$ for every $a\in G$. Under what conditions $|N|$ divides $|G|$ ?

Is there any particular context you care about? If, say, $G$ is abelian then every subset is normal... — lulu, Oct 18 '16 at 19:05
I mean especially when $G$ is non-abelian (i edited my question) — boaz, Oct 18 '16 at 19:07
@ThePortakal Yes, that's Lagrange's Theorem. Yet here it is given a subset, not a subgroup. Perhaps a typo, perhaps that's what the OP really means. — DonAntonio, Oct 18 '16 at 19:10
Though, again, if $G$ has a non-trivial center then any subset of that will work...so perhaps you meant to also assume that $Z(G)=e$ or something close to that? — lulu, Oct 18 '16 at 19:13
Well, If we take $N\subset Z(G)$ then of course $N$ commutes with every group element, so it is trivially normal. Now, if $|Z(G)|≤2$ this doesn't tell us much (since the subsets $N$ would have either one or two elements and we don't learn anything about $|G|$ either way). But if $|Z(G)|>2$ there are interesting subsets. — lulu, Oct 18 '16 at 19:21
But this is perhaps a condition of such N to exist, why $|N|$ divides $|G|$? — boaz, Oct 18 '16 at 19:30

score 1 · Accepted Answer · answered Oct 18 '16 at 19:19

$N$ is necessarily the union of entire conjugacy classes. The size of every conjugacy class divides $|G|$ (by the orbit stabiliser theorem) so if $N$ consists of only one conjugacy class, its size will divide the group order. If it contains more than one conjugacy class, not much can be said about divisibility in general. For example, in $S_n$ you could make $N$ the union of any set of classes of all permutations with certain cycle-types. So in $S_4$ the size of $N$ could be any of 1,3,8,6,6 or any of their sums.

A question about normal sets in a finite subgroup

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