Theorem NG(S)≤G.
proof: Clearly 1∈NG(S). Let a,b in NG(S) then Sab=(Sa)b=Sb=S so ab is too. Since the group is finite we have inverses too (make n big enough so that an=1 then a−1=an−1). Alternative (more sensible) argument: S1=Saa−1=Sa−1.
Theorem If H≤G then H⊴NG(H).
proof: We have already seen that it's a group, we just need to show ∀n∈NG(H),nH=Hn but that's immediate by the definition of normalizer.
Corollary When H≤G, G=NG(H) implies H⊴G.
Definition Another equivalence relation, conjugacy is defined by x∼y if there exists some g such that xg=y.
Lemma Central elements partition the group into trivial conjugacy classes.
proof: Let a∈Z(G) then ga=a−1ga=g for all g so every such conjugacy class is a single element.
Lemma The number of conjugates of x (including x) is [G:NG(x)].
proof: orbit stabilizer.
Theorem (Conjugacy Class Equation) |G|=|Z(G)|+∑x[G:NG(x)]
where the sum is taken over representatives of the non-singleton conjugacy classes.
proof: This is really just a corollary of the lemma plus the information that center elements have trivial conjugacy classes, but we discuss it a bit more anyway. In the abelian case G=Z(G) and there are non non-singleton conjugacy classes so the result is immediate. In the general case G∖Z(G) are the elements which the conjugacy relation partitions into non-singleton sets - the relation proved in the previous lemma gives the equality.
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