Definition The normalizer $N_G(S)$ of a subset $S$ of a group $G$ is the set of all $a \in G$ such that $S^a = S$.
Theorem $N_G(S) \le G$.
proof: Clearly $1 \in N_G(S)$. Let $a,b$ in $N_G(S)$ then $S^{ab} = (S^a)^b = S^b = S$ so $ab$ is too. Since the group is finite we have inverses too (make n big enough so that $a^n = 1$ then $a^{-1} = a^{n-1}$). Alternative (more sensible) argument: $S^{1} = S^{aa^{-1}}=S^{a^{-1}}$.
Theorem If $H \le G$ then $H \unlhd N_G(H)$.
proof: We have already seen that it's a group, we just need to show $\forall n \in N_G(H), nH = Hn$ but that's immediate by the definition of normalizer.
Corollary When $H \le G$, $G = N_G(H)$ implies $H \unlhd G$.
Definition Another equivalence relation, conjugacy is defined by $x \sim y$ if there exists some $g$ such that $x^{g} = y$.
Lemma Central elements partition the group into trivial conjugacy classes.
proof: Let $a \in Z(G)$ then $g^a = a^{-1}ga = g$ for all $g$ so every such conjugacy class is a single element.
Lemma The number of conjugates of $x$ (including $x$) is $[G:N_G(x)]$.
proof: orbit stabilizer.
Theorem (Conjugacy Class Equation) $$
\left|{G}\right| = \left|{Z \left({G}\right)}\right| + \sum_{x} \left[{G : N_G \left({x}\right)}\right]$$ 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 $G \setminus Z(G)$ 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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