Let \(G\) be a group and \(H\leq G\) a subgroup. The normalizer of \(H\) in \(G\) is \[ N_G(H) \coloneqq \theset{g\in G \suchthat gHg^{-1} = H}. \]
Let \(H\leq G\). Then \(N_G(H)\) is a subgroup of \(G\) in which \(H\) is normal, and every subgroup \(M\leq G\) with \(H\normal M\) is contained in \(N_G(H)\). Hence \[ N_G(H) = \bigcup_{M\in S} M, \qquad S \coloneqq \theset{M\leq G\suchthat H \normal M}. \]
If \(gHg^{-1}=H\) and \(g'Hg'^{-1}=H\), then \((gg')H(gg')^{-1}=H\) and \(g^{-1}Hg=H\), so \(N_G(H)\) is a subgroup; it contains \(H\), and \(H\) is normal in it by definition. If \(H\normal M\), then \(mHm^{-1}=H\) for every \(m\in M\), so \(M\subseteq N_G(H)\). The union of all such \(M\) therefore lies in \(N_G(H)\) and contains \(N_G(H)\in S\).
Conjugation defines a homomorphism \(N_G(H)\to\Aut(H)\), \(g\mapsto(h\mapsto ghg^{-1})\), whose kernel is the centralizer \(C_G(H)\). Thus \(C_G(H)\subseteq N_G(H)\), and elements of \(N_G(H)\) permute the elements of \(H\) by conjugation while elements of \(C_G(H)\) fix them.