Theorem If $G$ is a nilpotent group and $H < G$ then $H < N_G(H)$.
proof: Take its central series. Let $i$ be minimal such that $Z_i(G) \not \le H$ then certainly $i > 0$ and $Z_{i-1}(G) \le H$. We know $Z_{i-1} \lhd Z_i$, take $z \in Z_i(G) \setminus H$ then $Z_{i-1}(G)z$ lies in the center of $G/Z_{i-1}(G)$ (since $Z_i/Z_{i-1} \le Z(G/Z_{i-1}(G))$). For all $h \in H \le G$ we get $zh Z_{i-1}(G) = hz Z_{i-1}(G)$ and since $Z_{i-1}(G) \le H$ we have $[h,z] \in H$ for all $h$: by basic properties of commutators [and that $z$ is not in $H$] this shows that $z \in N_G(H) \setminus H$.
Proposition (This is a Frattini argument) If $H \unlhd G$ and $P$ is a Sylow $p$-subgroupof $H$ for some prime $p$ then $G = N_G(P) H$.
proof: Take $g \in G$ and consider that $P^g \le H^g = H$ is a Sylow $p$-group hence conjugate to $P$, i.e. there's some $h \in H$ such that $P^g = P^h$ which implies $gh^{-1} \in N_G(P)$ so each $g \in N_G(P) H$.
Theorem If $P$ is a Sylow $p$-subgroup of $G$ then $N_G(P) = N_G(N_G(P))$.
proof: $P$ is the only Sylow $p$-subgroup of $N_G(P)$! Any other would be conjugate (by Sylow's theorem) but $P^g = P$ by the definition of normalizer. $P \le N_G(P) \le N_G(N_G(P))$ so in fact $P$ is the only Sylow $p$-group of $N_G(N_G(P))$ too, we just need to show that $N_G(N_G(P)) \le N_G(P)$ now: let $x \in N_G(N_G(p))$ so $N_G(P)^x = N_G(P)$ and hence $P^x \le N_G(P)$ but we've already seen that conjugates of Sylow $p$-groups are all equal so $x \in N_G(P)$.
proof: You can prove it quicker with Frattini: $N_G(P) \unlhd N_G(N_G(P))$ so $N_G(N_G(P)) = N_G(P)N_{N_G(P)}(P)$ but $N_{N_G(P)}(P) \le N_G(P)$ so $N_G(N_G(P)) = N_G(P)$.
Theorem If $H$ is a subgroup of $G$ which contains the normalizer of a Sylow subgroup $P$, then $H=N_G(H)$.
proof: Since $H \unlhd N_G(H)$ the Frattini argument gives us $N_G(H) = N_{N_G(H)}(P) H$ but $N_{N_G(H)}(P) \le N_{G}(P) \le H$.
Proposition For $p$ prime, a nilpotent group is a direct product of its $p$-groups.
proof: We have already seen that a $p$-group is nilpotent and the direct product of nilpotent groups is nilpotent. In the other direction, let $G$ be some nilpotent group and take any $p$-Sylow subgroup of $G$ for some $p$ dividing $|G|$: let $N=N_G(P)$ we know $N < G$ implies $N < N_G(N)$ but this contradicts the previous result, so $N=G$ thus $P \unlhd G$, since all the $p$-groups have coprime order their direct product gives back the original group. (Note: $G = N_G(P)$ implies $P \unlhd G$ is proved in the properties about normalizers).
Theorem Normalizers always group in a nilpotent group.
proof:
Let $H$ be a subgroup of a nilpotent group $G$, then if $H$ doesn't
contain the center take some element of the center not in $H$: that'll
be in the normalizer. If $H$ is the center the normalizer will be the
whole group. If $H$ strictly contains the (nontrivial!) center $Z$ then by induction (which we can only do due to the nilpotent condition) the normalizer of $H/Z$ in $G/Z$ will grow and there will be some $nZ$ in the normalizer which isn't of the form $hZ$, and $(H/Z)^(nZ) = H/Z$ so $H^n=H$.
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