Minimal subgroups lie in the center so group is nilpotent Let $G$ be a group of odd order. If every minimal subgroup lies in the center, prove that $G$ is  nilpotent . Thanks!
 A: Proposition: Let $G$ be a finite group, $p$ an odd prime, such that every subgroup of order $p$ is contained in the center of $G$. Then $G$ is $p$-nilpotent.
Corollary: If $G$ has odd order and every subgroup of prime order is central, then $G$ is nilpotent.
Lemma: (Thompson) If $P \leq G$ is a $p$-group for $p$ an odd prime, and $Q \leq N_G(P)$ is a subgroup whose order is relatively prime to $p$, and $Q$ centralizes every element of $P$ of order $p$, then $Q$ centralizes $P$.
Proof: See Gorenstein's Finite Groups theorem 5.3.10, page 184. $\square$
Proof: (of the proposition) Let $P$ be a $p$-subgroup of $G$ and let $x \in N_G(P)$ have order relatively prime to $p$. Since every subgroup of order $p$ is central, $x$ centralizes every element of order $p$, and thus by the lemma, $x \in C_G(P)$. Hence $[N_G(P):C_G(P)]$ is a power of $p$. By Frobenius's normal $p$-complement theorem (Theorem 7.4.5, page 253), $G$ is $p$-nilpotent. $\square$
Proof: (of the corollary) Such a group is $p$-nilpotent for every odd prime $p$ by the proposition. Since the group itself has odd order, it is $p$-nilpotent for every prime $p$ dividing the order of $G$. Hence the group is nilpotent (the intersection of the normal Sylow $q$-complements for $q \neq p$ is a normal Sylow $p$-subgroup). $\square$
A: This is more like Huppert's proof, which is mostly on the bottom of page 434 to the top of page 435 (and then maybe in a previous chapter?)
Proposition: Let $G$ be a finite group, $p$ an odd prime, such that every subgroup of order $p$ is contained in the center of $G$. Then $G$ is $p$-nilpotent.
Corollary: If $G$ has odd order and every subgroup of prime order is central, then $G$ is nilpotent.
Proof: (of proposition) Let $G$ be a minimal counterexample. Every subgroup of $G$ still satisfies the hypothesis (that all elements of order p are central). Hence every proper subgroup is $p$-nilpotent, by induction. Let $P$ be a Sylow $p$-subgroup of $G$. If $N_G(P)$ is a proper subgroup of $G$, then it is $p$-nilpotent. That is, $N_G(P) = P \ltimes Q$ with $Q$ a normal Sylow $p$-complement. However, $Q \leq N_G(P)$, so we get that both $P$ and $Q$ are normal in $N_G(P)$ and $N_G(P) = Q \times P$ so that $Q \leq C_G(P)$. [Note how we've recovered part of the Thompson lemma.] 
... Now we prove $P$ has exponent $p$. ... [eek time to go]
Since every element of $P$ has order (dividing) $p$, we see that $P \leq Z(G)$, so that by Schur-Zassenhaus $G= Q \times P$ (for a slightly bigger $Q$, a Sylow $p$-complement of $G$). $\square$
