Schmidt group and sylow subgroup Let $G$ be Schmidt group. 
If $Q=\langle a\rangle$ is $q$-subgroup of $G$, then $a^q \in Z(G)$
 A: A Schmidt group is a finite non-nilpotent group in which every proper subgroup is nilpotent. For instance $S_3$ and $A_4$. Like these examples, it has order divisible by only two different primes, say $p$ and $q$, and like these examples the Sylow $p$-subgroup $P$ (say) is normal and the Sylow $q$-subgroup $Q=\langle a \rangle$ is cyclic.
An example where $a^q \neq 1$ is given by $Q= \langle (2,3)(4,5,6,7) \rangle$ and $P=\langle (1,2,3) \rangle$.
Proof: Now $\langle a^q \rangle P$ is a proper subgroup of $G$, so it is nilpotent. A finite nilpotent group is a direct product of its Sylow subgroups, so $\langle a^q \rangle P = \langle a^q \rangle \times P$ and so $a^q$ commutes both with $\langle a \rangle$ and $P$, so $a^q$ is in the center of $G= \langle a \rangle P$. $\square$

Sometimes $\langle a^q \rangle$ is written $\Phi(\langle a \rangle)$ or $\Phi(Q)$, so we have $\Phi(Q) \leq Z(G)$.
A few other useful facts: $\langle a^g : g \in G \rangle$ is generated by its Sylow $q$-subgroups ($\langle a^g \rangle$) and so is equal to $\langle a \rangle$ if it is nilpotent. However, that would mean that $\langle a \rangle$ was normal in $G$, and that $G$ was nilpotent, a contradiction. Hence $\langle a^g : g \in G \rangle =G$ is generated by its Sylow $q$-subgroups.
Similarly, if $N$ is a characteristic proper subgroup of $P$, then $\langle a \rangle N$ is a proper subgroup, so nilpotent, so $\langle a \rangle \times N$ is a direct product and $a$ centralizes $N$. Since $G$ is generated by the conjugates of $a$, and $a^g$ centralizes $N=N^g$, we get that $N \leq Z(G)$. In particular, $\Phi(P) \leq Z(G)$ as well.
These are all from Huppert's Endliche Gruppen Satz III.5.2 on page 281–282.
