TITLE: A question about Solvable Group.
QUESTION [5 upvotes]: I have a solvable group $G$ with a $p$-sylow subgroup $P$ such that $C_G(P)=Z(G)$ and for every $q$-sylow subgroup $Q$ of $G$ where $q\neq p$, $Z(G)\lvertneqq C_G(Q)$. Now I want to prove that $P$ is not normal in $G$ and the other sylow subgroups are normal.
Any idea?
Thanks in Advanced.

REPLY [5 votes]: If $G=P$ (is any non-abelian $p$-group) then you have a counterexample.
A less trivial counterexample is $G=\operatorname{SL}(2,3)$ of order 24 with $p=2$. Its Sylow 2-subgroup is $P=Q_8$, and its Sylow 3-subgroups $Q$ are all cyclic of order 3. The centralizer of $P$ is $Z(P) = Z(G)$, but of course the centralizer of $Q$ contains not only $Z(G)$ but also $Z(Q)=Q$. Contrary to your goals, $P$ is normal and $Q$ is not.
In case you got the normal versus not normal switched: the dihedral group of order 24 satisfies the hypothesis with $p=2$, but $Q$ is normal and $P$ is not.