Let $P$ be a nonabelian $p$-group, where $p$ is an odd prime. Let $Q$ be a nonabelian normal subgroup of $P$. Does there exist a normal noncyclic abelian subgroup of $P$ that is contained in $Q$?
Since $p$ is odd, I know that both $P$ and $Q$ contains a normal noncyclic abelian subgroup. (See Isaacs, Finite Group Theory, Th.6.12.) But I failed to get the result from this. Or is there any counterexample?
The answer is yes, there is one. This is a special case of Lemma 1.4 from Berkovich's Groups of Prime Power Order, which states that if $G$ is a $p$-group and $N$ is a normal subgroup of $G$, and if $N$ possesses no subgroup $C_p\times C_p$ that is normal in $G$, then $p=2$ and $G$ is dihedral, quaternion or semidihedral. A direct proof is below.
To prove this, use induction on $|Q|$. Notice that if $(P,Q)$ is a minimal counterexample, then any normal subgroup $R$ of $P$ contained in $Q$ is cyclic, as $R$ cannot be non-cyclic and abelian, and if it is non-abelian then induction states that there is a non-cyclic abelian normal subgroup of $P$ contained in $R$, hence in $Q$.
Now consider the set of maximal sugbroups of $Q$. Their number is congruent to $1$ modulo $p$ as they correspond to hyperplanes of a vector space. Thus (at least) one of them is normalized by $P$, hence it must be cyclic. In particular, $Q$ has a cyclic subgroup of index $p$, where $p$ is odd, and is non-abelian. Thus $Q$ is the 'modular group' that is a semidirect product of $C_{p^r}$ by $C_p$. This has exactly $p+1$ maximal subgroups, and they are all abelian. (This is because the order-$p$ automorphism of $C_{p^r}$ centralizes $C_{p^{r-1}}$.) Clearly there are normal subgroups $C_{p^r}$ and $C_{p^{r-1}}\times C_p$.
Now we have to decide whether the other $p-1$ subgroups are cyclic or not. In fact, they are cyclic, and so the non-cyclic one is normal in $P$.