Do elements of two abelian normal subgroups commute? So $H$ and $K$ are normal abelian subgroups of some group. Is it true for all $h \in H$ and for all $k \in K$ that $hk=kh$? I do not think the statement is valid but I am unable to find a (rather simple) counterexample.
 A: Let $G=\{\pm1,\pm i,\pm j,\pm k\}$ be the quaternion group of order $8$.
Consider $H=\{\pm1,\pm i\}$ and $K=\{\pm1,\pm j\}$.
A: The easiest counterexample is the dihedral group $D_8$, say generated by $a$ of order $4$ and $b$ of order $2$. Every element of $D_8$ lies in a normal subgroup of order $4$: $\{1,a,a^2,a^3\}$, $\{1,a^2,b,a^2b\}$ and $\{1,a^2,ab,a^3b\}$. These are of course all abelian, since they have order $4$. If your statement held, then $D_8$ would therefore be abelian, which is of course is not.
The example of $Q_8$ from the other two answers is perfectly valid, of course. In fact, if $G$ is any non-abelian group of order $p^3$ then every element lies in a subgroup of order $p^2$ (which is necessarily abelian and normal), and so every non-abelian group of order $p^3$ is a counterexample.
A: Any Hamiltonian group will give you a counterexample by definition, as any cyclic subgroup is abelian and normal, yet you can find two cyclic subgroups with generators that do not commute.
The smallest such  example are the quaternion group $Q_8$.
