If $G$ is an abelian group, every subgroup $H$ of $G$ is normal.
I searched for non-abelian finite groups $G$ , such that every subgroup is normal and GAP showed only the groups $G'\times Q_8$ , where $Q_8$ is the quaternion-group of order $8$ and $G'$ is an abelian group with the additional property that $C_4$ is not a subgroup of $G'$. The largest group I found with GAP was $C_{15}\times Q_8$.
Is it true that every non-abelian finite group $G$ with the property that every subgroup of $G$ is normal, is isomorphic to $G'\times Q_8$ with an abelian group $G'$ without $C_4$ as a subgroup ?
Hall's book "The Theory of Groups" proves (Theorem 12.5.4) that these groups, known as Hamiltonian groups, are all of the form $Q_8 \times A \times B$ where $A$ is an elementary $2$-group, $B$ is an abelian group where every element is of finite odd order, and $Q_8$ is the quaternion group. All groups of this form are Hamiltonian. Note there is no finiteness restriction on these $A$, $B$ groups, so there are also infinite group.