# Are all nilpotent groups hamiltonian?

Are all nilpotent groups hamiltonian? That is, is every subgroup of a nilpotent group normal?

I don't think so. Every Sylow subgroup of nilpotent group is normal and every nilpotent group is a direct product of its Sylow subgroups. But, are they hamiltonian?

And, would the converse be true? That is, is every hamiltonian group nilpotent? Any small counterexamples? Thanks beforehand.

• For the first question, check by hand the dihedral group of order $2^n\ge 8$. – YCor Jun 25 at 13:08

In addition to what has been answered: a subgroup $$H$$ of $$G$$ is called subnormal if there exists a series of subgroups $$H=H_0 \lhd H_1 \lhd \cdots \lhd H_s=G$$. A normal subgroup is obviously subnormal, but the converse is not true. Now, finite nilpotent groups are exaclty the finite groups in which every subgroup is subnormal. For a proof: see for example M.I. Isaacs, Finite Group Theory, Lemma 2.1.
A finite group $$G$$ is hamiltonian if and only if $$G\cong Q_8\times A\times B$$, where $$A$$ is an abelian group with odd order and $$B$$ is a direct product of $$\mathbb{Z}_2$$. Example: every $$p$$-group is nilpotent but it is not hamiltonian in general.
• How could we prove that if a group is hamiltonian, it can be written as $Q_8\times A\times B$? – vidyarthi Jun 25 at 14:23