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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.

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

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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.

Conversely, a finite group is nilpotent if and only if it is a direct product of Sylow subgroups. Hence if a group is hamiltonian, all of its Sylow subgroups are normal. Hence it is a direct product of Sylow subgroups.

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  • $\begingroup$ How could we prove that if a group is hamiltonian, it can be written as $Q_8\times A\times B$? $\endgroup$ – vidyarthi Jun 25 at 14:23
  • $\begingroup$ Try googling "Hamiltonian Groups". The first five links or so all give references to proofs. $\endgroup$ – verret Jun 26 at 1:31

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